GCSE Maths: Trigonometry and Pythagoras in three dimensions, and combined problems (Higher)
Part of the GCSE Maths pythagoras and trigonometry chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Trigonometry and Pythagoras in three dimensions, and combined problems
Many solids contain right-angled triangles. The trick for every 3D problem is the same: find a right-angled triangle inside the solid, draw it on its own as a flat picture, and use Pythagoras or SOH CAH TOA. Often you need to use two triangles one after the other.
The space diagonal of a cuboid. In the cuboid below, the line AG goes from one bottom corner to the opposite top corner. First find AC, the diagonal of the base, using triangle ABC (right angle at B). Then use triangle ACG (right angle at C, because CG is vertical) to find AG.
Putting the two steps together gives the formula AG2 = l2 + w2 + h2, where l, w and h are the length, width and height. You must still understand how it arises, and show the two-triangle method in an exam if the question asks you to “show that”.
The angle between a line and a plane. The angle between a line and a flat surface (a plane) is the angle between the line and its shadow (its projection) on the plane. It is the smallest angle that the line makes with the plane. To find the angle between AG and the base ABCD: the shadow of AG on the base is AC (look straight down on AG). So the angle is the angle GAC, in the right-angled triangle ACG.
Pyramids. A pyramid with a rectangular base has its apex V directly above the centre M of the base (this is the usual case). The centre M is halfway along both diagonals. The right-angled triangle V M A (right angle at M) contains the height VM, the half-diagonal AM and the sloping edge VA.
Combined problems. Some problems need several tools. A good plan: (1) draw each triangle separately and mark what you know; (2) decide, for each triangle, whether it is right-angled (Pythagoras, SOH CAH TOA) or not (sine rule, cosine rule, area formula); (3) work out an unknown that lets you move to the next triangle; (4) keep full calculator values until the end. Bearings, angles of elevation and 3D shapes all give you triangles of this kind.
Key ideas
- 3D: find a right-angled triangle, draw it flat, and label it with the lengths you know.
- Space diagonal of a cuboid: AG2 = l2 + w2 + h2 (found in two steps).
- The angle between a line and a plane = the angle between the line and its shadow on the plane.
- Pyramid: the apex is above the centre M of the base; use the triangle with the height VM.
- Not right-angled? Use the sine rule, the cosine rule or ½ab sin C.
Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. Use Pythagoras’ theorem twice: first across the base to find the base diagonal, then upwards with the height to find the 3D diagonal.
- Find the diagonal of the base first. AC lies along the base. Triangle ABC is right-angled at B, with AB = 8 cm and BC = 6 cm, so we can use Pythagoras.AC2 = 82 + 62 = 64 + 36 = 100
- Now use the vertical edge. CG stands straight up from C, so triangle ACG is right-angled at C. The 3D diagonal AG is its hypotenuse. We can use the exact value AC2 = 100 without finding AC first.AG2 = AC2 + CG2 = 100 + 52 = 100 + 25 = 125
- Take the square root and round to 1 d.p. We want AG, not AG2.AG = √125 = 11.18033…
AG = 11.2 cm
Check. The shortcut: AG2 = 82 + 62 + 52 = 64 + 36 + 25 = 125. ✓
Watch out. Do not add the three edges. Add their squares, and then take one square root at the end. Always find a right angle to work in, and draw the triangle separately.
Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.
Method. The angle between a line and a plane is the angle between the line and its shadow (projection) on the plane. Find the shadow, then use trigonometry in the right-angled triangle.
- Find the projection of the line onto the base. The angle between a line and a plane is the angle between the line and its shadow on the plane. The shadow of AG on the base is AC. The angle we want is angle GAC.Triangle ACG is right-angled at C.
- Find the length of the shadow, AC. Use Pythagoras in the base triangle.AC2 = 82 + 62 = 100
AC = 10 cm - Label triangle ACG for trigonometry. From angle GAC: the opposite side is CG = 5 cm and the adjacent side is AC = 10 cm. Opposite and adjacent means TOA.tan GAC = oppositeadjacent = 510
- Use the inverse tangent. The inverse button turns the tangent value into an angle.tan GAC = 0.5
angle GAC = tan−1(0.5) = 26.56505…° - Round to 1 d.p. Angle = 26.6°
Check. Use sine instead: sin GAC = 5 ÷ AG, where AG = √(64 + 36 + 25) = 11.18033…. 5 ÷ 11.18033… = 0.44721…, and sin 26.6° = 0.448. ✓
Watch out. The angle with the base is measured to the base diagonal AC, not to an edge of the base like AB.
Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. Draw the right-angled triangle inside the pyramid that contains the height: apex, centre of the base and a corner. Use Pythagoras’ theorem.
- Find the diagonal of the base. The apex is directly above the centre M of the base. M is the middle of the base diagonal AC. Triangle ABC is right-angled at B.AC2 = 102 + 102 = 100 + 100 = 200
AC = 14.14213… cm - Halve it to get AM. M is the midpoint of the diagonal.AM = 14.14213… ÷ 2 = 7.07106… cm
AM2 = 50 - Use the right-angled triangle VMA. VM is the vertical height, so the angle at M is a right angle. VA is the hypotenuse.VA2 = AM2 + VM2 = 50 + 122 = 50 + 144 = 194
- Take the square root and round to 1 d.p. We want VA, not VA2.VA = √194 = 13.92838…
VA = 13.9 cm
Check. The slant edge (13.9) is longer than the height (12) and longer than AM (7.1), as the hypotenuse should be. ✓
Watch out. AM is half the base diagonal, not half a side of the base.
Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.
Method. Find the shadow of the edge on the base (from the centre M to the corner), then use trigonometry in the right-angled triangle VMA.
- Find the shadow of VA on the base. The angle between the edge VA and the base is the angle between VA and its shadow AM (M is the centre of the base). The angle we want is angle VAM.Triangle VMA is right-angled at M.
- Find AM. AM is half of the base diagonal AC, and we find AC using Pythagoras.AC2 = 102 + 102 = 200, so AC = 14.14213… cm
AM = 14.14213… ÷ 2 = 7.07106… cm - Label the sides for trigonometry. From angle VAM, the opposite side is VM = 12 cm and the adjacent side is AM = 7.07106… cm. That is TOA.tan VAM = 127.07106… = 1.69705…
- Use the inverse tangent and round. The inverse button turns the tangent back into an angle.angle VAM = tan−1(1.69705…) = 59.49104…°
angle VAM = 59.5°
Check. Use sine instead: VA = √(50 + 144) = 13.92838…, and sin VAM = 12 ÷ 13.92838… = 0.86154…; sin 59.5° = 0.862. ✓
Watch out. The right angle is at M (the centre of the base), not at the corner.
Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.
Method. The angle between a sloping face and the base is found in the right-angled triangle formed by the height, the line from the centre M to the middle N of a base edge, and the line VN down the face.
- Find the right-angled triangle. N is the midpoint of AB. MN is half of the base edge, and VM is the height. The angle at M is a right angle. The angle between the face VAB and the base is the angle VNM.MN = 10 ÷ 2 = 5 cm, VM = 12 cm
- Label from the angle VNM. VM is opposite the angle and MN is adjacent. That is TOA.tan VNM = 125
- Use the inverse tangent. Turn the ratio into an angle.VNM = tan−1(2.4) = 67.38013…°
- Round. Angle = 67.4°
Check. The slant height VN = √(52 + 122) = 13, and sin VNM = 12 ÷ 13 = 0.92307…; sin 67.4° = 0.923. ✓
Watch out. Use the midpoint N of the base edge (half the side), not the corner A (half the diagonal).
Watch out
In 3D questions the picture is not drawn to scale, so do not measure from it. Draw the right-angled triangle separately, marking the right angle. And remember that the angle between a line and a plane is measured from the shadow of the line, not from an edge that is just nearby.Exercise 20M Trigonometry and Pythagoras in 3D
Calculator allowedKey words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. Use Pythagoras’ theorem twice: first across the base to find the base diagonal, then upwards with the height to find the 3D diagonal.
- Find the diagonal of the base first. AC lies along the base. Triangle ABC is right-angled at B, with AB = 3 cm and BC = 4 cm, so we can use Pythagoras.AC2 = 32 + 42 = 9 + 16 = 25
- Now use the vertical edge. CG stands straight up from C, so triangle ACG is right-angled at C. The 3D diagonal AG is its hypotenuse. We can use the exact value AC2 = 25 without finding AC first.AG2 = AC2 + CG2 = 25 + 122 = 25 + 144 = 169
- Take the square root. We want AG, not AG2.AG = √169 = 13 cm
Check. The shortcut: AG2 = 32 + 42 + 122 = 9 + 16 + 144 = 169. ✓
Watch out. Do not add the three edges. Add their squares, and then take one square root at the end. Always find a right angle to work in, and draw the triangle separately.
Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. Use Pythagoras’ theorem twice: first across the base to find the base diagonal, then upwards with the height to find the 3D diagonal.
- Find the diagonal of the base first. AC lies along the base. Triangle ABC is right-angled at B, with AB = 2 cm and BC = 3 cm, so we can use Pythagoras.AC2 = 22 + 32 = 4 + 9 = 13
- Now use the vertical edge. CG stands straight up from C, so triangle ACG is right-angled at C. The 3D diagonal AG is its hypotenuse. We can use the exact value AC2 = 13 without finding AC first.AG2 = AC2 + CG2 = 13 + 62 = 13 + 36 = 49
- Take the square root. We want AG, not AG2.AG = √49 = 7 cm
Check. The shortcut: AG2 = 22 + 32 + 62 = 4 + 9 + 36 = 49. ✓
Watch out. Do not add the three edges. Add their squares, and then take one square root at the end. Always find a right angle to work in, and draw the triangle separately.
Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. Use Pythagoras’ theorem twice: first across the base to find the base diagonal, then upwards with the height to find the 3D diagonal.
- Find the diagonal of the base first. AC lies along the base. Triangle ABC is right-angled at B, with AB = 5 cm and BC = 8 cm, so we can use Pythagoras.AC2 = 52 + 82 = 25 + 64 = 89
- Now use the vertical edge. CG stands straight up from C, so triangle ACG is right-angled at C. The 3D diagonal AG is its hypotenuse. We can use the exact value AC2 = 89 without finding AC first.AG2 = AC2 + CG2 = 89 + 92 = 89 + 81 = 170
- Take the square root and round to 1 d.p. We want AG, not AG2.AG = √170 = 13.03840…
AG = 13.0 cm
Check. The shortcut: AG2 = 52 + 82 + 92 = 25 + 64 + 81 = 170. ✓
Watch out. Do not add the three edges. Add their squares, and then take one square root at the end. Always find a right angle to work in, and draw the triangle separately.
Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. Use Pythagoras’ theorem twice: first across the base to find the base diagonal, then upwards with the height to find the 3D diagonal.
- Find the diagonal of the base first. AC lies along the base. Triangle ABC is right-angled at B, with AB = 40 cm and BC = 30 cm, so we can use Pythagoras.AC2 = 402 + 302 = 1600 + 900 = 2500
- Now use the vertical edge. CG stands straight up from C, so triangle ACG is right-angled at C. The 3D diagonal AG is its hypotenuse. We can use the exact value AC2 = 2500 without finding AC first.AG2 = AC2 + CG2 = 2500 + 1202 = 2500 + 14400 = 16900
- Take the square root. We want AG, not AG2.AG = √16900 = 130 cm
Check. The shortcut: AG2 = 402 + 302 + 1202 = 1600 + 900 + 14400 = 16900. ✓
Watch out. Do not add the three edges. Add their squares, and then take one square root at the end. Always find a right angle to work in, and draw the triangle separately.
Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.
Method. The angle between a line and a plane is the angle between the line and its shadow (projection) on the plane. Find the shadow, then use trigonometry in the right-angled triangle.
- Find the projection of the line onto the base. The angle between a line and a plane is the angle between the line and its shadow on the plane. The shadow of AG on the base is AC. The angle we want is angle GAC.Triangle ACG is right-angled at C.
- Find the length of the shadow, AC. Use Pythagoras in the base triangle.AC2 = 122 + 52 = 169
AC = 13 cm - Label triangle ACG for trigonometry. From angle GAC: the opposite side is CG = 6 cm and the adjacent side is AC = 13 cm. Opposite and adjacent means TOA.tan GAC = oppositeadjacent = 613
- Use the inverse tangent. The inverse button turns the tangent value into an angle.tan GAC = 0.46153…
angle GAC = tan−1(0.46153…) = 24.77514…° - Round to 1 d.p. Angle = 24.8°
Check. Use sine instead: sin GAC = 6 ÷ AG, where AG = √(144 + 25 + 36) = 14.31782…. 6 ÷ 14.31782… = 0.41905…, and sin 24.8° = 0.419. ✓
Watch out. The angle with the base is measured to the base diagonal AC, not to an edge of the base like AB.
Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.
Method. The angle between a line and a plane is the angle between the line and its shadow (projection) on the plane. Find the shadow, then use trigonometry in the right-angled triangle.
- Find the projection of the line onto the base. The angle between a line and a plane is the angle between the line and its shadow on the plane. The shadow of AG on the base is AC. The angle we want is angle GAC.Triangle ACG is right-angled at C.
- Find the length of the shadow, AC. Use Pythagoras in the base triangle.AC2 = 102 + 102 = 200
AC = 14.14213… cm - Label triangle ACG for trigonometry. From angle GAC: the opposite side is CG = 10 cm and the adjacent side is AC = 14.14213… cm. Opposite and adjacent means TOA.tan GAC = oppositeadjacent = 1014.14213…
- Use the inverse tangent. The inverse button turns the tangent value into an angle.tan GAC = 0.70710…
angle GAC = tan−1(0.70710…) = 35.26438…° - Round to 1 d.p. Angle = 35.3°
Check. Use sine instead: sin GAC = 10 ÷ AG, where AG = √(100 + 100 + 100) = 17.32050…. 10 ÷ 17.32050… = 0.57735…, and sin 35.3° = 0.578. ✓
Watch out. The angle with the base is measured to the base diagonal AC, not to an edge of the base like AB.
Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.
Method. The angle between the face diagonal AF and the base is the angle between AF and its shadow on the base. Look straight down: the shadow of AF is AB.
- Find the shadow. F is straight above B, so the shadow of F on the base is B. The shadow of AF is AB.The angle is angle FAB.
- Find the right-angled triangle. FB is vertical, so triangle ABF has a right angle at B. The opposite side to the angle at A is FB = 5 cm and the adjacent side is AB = 9 cm. TOA.tan FAB = 59 = 0.55555…
- Use the inverse tangent and round. FAB = tan−1(0.55555…) = 29.05460…° = 29.1°
Check. Use sine instead: AF = √(92 + 52) = √106 = 10.29563…, sin FAB = 5 ÷ 10.29563… = 0.48564…, and sin 29.1° = 0.486. ✓
Watch out. You do not need the base diagonal here. Pick the right-angled triangle that contains the line and its shadow.
Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. We know the space diagonal and two edges of the base, so work backwards: first the base diagonal, then the height.
- Find the diagonal of the base. Triangle ABC is right-angled at B.AC2 = 62 + 82 = 36 + 64 = 100, so AC = 10 cm
- Use triangle ACG. The right angle is at C and AG = 15 cm is the hypotenuse. We are finding a shorter side, so subtract.CG2 = AG2 − AC2 = 152 − 102 = 225 − 100 = 125
- Take the square root and round. CG = √125 = 11.18033…
CG = 11.2 cm
Check. Forwards: 62 + 82 + 11.22 = 225.4, close to 152 = 225. ✓
Watch out. The diagonal is the hypotenuse of the triangle ACG, so subtract to find the height.
Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. Draw the right-angled triangle inside the pyramid that contains the height: apex, centre of the base and a corner. Use Pythagoras’ theorem.
- Find the diagonal of the base. The apex is directly above the centre M of the base. M is the middle of the base diagonal AC. Triangle ABC is right-angled at B.AC2 = 82 + 82 = 64 + 64 = 128
AC = 11.31370… cm - Halve it to get AM. M is the midpoint of the diagonal.AM = 11.31370… ÷ 2 = 5.65685… cm
AM2 = 32 - Use the right-angled triangle VMA. VM is the vertical height, so the angle at M is a right angle. VA is the hypotenuse.VA2 = AM2 + VM2 = 32 + 102 = 32 + 100 = 132
- Take the square root and round to 1 d.p. We want VA, not VA2.VA = √132 = 11.48912…
VA = 11.5 cm
Check. The slant edge (11.5) is longer than the height (10) and longer than AM (5.7), as the hypotenuse should be. ✓
Watch out. AM is half the base diagonal, not half a side of the base.
Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.
Method. Find the shadow of the edge on the base (from the centre M to the corner), then use trigonometry in the right-angled triangle VMA.
- Find the shadow of VA on the base. The angle between the edge VA and the base is the angle between VA and its shadow AM (M is the centre of the base). The angle we want is angle VAM.Triangle VMA is right-angled at M.
- Find AM. AM is half of the base diagonal AC, and we find AC using Pythagoras.AC2 = 82 + 82 = 128, so AC = 11.31370… cm
AM = 11.31370… ÷ 2 = 5.65685… cm - Label the sides for trigonometry. From angle VAM, the opposite side is VM = 10 cm and the adjacent side is AM = 5.65685… cm. That is TOA.tan VAM = 105.65685… = 1.76776…
- Use the inverse tangent and round. The inverse button turns the tangent back into an angle.angle VAM = tan−1(1.76776…) = 60.50379…°
angle VAM = 60.5°
Check. Use sine instead: VA = √(32 + 100) = 11.48912…, and sin VAM = 10 ÷ 11.48912… = 0.87038…; sin 60.5° = 0.870. ✓
Watch out. The right angle is at M (the centre of the base), not at the corner.
Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. The apex is directly above the centre M, so VM is perpendicular to the base. Use the right-angled triangle VMA and subtract to find the shorter side VM.
- Find AM. M is the centre of the square base, so AM is half the diagonal AC. First find AC with Pythagoras.AC2 = 102 + 102 = 200, so AM2 = (AC 2) ÷ 4 = 50
AM = √50 = 7.07106… cm - Use triangle VMA. The right angle is at M and VA = 13 cm is the hypotenuse.VM2 = VA2 − AM2 = 132 − 50 = 169 − 50 = 119
- Take the square root and round. VM = √119 = 10.90871…
VM = 10.9 cm
Check. Forwards: 10.92 + 50 = 168.8, close to 132 = 169. ✓
Watch out. AM is half of the base diagonal (√50), not half of the side (5).
Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.
Method. Draw the right-angled triangle VMN, where N is the midpoint of a base edge. The angle between the face and the base is the angle VNM.
- Find MN. M is the centre of the square base. N is the midpoint of an edge, so MN is half the side.MN = 6 ÷ 2 = 3 cm
- Label from the angle VNM. VM = 4 cm is opposite the angle, and MN = 3 cm is adjacent. TOA.tan VNM = 43
- Use the inverse tangent. VNM = tan−1(1.33333…) = 53.13010…° = 53.1°
Check. The slant height VN = √(32 + 42) = 5 and sin VNM = 4 ÷ 5 = 0.8, sin 53.1° = 0.800. ✓
Watch out. This 3, 4, 5 triangle appears again and again.
Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. Draw the right-angled triangle inside the pyramid that contains the height: apex, centre of the base and a corner. Use Pythagoras’ theorem.
- Find the diagonal of the base. The apex is directly above the centre M of the base. M is the middle of the base diagonal AC. Triangle ABC is right-angled at B.AC2 = 32 + 32 = 9 + 9 = 18
AC = 4.24264… m - Halve it to get AM. M is the midpoint of the diagonal.AM = 4.24264… ÷ 2 = 2.12132… m
AM2 = 4.5 - Use the right-angled triangle VMA. VM is the vertical height, so the angle at M is a right angle. VA is the hypotenuse.VA2 = AM2 + VM2 = 4.5 + 2.42 = 4.5 + 5.76 = 10.26
- Take the square root and round to 1 d.p. We want VA, not VA2.VA = √10.26 = 3.20312…
VA = 3.2 m
Check. The slant edge (3.2) is longer than the height (2.4) and longer than AM (2.1), as the hypotenuse should be. ✓
Watch out. AM is half the base diagonal, not half a side of the base.
Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.
Method. Use Pythagoras’ theorem twice, first in the base and then in the vertical triangle, and substitute.
- Pythagoras in the base. Triangle ABC is right-angled at B.AC2 = AB2 + BC2 = l2 + w2
- Pythagoras in triangle ACG. CG is vertical, so the triangle is right-angled at C. CG = h.AG2 = AC2 + CG2 = AC2 + h2
- Substitute. Replace AC2 by l2 + w2.AG2 = l2 + w2 + h2
- State the result. So AG = √(l2 + w2 + h2).
Check. Test with 8, 6, 5: √(64 + 36 + 25) = √125 = 11.2, the same as the example. ✓
Watch out. A “show that” needs both triangles shown separately.
Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.
Method. Take a cube of side 1 (the answer does not depend on the size). Find the base diagonal, then the angle in triangle ACG.
- Find AC. AC is the diagonal of a square of side 1.AC2 = 12 + 12 = 2, so AC = √2
- Use triangle ACG. Right angle at C. The angle GAC has opposite side CG = 1 and adjacent side AC = √2. TOA.tan GAC = 1√2
- Find the angle. Use the inverse tangent.GAC = tan−1(0.70710…) = 35.26438…°
- Round. To 1 decimal place.GAC = 35.3°
Check. Another way: sin GAC = 1 ÷ AG = 1 ÷ √3 = 0.57735…, and sin 35.3° = 0.578. ✓
Watch out. For a cube of side a every length is multiplied by a, so the angle is the same.
Exercise 20N Combined problems
Calculator allowedKey words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. The two ships and the port make a triangle with two sides and the angle between them known, so use the cosine rule for AB. Then use the sine rule for the angle at A, and finally find the bearing.
- Find the angle at the port. The bearings from P are 040° and 110°, so the angle between PA and PB is the difference.Angle APB = 110° − 40° = 70°
- Use the cosine rule for AB. Two sides (12 and 9) and the angle between them (SAS).AB2 = 122 + 92 − 2 × 12 × 9 × cos 70° = 225 − 73.87635… = 151.12364…
AB = 12.29323… km - Use the sine rule for angle PAB. Now we have a matching pair: AB is opposite the 70° angle. PB = 9 km is opposite angle PAB. PB is shorter than PA, so angle PAB is acute.sin PAB = 9 × sin 70°12.29323… = 0.68795…
PAB = sin−1(0.68795…) = 43.46870…° - Find the bearing of B from A. The back bearing of P from A is 040° + 180° = 220°. Ship B is on the other side of the line AP, turned clockwise less from north, so subtract the angle PAB.Bearing = 220° − 43.46870…° = 176.53129…°, which is 177° to the nearest degree.
Check. Using coordinates from P: A = (7.71345…, 9.19253…), B = (8.45723…, −3.07818…). B is 0.74378… east and 12.27071… south of A, which gives a bearing of 177°. ✓
Watch out. The bearing of B from A starts with a north line at A, not at P.
Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. The sine rule: each side divided by the sine of its opposite angle gives the same number. Use one complete pair to find the missing side.
- Find angle C. The angles in a triangle add to 180°. AB = 9 cm is opposite C.C = 180° − 50° − 64° = 66°
- Match each side with the angle opposite it. The sine rule links a side with the angle opposite that side. We need one matching pair that we know completely, and one pair with the unknown.Known pair: side c = 9 cm is opposite angle C = 66°.
Wanted pair: side b is opposite angle B = 64°. - Write the sine rule with the unknown on top. Putting the side we want on top means we only need to multiply at the end.bsin B = csin C
bsin 64° = 9sin 66° - Multiply both sides by sin 64°. This gets b on its own.b = 9 × sin 64°sin 66°
- Use the calculator (degree mode). Find the two sines, then multiply and divide.sin 64° = 0.89879… and sin 66° = 0.91354…
b = 9 × 0.89879… ÷ 0.91354… = 8.85467… - Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.b = 8.9 cm
Check. Each side ÷ sine of its opposite angle should be the same number: 9 ÷ sin 66° = 9.9 and 8.9 ÷ sin 64° = 9.9. ✓
Watch out. Always pair a side with the angle opposite it (not next to it). Draw arrows from each angle to the side across from it if you are unsure.
Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. Now we know two sides (AB = 9 and AC) and the angle A = 50° between them, so use area = ½ ab sin C.
- Use the unrounded value of AC. Rounding AC first would make the area less accurate.AC = 8.85467… cm
- Write the area formula. The angle between AB and AC is A = 50°.Area = ½ × 9 × 8.85467… × sin 50° = 30.52382…
- Round. Area = 30.5 cm2
Check. Use another pair of sides: ½ × AB × BC × sin B with BC = 7.54686… gives 30.52382…. ✓
Watch out. Use the angle between the two sides you pick.
Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. Split the quadrilateral along the diagonal AC into two triangles. Triangle ABC is right-angled. Triangle ACD has three known sides once we know AC.
- Find AC in the right-angled triangle. AC is the hypotenuse of triangle ABC.AC2 = 62 + 82 = 100, so AC = 10 cm
- Use the cosine rule in triangle ACD. We want angle D, which is opposite AC = 10. The other sides are CD = 7 and AD = 9.cos D = 72 + 92 − 1022 × 7 × 9 = 30126 = 0.23809…
D = cos−1(0.23809…) = 76.22585…° - Round. D = 76.2°
Check. Substitute back: 72 + 92 − 2 × 7 × 9 × cos 76.2° = 99.9, which is close to AC2 = 100. ✓
Watch out. Name each triangle you use and mark its right angle or known sides separately.
Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.
Method. The two legs are on bearings that differ by 90°, so they are at right angles. Use Pythagoras for the distance and tan for the bearing.
- Show the angle at B. The first leg is on 050° and the second on 140°. The difference is a right angle.140° − 50° = 90°, so angle ABC = 90°
- Find AC. AC is the hypotenuse.AC2 = 1202 + 852 = 14 400 + 7225 = 21 625
AC = 147.05441… km - Find angle BAC. At A: opposite = BC = 85, adjacent = AB = 120. TOA.tan BAC = 85120 = 0.70833…
BAC = tan−1(0.70833…) = 35.31121…° - Find the bearing of C from A. The first leg is on 050°. Angle BAC is on the clockwise side (towards 140°), so add.Bearing = 50° + 35.3° = 85.3° = 085°
Check. Coordinates: B = (91.92533…, 77.13451…) and C = B + 85(sin 140°, cos 140°) = (146.56227…, 12.02073…); the bearing of C from A is 85.3°. ✓
Watch out. When the legs are perpendicular you can use right-angled trigonometry. Otherwise use the cosine rule.
Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. The cosine rule finds the third side of a triangle when two sides and the angle between them are known: a2 = b2 + c2 − 2bc cos A.
- Name the sides. Call the side opposite angle P the side q (that is QR), and the two sides next to P: b = PQ = 80 m and c = PR = 65 m.QR is opposite angle P = 58°.
- Spot the angle and the two sides around it. The cosine rule is used when we know two sides and the angle between them. The side we want is the one opposite that angle.Known: b = 80 m, c = 65 m, angle P = 58° between them.
Wanted: q, opposite angle P. - Write the cosine rule and put the numbers in. Square the wanted side on the left; square the other two sides on the right, then take away 2 × side × side × cos(angle).q2 = b2 + c2 − 2bc cos P
q2 = 802 + 652 − 2 × 80 × 65 × cos 58° - Work out the two squares and add. This is the first part, b2 + c2.802 + 652 = 6400 + 4225 = 10625
- Work out the last part on its own. Multiply everything together first. Do this before subtracting (BIDMAS: multiplication comes before subtraction).cos 58° = 0.52991…
2 × 80 × 65 × 0.52991… = 5511.16034… - Subtract. This gives q2.q2 = 10625 − 5511.16034… = 5113.83965…
- Square root, then round to 1 d.p.. We want q, not q2.q = √5113.83965… = 71.51111…
q = 71.5 m
Check. One side of a triangle is between the difference and the sum of the other two: 15 is less than 71.5, which is less than 145. Angle P is less than 90°, so q2 should be less than b2 + c2: 5113.83965… is less than 10625. ✓
Watch out. Do 2bc cos A first and then subtract it. A very common mistake is to work out (b2 + c2 − 2bc) first. The rule has a minus sign, even for an obtuse angle (cos of an obtuse angle is negative).
Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. Area of any triangle = ½ × a × b × sin C, where C is the angle between sides a and b.
- Spot the two sides and the angle between them. The formula needs two sides and the included angle (the angle between those two sides).Sides: 80 m and 65 m. Included angle C = 58°.
- Write the formula and put the numbers in. Area = half × one side × the other side × sine of the angle between them.Area = ½ ab sin C
Area = ½ × 80 × 65 × sin 58° - Use the calculator. Find the sine first, then multiply everything together.sin 58° = 0.84804…
Area = 0.5 × 80 × 65 × 0.84804… = 2204.92505… - Round to 1 d.p. and write the units. Area is measured in square units.Area = 2204.9 m2
Check. Another way: take 80 as the base. The height is 65 × sin 58° = 55.12312…. Area = ½ × 80 × 55.12312… = 2204.92505…. ✓
Watch out. The angle must be the one between the two sides you use. If it is not between them, find the angle you need first (angles in a triangle add up to 180°).
Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. Use the cosine rule for the largest angle (opposite the longest side 12 cm), then the sine rule for a second angle, and the angle sum for the third.
- Largest angle first. The largest angle is opposite 12 cm. Call it C: c = 12, and the other sides are 7 and 9. This is SSS, so use the cosine rule. The cosine rule gives the obtuse angle correctly.cos C = 72 + 92 − 1222 × 7 × 9 = −14126 = −0.11111…
C = 96.37937…° - Use the sine rule for angle A (opposite 7). Now there is a matching pair: 12 cm is opposite C. The angle A is opposite 7 cm, the shortest side, so A is acute.sin A = 7 × sin 96.37937…°12 = 0.57972…
A = 35.43094…° - Third angle. Angles in a triangle add to 180°.B = 180° − 96.4° − 35.4° = 48.2°
Check. Use the cosine rule for the third angle too: cos B = (72 + 122 − 92) ÷ (2 × 7 × 12) = 0.66666…, B = 48.18968…°. ✓
Watch out. Use the cosine rule first, because the sine rule could give the wrong (acute) value for an obtuse angle.
Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. Write the area of the triangle in two ways, using the same triangle but different pairs of sides, then compare.
- Area using sides b and c and angle A. The angle A is between sides b and c.Area = ½ bc sin A
- Area using sides a and c and angle B. The angle B is between sides a and c.Area = ½ ac sin B
- Equate. Both are the area of the same triangle.½ bc sin A = ½ ac sin B
- Simplify. Multiply both sides by 2 and divide both sides by c.b sin A = a sin B
- Rearrange. Divide both sides by sin A sin B.asin A = bsin B. This is the sine rule.
Check. Numerical test: a = 6, b = 8, A = 36.87°, B = 53.13°: 6 ÷ 0.6 = 10 and 8 ÷ 0.8 = 10. ✓
Watch out. Pick two pairs of sides, and the angle that is between each pair.
Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. Triangle AVB is isosceles with VA = VB = 13 cm and AB = 10 cm. We know all three sides, so use the cosine rule for the angle at V.
- Identify the triangle. The two sloping edges VA and VB are both 13 cm. The base edge AB is 10 cm.VA = VB = 13 cm, AB = 10 cm.
- Use the cosine rule. The angle at V is opposite the 10 cm side.cos AVB = 132 + 132 − 1022 × 13 × 13 = 238338 = 0.70414…
- Use the inverse cosine. AVB = cos−1(0.70414…) = 45.23972…° = 45.2°
Check. Split the isosceles triangle: sin(½ AVB) = 5 ÷ 13 = 0.38461…, so ½ AVB = 22.61986…° and AVB = 45.23972…°. ✓
Watch out. You do not need the height of the pyramid: all three sides of triangle AVB are known.
Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. The three sides of the roof truss are known, so use the cosine rule for the angle at the apex (opposite the 8 m span).
- Name the sides. Call the angle at the top C. The side opposite C is the span, 8 m. The other sides are 5.5 m and 6.5 m.c = 8, a = 5.5, b = 6.5
- Use the cosine rule. Rearranged for cos C.cos C = 5.52 + 6.52 − 822 × 5.5 × 6.5 = 8.571.5 = 0.11888…
- Use the inverse cosine and round. C = cos−1(0.11888…) = 83.17246…° = 83.2°
Check. The angle at the top is much smaller than 90° for a long, shallow roof. Also the angles A, B and C must add to 180°; for this truss the base angle at the end of the 5.5 m rafter is 53.8°. ✓
Watch out. The angle is opposite the 8 m span, not between the 8 m span and a rafter.
Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.
Method. The known angle is A = 35° with BC = 7 cm opposite it, and AB = 10 cm is opposite the unknown angle C. This is the ambiguous case, so check for two triangles.
- Use the sine rule for C. Matching pair: BC = 7 opposite A = 35°. Side AB = 10 is opposite C.sin C = 10 × sin 35°7 = 0.81939…
C = sin−1(0.81939…) = 55.02426…° - Find the second possible angle. The obtuse angle with the same sine is 180° minus the first. It fits if the angles still add up to less than 180°.C = 180° − 55.02426…° = 124.97573…°
124.97573…° + 35° = 159.97573…° which is less than 180°, so both are possible. - Find angle B in each triangle. Angles in a triangle add up to 180°.First triangle: B = 180° − 35° − 55.02426…° = 89.97573…°
Second triangle: B = 180° − 35° − 124.97573…° = 20.02426…° - Use the sine rule for AC (opposite B). AC = b = BC × sin B ÷ sin A.First: AC = 7 × sin 89.97573…° ÷ sin 35° = 12.20412… cm
Second: AC = 7 × sin 20.02426…° ÷ sin 35° = 4.17891… cm - Round. AC = 12.2 cm or 4.2 cm
Check. Check the first triangle with the cosine rule: 72 = 102 + 12.22 − 2 × 10 × 12.2 cos 35° = 49.0; this is close to 49. ✓
Watch out. Whenever the known side opposite the angle is shorter than the other known side, check for a second triangle.