Trigonometry Worked Examples for IGCSE Maths

Question

In triangle ABC, angle B = 90 degrees, angle A = 35 degrees, and BC = 8 cm. Find the length AC.

Solution

1. 1

   Identify the sides relative to angle A

   BC is opposite angle A (across from it), AC is the hypotenuse (opposite the right angle)

   We need to label sides as opposite, adjacent, or hypotenuse relative to the angle we are using.

2. 2

   Choose the correct ratio

   We have opposite (BC = 8) and need hypotenuse (AC), so use sin A = opposite / hypotenuse

   SOH: sin = opposite / hypotenuse matches the sides we have and need.

3. 3

   Substitute and solve

   sin 35 = 8 / AC, so AC = 8 / sin 35 = 8 / 0.5736 = 13.9 cm (3 s.f.)

   Rearrange to make AC the subject by dividing both sides, then use your calculator in degree mode.

Question

In triangle PQR, PQ = 9 cm, QR = 12 cm, and angle PQR = 110 degrees. Find the length PR.

Solution

1. 1

   Identify this as a cosine rule problem

   We have two sides and the included angle (SAS), so we need the cosine rule.

   The cosine rule is used when we have two sides and the angle between them, or all three sides.

2. 2

   Write the cosine rule

   PR squared = PQ squared + QR squared - 2(PQ)(QR)cos(angle PQR)

   The side we want to find goes on the left, and the angle used must be the one between the two known sides.

3. 3

   Substitute values

   PR squared = 9 squared + 12 squared - 2(9)(12)cos 110

   Substitute all known values carefully.

4. 4

   Calculate

   PR squared = 81 + 144 - 216 cos 110 = 81 + 144 - 216(-0.3420) = 225 + 73.87 = 298.87

   cos 110 is negative because the angle is obtuse, so subtracting a negative gives addition.

5. 5

   Find PR

   PR = sqrt(298.87) = 17.3 cm (3 s.f.)

   Take the square root to find the length.

Question

Town B is 15 km from town A on a bearing of 070 degrees. Town C is 20 km from town A on a bearing of 140 degrees. Find the distance BC and the bearing of C from B.

Solution

1. 1

   Find angle BAC

   Angle BAC = 140 - 70 = 70 degrees

   The angle between the two bearings from A gives us the angle at A in triangle ABC.

2. 2

   Use cosine rule to find BC

   BC squared = 15 squared + 20 squared - 2(15)(20)cos 70 = 225 + 400 - 600(0.3420) = 625 - 205.2 = 419.8

   We have SAS so we use the cosine rule.

3. 3

   Calculate BC

   BC = sqrt(419.8) = 20.5 km (3 s.f.)

   Take the square root.

4. 4

   Use sine rule to find angle ABC

   sin(ABC) / 20 = sin 70 / 20.49, sin(ABC) = 20 sin 70 / 20.49 = 0.9172, angle ABC = 66.5 degrees

   Now use the sine rule to find an angle at B.

5. 5

   Find the bearing of C from B

   Back bearing of A from B = 070 + 180 = 250 degrees. Angle ABC = 66.5 degrees is measured from BA towards BC. From the diagram, C is to the south of the line BA, so bearing of C from B = 250 - 66.5 = 183.5 degrees, which rounds to 184 degrees (3 s.f.).

   The back bearing gives the direction from B to A. Subtract the angle ABC to find the bearing from B to C.

Question

In triangle XYZ, XY = 14 cm, YZ = 9 cm, and angle XZY = 80 degrees. Find angle XYZ.

Solution

1. 1

   Identify sine rule setup

   We have a complete pair: side XY = 14 is opposite angle XZY = 80. Side YZ = 9 is opposite angle YXZ which we do not need directly.

   The sine rule needs at least one complete pair (side and its opposite angle).

2. 2

   Apply the sine rule

   sin(XYZ) / XZ... actually, YZ / sin(YXZ) = XY / sin(XZY). But we want angle XYZ. Let us use: YZ / sin(YXZ) does not help directly. Use: sin(XYZ) / 9... We need to reconsider. XY is opposite Z, YZ is opposite X. We want angle Y. So we need XZ, which is opposite Y. We do not have XZ. Instead: sin X / YZ = sin Z / XY gives sin X / 9 = sin 80 / 14, so sin X = 9 sin 80 / 14 = 0.6326, X = 39.2 degrees. Then Y = 180 - 80 - 39.2 = 60.8 degrees.

   Find the third angle using the sine rule for one unknown angle, then use angles in a triangle = 180.

3. 3

   Calculate angle Y

   Angle XYZ = 180 - 80 - 39.2 = 60.8 degrees

   Angles in a triangle sum to 180 degrees.