Example 1: two angles and a side

• In triangle ABC, a = 8 cm, A = 30° and B = 45°. Find b and c.
• Solution C = 180° − 30° − 45° = 105°. From the rule, b = a × sin B / sin A = 8 × sin 45° / sin 30° = 8 × 0.7071 / 0.5 = 8√2 ≈ 11.31 cm.
• Also c = a × sin C / sin A = 8 × sin 105° / 0.5 = 16 × 0.9659 ≈ 15.45 cm.

Example 2: finding the diameter of the circumscribed circle

• In a triangle, a = 7 cm and A = 30°. Find the radius R of the circle passing through the vertices.
• Solution a / sin A = 2R, so 2R = 7 / 0.5 = 14, and R = 7 cm.

Example 3: finding an angle with the rule

• In a triangle, a = 6 cm, b = 8 cm and A = 30°. Find B.
• Solution sin B = b sin A / a = 8 × 0.5 / 6 = 2/3. So B ≈ 41.8° or B ≈ 180° − 41.8° = 138.2°.

• Since a < b, both values can give a triangle if the sum of the two known angles is less than 180°.
• Check: 30° + 41.8° = 71.8°, and 30° + 138.2° = 168.2°.
• Both are less than 180°, so there are two triangles:

Example 4: a case with two solutions

• In a triangle, a = 7 cm, b = 10 cm and A = 40°. Find the two possible values of B and C.
• Solution sin B = 10 × sin 40° / 7 = 10 × 0.6428 / 7 ≈ 0.9183. So B ≈ 66.7° or B ≈ 113.3°.
• Both are possible because 40° + 113.3° = 153.3° is less than 180°.
• The first solution gives C ≈ 73.3° and c ≈ 10.4 cm.
• The second gives C ≈ 26.7° and c ≈ 4.9 cm.

Example 5: a real-life problem

• Two points A and B on the ground are 200 metres apart.
• An object C is observed so that the angle at A is 50° and the angle at B is 70°.
• How far is the object from B?