Given: A = 50°, B = 70°, a = 9 cm. Find b.
Apply sine rule: b/sin B = a/sin A
b/sin 70° = 9/sin 50°
b = 9 × sin 70° / sin 50°
b = 9 × 0.9397 / 0.7660
b ≈ 11.0 cm
Given: P = 35°, Q = 80°, q = 14 m. Find p.
Apply sine rule: p/sin P = q/sin Q
p/sin 35° = 14/sin 80°
p = 14 × sin 35° / sin 80°
p = 14 × 0.5736 / 0.9848
p ≈ 8.16 m
Given: a = 6, b = 9, A = 30°. Find angle B.
sin B / b = sin A / a
sin B = 9 × sin 30° / 6 = 9 × 0.5 / 6 = 0.75
B1 = sin−1(0.75) ≈ 48.6°
Check: C = 180 − 30 − 48.6 = 101.4° > 0° ✓ Valid
Area = ½ab sin C = ½ × 8 × 11 × sin 55°
= 44 × 0.8192
≈ 36.0 cm²
Given: B = 62°, C = 48°, b = 15 m.
A = 180 − 62 − 48 = 70°
c/sin 48° = 15/sin 62°
c = 15 × sin 48° / sin 62° = 15 × 0.7431 / 0.8829 ≈ 12.63 m
Area = ½ × b × c × sin A = ½ × 15 × 12.63 × sin 70°
= ½ × 15 × 12.63 × 0.9397 ≈ 88.9 m²
Since X = 120° is obtuse, any solution for Y must be acute (Y + 120 must be less than 180, so Y < 60°). The supplementary angle 180 − Y would exceed 60°, making the angle sum exceed 180°. Therefore only one triangle is possible.
sin Y / y = sin X / x
sin Y = 6 × sin 120° / 10 = 6 × 0.8660 / 10 = 0.5196
Y = sin−1(0.5196) ≈ 31.3°
In triangle ACB: angle ACB = 62°, angle CAB = 74°
Angle ABC = 180 − 62 − 74 = 44°
Using sine rule: AB/sin(ACB) = AC/sin(ABC)
AB/sin 62° = 80/sin 44°
AB = 80 × sin 62° / sin 44°
AB = 80 × 0.8829 / 0.6947
AB ≈ 101.6 m
Given: a = 5, b = 7, A = 45°. A is acute and a < b, so check for ambiguous case.
sin B = 7 × sin 45° / 5 = 7 × 0.7071 / 5 = 0.9900
B1 = sin−1(0.9900) ≈ 81.9°; C1 = 180 − 45 − 81.9 = 53.1° ✓
B2 = 180 − 81.9 = 98.1°; C2 = 180 − 45 − 98.1 = 36.9° ✓
Both triangles are valid.
Area1 = ½ × 5 × 7 × sin(53.1°) = 17.5 × 0.7999 ≈ 14.0 cm²
Area2 = ½ × 5 × 7 × sin(36.9°) = 17.5 × 0.6004 ≈ 10.5 cm²
Draw diagram: angle at P between bearings 040° and 110° is 70°.
Angle at A in triangle PAB: the bearing of AB from A. The interior angle at A = 35° (given as angle PAB).
Angle at B = 180 − 70 − 35 = 75°
PA = 12 km (given)
Using sine rule: AB/sin(angle at P) = PA/sin(angle at B)
AB/sin 70° = 12/sin 75°
AB = 12 × sin 70° / sin 75° = 12 × 0.9397 / 0.9659 ≈ 11.7 km
Area = ½ab sin C
24 = ½ × 8 × 7 × sin C
sin C = 48/56 = 6/7 ≈ 0.8571
C1 ≈ 59.0° or C2 ≈ 121.0°
For C1 = 59.0°: find c using cosine rule:
c² = 8² + 7² − 2(8)(7)cos 59° = 64 + 49 − 112 × 0.5150 = 113 − 57.7 = 55.3
c ≈ 7.4 cm
For C2 = 121.0°: c² = 64 + 49 − 112 × cos 121° = 113 − 112 × (−0.5150) = 113 + 57.7 = 170.7
c ≈ 13.1 cm (also valid — both triangles have the same area by construction)