The sine rule: the ambiguous case Problem set B Worked solutions

Triangle ABC has angle A = 58°, side BC = 17.3 cm and side AC = 13.6 cm. Find angle B.

• \(\frac{\sin B}{13.6}=\frac{\sin58^\circ}{17.3}\)
• \(\sin B=\frac{13.6\sin58^\circ}{17.3}=0.6667,\;\;B=41.81^\circ\)
• \(180^\circ-41.81^\circ=138.2^\circ\text{; }138.2+58>180\text{, so reject it}\)
• \(B=41.8^\circ\)

A triangle ABC has angle A = 105°, BC = 14.8 cm and AC = 8.9 cm. Find angle B.

• \(\frac{\sin B}{8.9}=\frac{\sin105^\circ}{14.8}\)
• \(\sin B=\frac{8.9\sin105^\circ}{14.8}=0.5809,\;\;B=35.51^\circ\)
• \(180^\circ-35.51^\circ=144.5^\circ\text{; }144.5+105>180\text{, so reject it}\)
• \(B=35.5^\circ\)

A triangle ABC has angle A = 36°, BC = 7.9 cm and AC = 11.6 cm. Find the two possible values of angle B.

• \(\sin B=\frac{11.6\sin36^\circ}{7.9}=0.8631\)
• \(B=59.66^\circ\text{ or }B=180^\circ-59.66^\circ=120.3^\circ\)
• \(120.3+36=156.3<180\text{, so both fit}\)

A triangle ABC has angle A = 52°, BC = 10.4 cm and AC = 12.6 cm. Angle B can have two values. Find

1. (a) both possible values of angle B

   • \(\sin B=\frac{12.6\sin52^\circ}{10.4}=0.9547\)
   • \(B=72.69^\circ\text{ or }107.3^\circ\)

2. (b) the angle C in each case

   • \(C=180^\circ-52^\circ-72.69^\circ=55.31^\circ\)
   • \(C=180^\circ-52^\circ-107.3^\circ=20.69^\circ\)

A triangle ABC has angle A = 40° and AC = 12 cm. How many different triangles are there when BC is

1. (a) a = 6 cm

   • \(h=12\sin40^\circ=7.7135\); a = 6: a < h: no triangle

2. (b) a = 10 cm

   • \(h=12\sin40^\circ=7.7135\); a = 10: h < a < b: two triangles

3. (c) a = 14 cm

   • \(h=12\sin40^\circ=7.7135\); a = 14: a ≥ b: one triangle

4. (d) a = 12 cm

   • \(h=12\sin40^\circ=7.7135\); a = 12: a ≥ b: one triangle

A walker leaves a hut H and walks 12 km to a summit S. A lake L is 9.5 km from the summit and angle LHS = 35°. Find

1. (a) the two possible sizes of angle HLS

   • \(\frac{\sin L}{12}=\frac{\sin35^\circ}{9.5}\;\Rightarrow\;\sin L=0.7245\)
   • \(L=46.43^\circ\text{ or }133.6^\circ\;(180^\circ-46.43^\circ)\)

2. (b) the two possible distances HL

   • \(\text{angle at }S=180^\circ-35^\circ-46.43^\circ=98.57^\circ,\;\;HL=\frac{9.5\sin98.57^\circ}{\sin35^\circ}=16.4\)
   • \(\text{angle at }S=180^\circ-35^\circ-133.6^\circ=11.43^\circ,\;\;HL=3.28\)

A pilot at P sees a control tower T and a radio mast R. PR = 29.5 km, TP = 21.7 km and angle PTR = 71°. Find angle PRT.

• \(\frac{\sin R}{21.7}=\frac{\sin71^\circ}{29.5}\)
• \(\sin R=0.6955,\;\;R=44.07^\circ\)
• \(180^\circ-44.07^\circ=135.9^\circ\text{ is too big with }71^\circ\text{, so there is one answer}\)
• 44.1°

A joiner makes a triangular brace ABC with AB = 18 cm and angle A = 40°. The side BC has length x cm. Find the values of x for which two different braces can be made.

• Two triangles need the side opposite A to be longer than the height h but shorter than the other known side.
• \(h=18\sin40^\circ=11.57\)
• \(11.6<x<18\)

A triangle ABC has angle A = 34°, BC = 8.4 cm and AC = 14.2 cm. There are two possible triangles. Find

1. (a) both possible values of angle B

   • \(\sin B=\frac{14.2\sin34^\circ}{8.4}=0.9453\)
   • \(B=70.96^\circ\text{ or }109.0^\circ\)

2. (b) the area of each triangle

   • \(C=75.04^\circ\text{ or }36.96^\circ\)
   • \(\text{area}=\frac{1}{2}ab\sin C=\frac{1}{2}\times8.4\times14.2\times\sin75.04^\circ=57.6\)
   • \(\text{or }\frac{1}{2}\times8.4\times14.2\times\sin36.96^\circ=35.9\)