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

In triangle ABC, angle A = 72°, BC = 14.2 cm and AC = 11.5 cm. Find angle B.

• \(\frac{\sin B}{11.5}=\frac{\sin72^\circ}{14.2}\)
• \(\sin B=\frac{11.5\sin72^\circ}{14.2}=0.7702,\;\;B=50.37^\circ\)
• \(180^\circ-50.37^\circ=129.6^\circ\text{; }129.6+72>180\text{, so reject it}\)
• \(B=50.4^\circ\)

In triangle ABC, angle A = 112°, BC = 16.5 cm and AC = 9.8 cm. Find angle B.

• \(\frac{\sin B}{9.8}=\frac{\sin112^\circ}{16.5}\)
• \(\sin B=\frac{9.8\sin112^\circ}{16.5}=0.5507,\;\;B=33.41^\circ\)
• \(180^\circ-33.41^\circ=146.6^\circ\text{; }146.6+112>180\text{, so reject it}\)
• \(B=33.4^\circ\)

In triangle ABC, angle A = 32°, BC = 8.5 cm and AC = 12.4 cm. Find the two possible values of angle B.

• \(\sin B=\frac{12.4\sin32^\circ}{8.5}=0.7731\)
• \(B=50.63^\circ\text{ or }B=180^\circ-50.63^\circ=129.4^\circ\)
• \(129.4+32=161.4<180\text{, so both fit}\)

In triangle ABC, angle A = 46°, BC = 9.2 cm and AC = 11.8 cm. Angle B can have two values. Find

1. (a) both possible values of angle B

   • \(\sin B=\frac{11.8\sin46^\circ}{9.2}=0.9226\)
   • \(B=67.31^\circ\text{ or }112.7^\circ\)

2. (b) the angle C in each case

   • \(C=180^\circ-46^\circ-67.31^\circ=66.69^\circ\)
   • \(C=180^\circ-46^\circ-112.7^\circ=21.31^\circ\)

In triangle ABC, angle A = 50° and AC = 10 cm. How many different triangles are there when BC is

1. (a) a = 7 cm

   • \(h=10\sin50^\circ=7.6604\); a = 7: a < h: no triangle

2. (b) a = 9 cm

   • \(h=10\sin50^\circ=7.6604\); a = 9: h < a < b: two triangles

3. (c) a = 12 cm

   • \(h=10\sin50^\circ=7.6604\); a = 12: a ≥ b: one triangle

4. (d) a = 10 cm

   • \(h=10\sin50^\circ=7.6604\); a = 10: a ≥ b: one triangle

A boat leaves a harbour H and sails 14 km to a buoy B. A lighthouse L is 11 km from the buoy and angle LHB = 40°. Find

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

   • \(\frac{\sin L}{14}=\frac{\sin40^\circ}{11}\;\Rightarrow\;\sin L=0.8181\)
   • \(L=54.89^\circ\text{ or }125.1^\circ\;(180^\circ-54.89^\circ)\)

2. (b) the two possible distances HL

   • \(\text{angle at }B=180^\circ-40^\circ-54.89^\circ=85.11^\circ,\;\;HL=\frac{11\sin85.11^\circ}{\sin40^\circ}=17.1\)
   • \(\text{angle at }B=180^\circ-40^\circ-125.1^\circ=14.89^\circ,\;\;HL=4.40\)

Radar stations R and S track a plane P. PS = 52 km, RP = 38 km and angle PRS = 62°. Find angle PSR.

• \(\frac{\sin S}{38}=\frac{\sin62^\circ}{52}\)
• \(\sin S=0.6452,\;\;S=40.18^\circ\)
• \(180^\circ-40.18^\circ=139.8^\circ\text{ is too big with }62^\circ\text{, so there is one answer}\)
• 40.2°

A designer makes a triangular frame ABC with AB = 15 cm and angle A = 36°. The side BC has length x cm. Find the values of x for which two different frames can be made.

• Two triangles need the side opposite A to be longer than the height h but shorter than the other known side.
• \(h=15\sin36^\circ=8.817\)
• \(8.82<x<15\)

In triangle ABC, angle A = 28°, BC = 7.2 cm and AC = 13.5 cm. There are two possible triangles. Find

1. (a) both possible values of angle B

   • \(\sin B=\frac{13.5\sin28^\circ}{7.2}=0.8803\)
   • \(B=61.67^\circ\text{ or }118.3^\circ\)

2. (b) the area of each triangle

   • \(C=90.33^\circ\text{ or }33.67^\circ\)
   • \(\text{area}=\frac{1}{2}ab\sin C=\frac{1}{2}\times7.2\times13.5\times\sin90.33^\circ=48.6\)
   • \(\text{or }\frac{1}{2}\times7.2\times13.5\times\sin33.67^\circ=26.9\)