Sine rule and cosine rule Problem set A One question at a time

In triangle ABC, angle A = 52°, angle B = 71° and BC = 8.3 cm. Find the length of CA.

• BC is opposite A, and CA is opposite B
• \(\frac{x}{\sin 71°}=\frac{8.3}{\sin 52°}\)
• \(x=\frac{8.3\times\sin 71°}{\sin 52°}\), so from the calculator CA = 9.96 cm

In triangle PQR, angle P = 67°, QR = 14.5 cm and RP = 10.6 cm. Find angle Q, which is acute.

• QR is opposite P, and RP is opposite Q
• \(\frac{\sin\theta}{10.6}=\frac{\sin 67°}{14.5}\)
• \(\sin\theta=\frac{10.6\times\sin 67°}{14.5}\)
• From the calculator: θ = sin⁻¹(0.6729) = 42.3°

In triangle ABC, angle B = 68°, CA = 12.1 cm and AB = 8.7 cm.

1. (a) Find angle C, which is acute.

   • AB is opposite C, and CA is opposite B
   • \(\frac{\sin C}{8.7}=\frac{\sin 68°}{12.1}\)
   • From the calculator: C = sin⁻¹(0.6667) = 41.8°

2. (b) Find angle A.

   • Angles in a triangle add to 180°
   • A = 180° − 68° − 41.8° = 70.2°

In triangle ABC, CA = 9.7 cm, AB = 6.4 cm and angle A = 108°. Find the length of BC.

• Two sides and the angle between them: use the cosine rule
• \(x^2=9.7^2+6.4^2-2\times9.7\times6.4\times\cos 108°\)
• \(x^2=173.4\)
• x = 13.2 cm

In triangle ABC, BC = 8.2 cm, CA = 10.5 cm and AB = 13.1 cm. Find the largest angle.

• The largest angle is opposite the longest side, AB: it is angle C
• \(\cos\theta=\frac{8.2^2+10.5^2-13.1^2}{2\times8.2\times10.5}\) = 0.03415
• θ = cos⁻¹(0.03415) = 88.0°

The diagram shows triangle PQR. The perpendicular from R meets PQ at S, and RS = h. The side opposite P is p and the side opposite Q is q. Show that \(\frac{p}{\sin P}=\frac{q}{\sin Q}\).

• In triangle PSR: \(\sin P=\frac{h}{q}\), so \(h=q\sin P\)
• In triangle QSR: \(\sin Q=\frac{h}{p}\), so \(h=p\sin Q\)
• Both equal h: \(q\sin P=p\sin Q\)
• Divide both sides by \(\sin P\sin Q\): \(\frac{p}{\sin P}=\frac{q}{\sin Q}\)

Two surveyors, A and B, stand 120 m apart on one bank of a lake. They look at a tree C on the other bank. Angle CAB = 62° and angle CBA = 71°. Find the distance AC.

• Angle C = 180° − 62° − 71° = 47°
• Angle C is opposite the known side AB = 120 m
• \(\frac{x}{\sin 71°}=\frac{120}{\sin 47°}\)
• \(x=\frac{120\times\sin 71°}{\sin 47°}\) = 155 m

Two ships leave port P at the same time. Ship A sails 18 km on one course and ship B sails 25 km on another. The angle between the two courses is 52°. Find the distance between the ships.

• The two sides and the angle between them are known: cosine rule
• \(x^2=25^2+18^2-2\times25\times18\times\cos 52°\)
• \(x^2=394.9\)
• x = 19.9 km

A triangular field has sides of 85 m, 112 m and 140 m. Find the largest angle of the field.

• The largest angle is opposite the longest side
• \(\cos\theta=\frac{85^2+112^2-140^2}{2\times85\times112}\) = 0.008876
• θ = cos⁻¹(0.008876) = 89.5°

In triangle ABC, AB = 9.4 cm, AC = 12.7 cm and angle A = 41°.

1. (a) Find BC.

   • \(a^2=12.7^2+9.4^2-2\times12.7\times9.4\times\cos 41°\) = 69.46
   • BC = 8.33 cm

2. (b) Find angle BCA.

   • Sine rule with the side just found: \(\frac{\sin C}{9.4}=\frac{\sin 41°}{8.334}\)
   • C = sin⁻¹(0.7400) = 47.7°
   • The side opposite C (9.4 cm) is shorter than 12.7 cm, so C is acute

In the quadrilateral ABCD, AB = 6.2 cm, BC = 8.4 cm, CD = 9.0 cm, DA = 7.1 cm and angle ABC = 70°.

1. (a) Find the length of AC.

   • \(AC^2=6.2^2+8.4^2-2\times6.2\times8.4\times\cos 70°\) = 73.38
   • AC = 8.57 cm

2. (b) Find angle ADC.

   • \(\cos\theta=\frac{7.1^2+9.0^2-8.566^2}{2\times7.1\times9.0}\) = 0.4541
   • θ = cos⁻¹(0.4541) = 63.0°