Sine and cosine rules

Geometry and measures

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A triangle has two sides of $8$ cm and $10$ cm with an included angle of $30°$. Find its area.

2 marks
In a triangle, angle $A = 30°$, the side opposite $A$ is $6$ cm and angle $B = 45°$. Find the side opposite $B$, in exact form.

2 marks
Angle $A = 40°$, angle $B = 60°$ and the side opposite $A$ is $10$ cm. Find the side opposite $B$ to $3$ significant figures.

3 marks
Two sides of a triangle are $5$ cm and $7$ cm, and the included angle is $60°$. Find the exact length of the third side.

3 marks
A triangle has sides $6$ cm, $7$ cm and $8$ cm. Find the angle opposite the $8$ cm side to $1$ decimal place.

4 marks
Two sides are $12$ cm and $9$ cm, and the included angle is $40°$. Find the area to $3$ significant figures.

4 marks
In triangle $ABC$, angle $A = 50°$, angle $B = 65°$ and the side opposite $A$ is $8$ cm. Find the side opposite $B$ to $3$ significant figures.

5 marks
A triangle has sides $7$ cm, $8$ cm and $9$ cm. Use the cosine rule to find the angle opposite $9$ cm to $1$ decimal place. Then find the exact area of the triangle.

Triangle with sides 7 cm, 8 cm and 9 cm. Angle C is opposite the 9 cm side.

6 marks

Mark scheme

Answer. $20$ cm$^{2}$
- $\dfrac{1}{2} \times 8 \times 10 \times \sin 30°$
- $20$ cm$^{2}$
Answer. $6\sqrt{2}$ cm
- $\dfrac{b}{\sin 45°} = \dfrac{6}{\sin 30°}$
- $b = 6 \times \dfrac{\sqrt{2}/2}{1/2} = 6\sqrt{2}$ cm
Answer. $13.5$ cm
- $b = \dfrac{10 \sin 60°}{\sin 40°}$
- $b = 13.473\ldots = 13.5$ cm to $3$ s.f.
Answer. $\sqrt{39}$ cm
- $c^{2} = 5^{2} + 7^{2} - 2 \times 5 \times 7 \times \cos 60°$
- $c^{2} = 74 - 35 = 39$, so $c = \sqrt{39}$ cm.
Answer. $75.5°$
- $\cos C = \dfrac{6^{2} + 7^{2} - 8^{2}}{2 \times 6 \times 7} = 0.25$
- $C = \cos^{-1}(0.25) = 75.5°$ to $1$ d.p.
Answer. $34.7$ cm$^{2}$
- $\dfrac{1}{2} \times 12 \times 9 \times \sin 40° = 54 \sin 40°$
- $34.711\ldots = 34.7$ cm$^{2}$ to $3$ s.f.
Answer. $9.46$ cm
- Angle $C = 65°$ as well, but the required side is opposite $65°$.
- $b = \dfrac{8 \sin 65°}{\sin 50°} = 9.4648\ldots$
- $9.46$ cm to $3$ s.f.
Answer. $73.4°$, and the area is $12\sqrt{5}$ cm$^{2}$
- $\cos C = \dfrac{7^{2} + 8^{2} - 9^{2}}{2 \times 7 \times 8} = \dfrac{2}{7}$
- $C = 73.4°$ to $1$ d.p.
- $\sin C = \dfrac{3\sqrt{5}}{7}$
- Area $= \dfrac{1}{2} \times 7 \times 8 \times \dfrac{3\sqrt{5}}{7} = 12\sqrt{5}$ cm$^{2}$.