Radians and triangles

Trigonometry

A-level
Non-calculator
8 questions
37 marks

These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.

Convert $150^\circ$ to radians, and convert $\dfrac{3\pi}{4}$ to degrees.

3 marks
A sector of a circle has radius $6$ cm and angle $\dfrac{2\pi}{3}$ radians. Find the exact arc length and the exact area of the sector.

3 marks
Write down the exact values of $\sin \dfrac{\pi}{3}$, $\cos \dfrac{\pi}{6}$, $\tan \dfrac{\pi}{4}$ and $\sin \dfrac{5\pi}{6}$.

4 marks
In triangle $ABC$, $AB = c = 8$, $AC = b = 7$ and angle $A = 60^\circ$. Find the exact area of the triangle.

4 marks
In triangle $ABC$, $a = 8$, $b = 7$ and $c = 5$. Find $\cos A$, where $A$ is the angle opposite the side of length $8$.

5 marks
In triangle $ABC$, angle $A = 30^\circ$, angle $B = 45^\circ$ and side $a = 6$. Find the exact length of side $b$.

5 marks
A sector has arc length $8$ cm and area $24$ cm$^2$. Find the radius and the angle, in radians, at the centre.

6 marks
In triangle $ABC$, $AB = c = 7$, $AC = b = 9$ and angle $A = 60^\circ$. Find the exact length of $BC$ and the exact area of the triangle.

Triangle ABC, not to scale

7 marks

Mark scheme

Answer. $150^\circ = \dfrac{5\pi}{6}$ and $\dfrac{3\pi}{4} = 135^\circ$.
- $150 \times \dfrac{\pi}{180} = \dfrac{5\pi}{6}$.
- $\dfrac{3\pi}{4} \times \dfrac{180}{\pi} = 135$.
Answer. The arc length is $4\pi$ cm and the area is $12\pi$ cm$^2$.
- Arc length $= 6 \times \dfrac{2\pi}{3} = 4\pi$.
- Area $= \dfrac{1}{2} \times 6^2 \times \dfrac{2\pi}{3} = 12\pi$.
Answer. $\dfrac{\sqrt{3}}{2}$, $\dfrac{\sqrt{3}}{2}$, $1$ and $\dfrac{1}{2}$.
- $\sin \dfrac{\pi}{3} = \cos \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$.
- $\tan \dfrac{\pi}{4} = 1$ and $\sin \dfrac{5\pi}{6} = \dfrac{1}{2}$.
Answer. $14\sqrt{3}$
- Area $= \dfrac{1}{2}bc\sin A$.
- $\dfrac{1}{2} \times 7 \times 8 \times \dfrac{\sqrt{3}}{2} = 14\sqrt{3}$.
Answer. $\cos A = \dfrac{1}{7}$
- The largest side is opposite $A$, so use the cosine rule.
- $\cos A = \dfrac{7^2 + 5^2 - 8^2}{2 \times 7 \times 5} = \dfrac{10}{70} = \dfrac{1}{7}$.
Answer. $6\sqrt{2}$
- Angle $C = 105^\circ$, but side $b$ is found from the sine rule using angles $A$ and $B$.
- $\dfrac{b}{\sin 45^\circ} = \dfrac{6}{\sin 30^\circ}$.
- $b = 6 \times \dfrac{\sqrt{2}/2}{1/2} = 6\sqrt{2}$.
Answer. The radius is $6$ cm and the angle is $\dfrac{4}{3}$ radians.
- $r\theta = 8$ and $\dfrac{1}{2}r^2\theta = 24$.
- Substitute $\theta = \dfrac{8}{r}$ to get $4r = 24$.
- $r = 6$ and $\theta = \dfrac{4}{3}$.
Answer. $BC = \sqrt{67}$ and the area is $\dfrac{63\sqrt{3}}{4}$.
- By the cosine rule, $BC^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \dfrac{1}{2} = 67$.
- Area $= \dfrac{1}{2} \times 7 \times 9 \times \dfrac{\sqrt{3}}{2} = \dfrac{63\sqrt{3}}{4}$.