Solving Triangles — A-Level Maths practice questions

Question 1

In $	riangle ABC$, $a = 5$, $b = 6$, $c = 7$. The inradius $r$ is closest to:

Accepted Answer

$s = 9$. By Heron, Area $= \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = 6\sqrt{6} \approx 14.70$. So $r \approx \dfrac{14.70}{9} \approx 1.63$.

Answer

$1.4$

Answer

$1.6$

Answer

$2.4$

Answer

$3.0$

Question 2

In a triangle, $a$ is opposite $\angle A = 30^{\circ}$ and $b$ is opposite $\angle B = 90^{\circ}$. If $a = 5$, then $b$ equals:

Accepted Answer

$\dfrac{5}{\sin 30^{\circ}} = \dfrac{b}{\sin 90^{\circ}}$.

$\sin 30^{\circ} = 	frac{1}{2}$ and $\sin 90^{\circ} = 1$, so $\dfrac{5}{1/2} = \dfrac{b}{1}$.

$b = 10$.

Answer

$5$

Answer

$\dfrac{5}{2}$

Answer

$10$

Answer

$5\sqrt{3}$

Question 3

In a triangle, two sides have lengths $4$ and $5$ and the included angle is $60^{\circ}$. The length of the third side is:

Accepted Answer

$c^{2} = 4^{2} + 5^{2} - 2(4)(5)\cos 60^{\circ} = 16 + 25 - 40 \cdot 	frac{1}{2} = 41 - 20 = 21$.

So $c = \sqrt{21}$.

Answer

$\sqrt{41}$

Answer

$\sqrt{21}$

Answer

$\sqrt{61}$

Answer

$3$

Question 4

The area of a triangle with sides $5$, $6$, $7$ is closest to:

Accepted Answer

$s = \dfrac{5 + 6 + 7}{2} = 9$. $A = \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = \sqrt{216} = 6\sqrt{6} \approx 14.70$.

Answer

$10$

Answer

$12$

Answer

$14.7$

Answer

$21$

Question 5

In $	riangle ABC$, $\angle A = \dfrac{\pi}{6}$, $\angle B = \dfrac{\pi}{4}$, and $a = 4$. The side $b$ equals:

Accepted Answer

$\dfrac{4}{\sin(\pi/6)} = \dfrac{b}{\sin(\pi/4)} \Rightarrow \dfrac{4}{1/2} = \dfrac{b}{\sqrt{2}/2} \Rightarrow 8 = \dfrac{b}{\sqrt{2}/2} \Rightarrow b = 8 \cdot \dfrac{\sqrt{2}}{2} = 4\sqrt{2}$.

Answer

$2\sqrt{2}$

Answer

$4$

Answer

$4\sqrt{2}$

Answer

$8\sqrt{2}$

Question 6

In $	riangle ABC$, $a = 6$ and $\angle A = \dfrac{\pi}{6}$. The circumradius $R$ equals:

Accepted Answer

$2R = \dfrac{a}{\sin A} = \dfrac{6}{1/2} = 12 \Rightarrow R = 6$.

Answer

$3$

Answer

$\sqrt{3}$

Answer

$6$

Answer

$12$

Question 7

In $	riangle ABC$, $AB = 5$, $BC = 12$, $AC = 13$. The area equals:

Accepted Answer

Since $5^2 + 12^2 = 13^2$, $	riangle ABC$ is right-angled at $B$. Area $= \dfrac{1}{2} \cdot 5 \cdot 12 = 30$.

Answer

$20$

Answer

$24$

Answer

$30$

Answer

$60$

Question 8

In $	riangle ABC$, $AB = 6$, $AC = 8$, and $\angle A = \dfrac{\pi}{6}$. The area equals:

Accepted Answer

Area $= \dfrac{1}{2} \cdot 6 \cdot 8 \cdot \sin\dfrac{\pi}{6} = 24 \cdot \dfrac{1}{2} = 12$.

Answer

$48$

Answer

$24$

Answer

$12$

Answer

$6$

Question 9

A right triangle has legs $3$ and $4$. Its area equals:

Accepted Answer

Area $= \dfrac{1}{2} \cdot 3 \cdot 4 = 6$.

Answer

$5$

Answer

$6$

Answer

$7$

Answer

$12$

Question 10

In $	riangle ABC$, $a = 7$, $b = 8$, $c = 5$. The value of $\cos C$ is:

Accepted Answer

$25 = 49 + 64 - 112\cos C \Rightarrow 112\cos C = 88 \Rightarrow \cos C = \dfrac{88}{112} = \dfrac{11}{14}$.

Answer

$\dfrac{1}{14}$

Answer

$\dfrac{11}{16}$

Answer

$\dfrac{11}{14}$

Answer

$\dfrac{5}{8}$

Solving Triangles practice questions — A-Level Maths

Problems & worked solutions

Problem 1 — Solving Triangles

Problem 2 — Solving Triangles

Problem 3 — Solving Triangles

Problem 4 — Solving Triangles

Problem 5 — Solving Triangles

Problem 6 — Solving Triangles

Problem 7 — Solving Triangles

Problem 8 — Solving Triangles

Problem 9 — Solving Triangles

Problem 10 — Solving Triangles