Properties of Triangles 1 Question 1
1. The angles $A, B$ and $C$ of a $\triangle A B C$ are in $AP$ and $a: b=1: \sqrt{3}$. If $c=4 cm$, then the area (in $sq cm$ ) of this triangle is
(2019 Main, 10 April II)
(a) $\frac{2}{\sqrt{3}}$
(b) $4 \sqrt{3}$
(c) $2 \sqrt{3}$
(d) $\frac{4}{\sqrt{3}}$
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Answer:
Correct Answer: 1. (c)
Solution:
- It is given that angles of a $\triangle A B C$ are in $A P$.
So, $\quad \angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow \angle B-d+\angle B+\angle B+d=180^{\circ}$
[if $\angle A, \angle B$ and $\angle C$ are in AP, then it taken as $\angle B-d$, $\angle B, \angle B+d$ respectively, where $d$ is common difference of AP]
$$ \begin{aligned} & \Rightarrow \quad 3 \angle B=180^{\circ} \Rightarrow \angle B=60^{\circ} \\ & \text { and } \quad \frac{a}{b}=\frac{1}{\sqrt{3}} \\ & \Rightarrow \quad \frac{\sin A}{\sin B}=\frac{1}{\sqrt{3}} \\ & \Rightarrow \quad \frac{\sin A}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}} \\ & \Rightarrow \quad \sin A=\frac{1}{2} \Rightarrow \angle A=30^{\circ} \\ & \text { [given. } \\ & \text { So, } \quad \angle C=90^{\circ} \end{aligned} $$
$\therefore$ From sine rule,
$$ \begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ & \Rightarrow \quad \frac{a}{\frac{1}{2}}=\frac{b}{\frac{\sqrt{3}}{2}}=\frac{4}{1} \\ & \Rightarrow \quad a=2 cm, b=2 \sqrt{3} cm \\ & \therefore \text { Area of } \triangle A B C=\frac{1}{2} a b \sin C=\frac{1}{2} \times 2 \times 2 \sqrt{3} \times 1 \\ & =2 \sqrt{3} sq \cdot cm \end{aligned} $$
