Question

The angles $A,B,C$ of a triangle $ABC$ are in $A.P$ and $a:b=1:\sqrt{3}$ . If $c=4cm$, then the area (in $sq.cm$) of this triangle is

Answer 1

Given that the angles $A,B,C$ of a triangle $ABC$ are in $A.P$

$\Rightarrow 2B=A+C------eq\left(i\right)$

Step1 - Finding the value of $\angle B$

We know

Angle sum property of a triangle, the sum of all angles in a triangle is $=180°$

$\Rightarrow \angle A+\angle B+\angle C=180°$

$$\begin{gathered} \Rightarrow 3B=180-------(byeq(i\left)\right) \\ \Rightarrow B=\frac{180}{3} \\ \Rightarrow \angle B=60° \end{gathered}$$

Step 2 - Finding the value of $\angle A$

Using sine formula -

$$\begin{gathered} \Rightarrow \frac{a}{\sin \left(A\right)}=\frac{b}{\sin \left(B\right)} \\ \Rightarrow \frac{a}{b}=\frac{\sin \left(A\right)}{\sin \left(B\right)} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{\sin \left(A\right)}{\sin \left(60°\right)}\because \left(\angle B=60°and\frac{a}{b}=\frac{1}{\sqrt{3}}\right) \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{\sin \left(A\right)}{{\displaystyle \frac{\sqrt{3}}{2}}}\because \left(\sin \left(60°\right)=\frac{\sqrt{3}}{2}\right) \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{2\sin \left(A\right)}{\sqrt{3}} \\ \Rightarrow 1=2\sin \left(A\right) \\ \Rightarrow \sin \left(A\right)=\frac{1}{2} \\ \Rightarrow \sin \left(A\right)=\sin \left(30°\right)\because \left(\sin \left(30°\right)=\frac{1}{2}\right) \\ \Rightarrow A=30° \end{gathered}$$

Step 3- Finding the value of $\angle C$

Again, using the angle sum property of the triangle

$$\begin{gathered} \Rightarrow \angle A+\angle B+\angle C=180° \\ \Rightarrow 30+60+\angle C=180 \\ \Rightarrow 90+\angle C=180 \\ \Rightarrow \angle C=180-90 \\ \Rightarrow \angle C=90° \end{gathered}$$

Step 4 - Finding the area of a triangle

Using sine formula

$$\begin{gathered} \Rightarrow \frac{a}{\sin \left(A\right)}=\frac{c}{\sin \left(C\right)} \\ \Rightarrow \frac{a}{c}=\frac{\sin \left(A\right)}{\sin \left(C\right)} \\ \Rightarrow \frac{a}{4}=\frac{\sin \left(30°\right)}{\sin \left(90°\right)}\because \left(c=4cm,\angle A=30°,\angle C=90°\right) \\ \Rightarrow \frac{a}{4}=\frac{{\displaystyle \frac{1}{2}}}{1}\because \left(\sin \left(30°\right)=\frac{1}{2},\sin \left(90°\right)=1\right) \\ \Rightarrow \frac{a}{4}=\frac{1}{2} \\ \Rightarrow a=2cm \end{gathered}$$

Also,

$$\begin{gathered} \Rightarrow \frac{a}{b}=\frac{1}{\sqrt{3}}\because given \\ \Rightarrow \frac{2}{b}=\frac{1}{\sqrt{3}} \\ \Rightarrow b=2\sqrt{3}cm \end{gathered}$$

Now,

Area of triangle $=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times a\times b$

$$\begin{gathered} =\left(\frac{1}{2}\times 2\times 2\sqrt{3}\right){cm}^2 \\ =2\sqrt{3}{cm}^2 \end{gathered}$$

Hence, the area of a triangle is $2\sqrt{3}{cm}^2$