Question

In $∆ABC$, if $AB=5,\angle B={\cos }^{-1}\left(\frac{3}{5}\right)$ and the radius of circumcircle of triangle is $5$.

Then the area of $∆ABC$ is

• A. $6+8\sqrt{3}$
• B. $3+4\sqrt{3}$
• C. $3+8\sqrt{3}$
• D. $6+8\sqrt{3}$

Answer 1

The correct option is D

$6+8\sqrt{3}$

Explanation for the correct option

Finding the area of $∆ABC$

$$\begin{gathered} \because \cos B=\frac{3}{5} \\ =\frac{\text{Base}}{\text{Hypotenuse}} \\ \Rightarrow \sin B=\frac{\text{Perpendicular}}{\text{Hypotenuse}} \\ =\frac{\sqrt{5^2-3^2}}{5} \\ =\frac{4}{5};\left[\because \left(\text{Hypotenuse}\right)²=\left(\text{Perpendicular}\right)²+\left(\text{Base}\right)²\right] \end{gathered}$$

$$\begin{gathered} \because b=2R\sin B \\ =2\times 5\times \frac{4}{5};\left[\because R=5isgiven\&sinB=\frac{4}{5}\right] \\ =8 \end{gathered}$$

Applying cosine rule in $∆ABC$ having side$a,b,c$

$$\begin{gathered} \Rightarrow \cos B=\frac{a^2+c^2-b^2}{2ac} \\ \Rightarrow \cos B=\frac{3}{5};\mathbf{\left[}\mathbf{\because }\mathbf{AB}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{}\mathbf{=}\mathbf{}\mathbf{c}\mathbf{}\mathbf{\&}\mathbf{}\mathbf{\angle }\mathbf{B}\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{}\left(\frac{3}{5}\right)\mathbf{}\mathbf{\Rightarrow }\mathbf{}\mathbf{cosB}\mathbf{}\mathbf{=}\frac{\mathbf{3}}{\mathbf{5}}\mathbf{}\mathbf{\right]} \\ \Rightarrow \frac{a^2+25-64}{2a\times 5}=\frac{3}{5} \\ \Rightarrow a^2-39=6a \\ \Rightarrow a^2-6a-39=0 \\ \Rightarrow a=\frac{6\pm \sqrt{36+156}}{2} \\ \Rightarrow a=\frac{6\pm \sqrt{192}}{2} \\ \Rightarrow a=\frac{\left(6\pm 8\sqrt{3}\right)}{2} \\ \Rightarrow a=3\pm 4\sqrt{3} \\ \Rightarrow a=3+4\sqrt{3}\mathbf{[}\mathbf{S}\mathbf{i}\mathbf{n}\mathbf{c}\mathbf{e}\mathbf{,}\mathbf{}\mathbf{a}\mathbf{}\mathbf{i}\mathbf{s}\mathbf{}\mathbf{l}\mathbf{e}\mathbf{n}\mathbf{g}\mathbf{t}\mathbf{h}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{i}\mathbf{t}\mathbf{}\mathbf{c}\mathbf{a}\mathbf{n}\mathbf{n}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{b}\mathbf{e}\mathbf{}\mathbf{n}\mathbf{e}\mathbf{g}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{}\mathbf{]} \end{gathered}$$

We know that the area of a triangle is,

$$\begin{gathered} △=\frac{abc}{4R} \\ =\frac{\left(3+4\sqrt{3}\right)\times 8\times 5}{4\times 5} \\ =6+8\sqrt{3} \end{gathered}$$

Therefore, the area of $∆ABC=$$6+8\sqrt{3}$

Hence, the correct answer is option (D).