The area (in square units) of $ riangle A B C$ if $\angle A=75^{\circ}, \angle B=45^{\circ}$ and $a=2(\sqrt{3}+1)$ is

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The area (in square units) of $	riangle A B C$ if $\angle A=75^{\circ}, \angle B=45^{\circ}$ and $a=2(\sqrt{3}+1)$ is

Aicharya · Accepted Answer

Given, $\angle A=75^{\circ}, \angle B=45^{\circ}$ and $a=2(\sqrt{3}+1)$ In $\triangle A O C, \quad \tan 60^{\circ}=\frac{x}{y} \Rightarrow \sqrt{3}=\frac{x}{y}$

$ \Rightarrow \quad x=\sqrt{3} y $ Now, $\quad x+y=2(\sqrt{3}+1)$ $ \begin{aligned} & \Rightarrow \quad \sqrt{3} y+y=2(\sqrt{3}+1) \\ & \Rightarrow \quad y(\sqrt{3}+1)=2(\sqrt{3}+1) \\ & \Rightarrow \quad y=2 \Rightarrow x=2 \sqrt{3} \\ & \end{aligned} $ Now, area of $\triangle A B C=$ area of $\triangle A O B+$ area of $\triangle A O C$ $ \begin{aligned} & =\frac{1}{2} \times x \times x+\frac{1}{2} \times x \times y=\frac{1}{2} x[x+y] \\ & =\frac{1}{2} \times 2 \sqrt{3} \times 2(\sqrt{3}+1) \\ & =2 \sqrt{3}(\sqrt{3}+1) \\ & =6+2 \sqrt{3} \text { sq units } \end{aligned} $

The area (in square units) of $\triangle A B C$ if $\angle A=75^{\circ}, \angle B=45^{\circ}$ and…

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