Question

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

• A. $6$
• B. $2\sqrt{3}$
• C. $6-2\sqrt{3}$
• D. $6+2\sqrt{3}$

Answer 1

Answer: (D)

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

$\Rightarrow x=\sqrt{3}y$
Now, $x+y=2(\sqrt{3}+1)$
$\Rightarrow \sqrt{3}y+y=2(\sqrt{3}+1)$
$\Rightarrow y(\sqrt{3}+1)=2(\sqrt{3}+1)$
$\Rightarrow y=2$
$\Rightarrow x=2\sqrt{3}$
Now, area of $\Delta ABC=$ area of $\Delta AOB+$ area of $\Delta AOC$
$=\frac{1}{2}\times x\times x+\frac{1}{2}\times x\times y=\frac{1}{2}x[x+y]$
$=\frac{1}{2}\times 2\sqrt{3}\times 2(\sqrt{3}+1)$
$=2\sqrt{3}(\sqrt{3}+1)$
$=6+2\sqrt{3}$ sq units