In a triangle P Q R ,   l e t   ∠ P Q R = 30 o and the sides P Q   a n d   Q R…

1. $\angle QPR={45}^o$
2. The area of the triangle $PQR \mathrm{i}\mathrm{s} 25\sqrt{3} \mathrm{a}\mathrm{n}\mathrm{d} \angle QRP={120}^o$
3. The radius of the incircle of the triangle $PQR \mathrm{i}\mathrm{s} 10\sqrt{3}-15$
4. The area of the circumcircle of the triangle $PQR \mathrm{i}\mathrm{s} 100\pi$

Solution

$\cos {30}^o=\frac{{\left(10\sqrt{3}\right)}^2+{\left(10\right)}^2-{\left(PR\right)}^2}{2\times 10\sqrt{3}\times 10}$
$\Rightarrow PR=10$
$\because QR=PR \Rightarrow \angle PQR=\angle QPR$
$\angle QPR={30}^o$
$\left(B\right)$ Area of $∆PQR=\frac{1}{2}\times 10\sqrt{3}\times 10\times \sin {30}^o=\frac{1}{2}\times 10\times 10\sqrt{3}\times \frac{1}{2}$
$= 25\sqrt{3}$
$\angle QRP={180}^o-\left({30}^o+{30}^o\right)={120}^o$
$\left(C\right)$ $r=\frac{∆}{S}=\frac{25\sqrt{3}}{\left(\frac{10+10+10\sqrt{3}}{2}\right)}=\frac{25\sqrt{3}}{10+5\sqrt{3}}$
$= 5\sqrt{3}.\left(2-\sqrt{3}\right)=10\sqrt{3}-15$
$\left(D\right)$ $R=\frac{a}{2\sin A}=\frac{10}{2\sin {30}^o}=10$
$\therefore \mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}=\pi R^2=100\pi$

Asked in: JEE Advanced 2018 (Paper 1)

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