In the figure, theta_1+theta_2=frac{pi}{2} an | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\sqrt{3} BE =4 AB$

$Ar (	riangle CAB )=2 \sqrt{3}-3$

$\frac{1}{2} x ^2 	an 	heta_1=2 \sqrt{3}-3$

$BE = BD + DE$

$= x \left(	an 	heta

Vedclass · Accepted Answer

$6$

Vedclass · Answer

$\sqrt{3} BE =4 AB$

$Ar (	riangle CAB )=2 \sqrt{3}-3$

$\frac{1}{2} x ^2 	an 	heta_1=2 \sqrt{3}-3$

$BE = BD + DE$

$= x \left(	an 	heta_1+	an 	heta_2ight)$

$BE = AB \left(	an 	heta_1+\cot 	heta_1ight)$

$\frac{4}{\sqrt{3}} 	an 	heta_1+\cot 	heta_1 \Rightarrow 	an 	heta_1=\sqrt{3}, \frac{1}{\sqrt{3}}$

$	heta_1=\frac{\pi}{6}$

$	heta_1=\frac{\pi}{3} \quad 	heta_2=\frac{\pi}{3}$

$as \frac{	heta_2}{	heta_1} 	ext { is } \operatorname{largest} 	herefore 	heta_1=\frac{\pi}{6} 	heta_2=\frac{\pi}{3}$

$	herefore x ^2=\frac{(2 \sqrt{3}-3) 	imes 2}{	an 	heta_1}=\frac{\sqrt{3}(2-\sqrt{3}) 	imes 2}{	an \frac{\pi}{6}}$

$x ^2=12-6 \sqrt{3}=(3-\sqrt{3})^2$

$x =3-\sqrt{3}$

Perimeter of $	riangle C E D$

$= CD + DE + CE$

$=3 \sqrt{3}+(3-\sqrt{3}) \sqrt{3}+(3-\sqrt{3}) 	imes 2=6$

In the figure, $\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(B E)=4(A B)$. If the area of $\triangle CAB$ is $2 \sqrt{3}-3$ unit $^2$, when $\frac{\theta_2}{\theta_1}$ is the largest, then the perimeter (in unit) of $\triangle CED$ is equal to $...........$.

Similar Questions

The value of $\left( {1 + \cos \frac{\pi }{9}} \right)\left( {1 + \cos \frac{{3\pi }}{9}} \right)\left( {1 + \cos \frac{{5\pi }}{9}} \right)\left( {1 + \cos \frac{{7\pi }}{9}} \right)$ is

The value of $\sum_{r-1}^{18} cos^2(5r)^o,$ where $x^o $ denotes the $x$ degree, is equals to

If $a{\sin ^2}x + b{\cos ^2}x = c,\,\,$$b\,{\sin ^2}y + a\,{\cos ^2}y = d$ and $a\,\tan x = b\,\tan y,$ then $\frac{{{a^2}}}{{{b^2}}}$ is equal to

In a triangle $ABC,$ the value of $\sin A + \sin B + \sin C$ is

$2{\cos ^2}\theta - 2{\sin ^2}\theta = 1$, then $\theta =$ .......$^o$