JEE Main 2023 (Online) 10th April Evening ShiftQuestion 1
In the figure, $\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(BE)=4(AB)$. 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 $\_\_\_\_\_\_$ . [FIGURE NOT TRANSCRIBED: right angles at $A$ and $D$; $CD\parallel AB$; $BDE$ vertical; angle $\theta_2$ at $C$ between $CD$ and $CE$, angle $\theta_1$ at $B$ between $BA$ and $BC$; see the paper image]
Trigonometric Ratio & Identities
Answer$6$
Solution
Concept. In the figure $ABDC$ is a rectangle ($AB=CD=w$, $BD=h$), $B,D,E$ are on one vertical line, $\theta_1=\angle ABC$ and $\theta_2=\angle DCE$. The figure is drawn for the case $\theta_1=30^\circ$ found below.Formulas used. (1) $\tan\theta_1=\frac hw$, $\tan\theta_2=\frac{DE}{w}$, and $\theta_1+\theta_2=\frac\pi2\Rightarrow\tan\theta_2=\cot\theta_1$Step 1 - $DE$. By (1), $\frac{DE}{w}=\frac wh$, so $DE=\frac{w^2}{h}$ and $BE=h+\frac{w^2}{h}$. Step 2 - Use $\sqrt3\,BE=4\,AB$. With $t=\frac hw=\tan\theta_1$: $\sqrt3(t^2+1)=4t$, so $t=\sqrt3$ or $\frac{1}{\sqrt3}$. Step 3 - Largest $\frac{\theta_2}{\theta_1}$. $\theta_1=30^\circ$, $\theta_2=60^\circ$ gives $2$ (the other case gives $\frac12$). So $h=\frac{w}{\sqrt3}$. Step 4 - Find $w$. Area of $\triangle CAB$ (right angle at $A$) is $\frac{wh}{2}=\frac{w^2}{2\sqrt3}=2\sqrt3-3$, so $w^2=12-6\sqrt3=(3-\sqrt3)^2$ and $w=3-\sqrt3$. Step 5 - $\triangle CED$. Right angle at $D$, $\angle C=60^\circ$: $CD=w$, $DE=\sqrt3w$, $CE=2w$. $$\text{perimeter}=(3+\sqrt3)w=(3+\sqrt3)(3-\sqrt3)=6.$$ Answer: 6