In a right-angled triangle,the hypotenuse is | vedclass

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

Question

(C) Let the right-angled triangle have sides $a$ and $b$ and hypotenuse $c$. Let $p$ be the length of the perpendicular from the right-angle vertex to

Vedclass · Accepted Answer

$\frac{\pi}{8}, \frac{3\pi}{8}$

Vedclass · Answer

(C) Let the right-angled triangle have sides $a$ and $b$ and hypotenuse $c$. Let $p$ be the length of the perpendicular from the right-angle vertex to the hypotenuse.We know that $p = \frac{ab}{c}$.Given $c = 2\sqrt{2}p$,we have $c = 2\sqrt{2} \frac{ab}{c}$,so $c^2 = 2\sqrt{2}ab$.Since $c^2 = a^2 + b^2$,we have $a^2 + b^2 = 2\sqrt{2}ab$.Dividing by $ab$,we get $\frac{a}{b} + \frac{b}{a} = 2\sqrt{2}$.Let $	heta$ be one of the acute angles,so $	an 	heta = \frac{b}{a}$.Then $\cot 	heta + 	an 	heta = 2\sqrt{2}$.$\frac{1}{	an 	heta} + 	an 	heta = 2\sqrt{2} \Rightarrow \frac{1 + 	an^2 	heta}{	an 	heta} = 2\sqrt{2}$.$\frac{\sec^2 	heta}{	an 	heta} = 2\sqrt{2}$ $\Rightarrow \frac{1}{\sin 	heta \cos 	heta} = 2\sqrt{2}$ $\Rightarrow \frac{2}{\sin 2	heta} = 2\sqrt{2}$.$\sin 2	heta = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$.Thus,$2	heta = \frac{\pi}{4}$ or $2	heta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.So,$	heta = \frac{\pi}{8}$ or $	heta = \frac{3\pi}{8}$.The two acute angles are $\frac{\pi}{8}$ and $\frac{3\pi}{8}$.

In a right-angled triangle,the hypotenuse is $2\sqrt{2}$ times the length of the perpendicular drawn from the opposite vertex to the hypotenuse. Then,the other two angles are:

Similar Questions

The number of values of $\theta$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\theta \neq \frac{n \pi}{5}$ for $n=0, \pm 1, \pm 2$ and $\tan \theta = \cot 5 \theta$ as well as $\sin 2 \theta = \cos 4 \theta$ is

If $P_1, P_2$ and $P_3$ are the lengths of the altitudes drawn from the vertices $A, B$ and $C$ of $\triangle ABC$ respectively,then $\frac{\cos A}{P_1} + \frac{\cos B}{P_2} + \frac{\cos C}{P_3} =$

If $2 \cos \theta + \sin \theta = 1$,then the value of $4 \cos \theta + 3 \sin \theta$ is equal to

If the equation $2 \sin^2 x + \frac{\sin 2x}{2} = k$ has at least one real solution,then the sum of all integral values of $k$ is

The sum of all values of $\theta \in [0, 2\pi]$ satisfying $2 \sin^2 \theta = \cos 2\theta$ and $2 \cos^2 \theta = 3 \sin \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module