ABCD is a trapezium such that AB and CD are p | vedclass

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

Question

(A) Let $AB = x$. Drop a perpendicular from $D$ to $AB$ at point $M$. Then $DM = BC = p$ and $MB = CD = q$. Thus,$AM = AB - MB = x - q$.In $	riangle

Vedclass · Accepted Answer

$\frac{(p^2 + q^2) \sin 	heta}{p \cos 	heta + q \sin 	heta}$

Vedclass · Answer

(A) Let $AB = x$. Drop a perpendicular from $D$ to $AB$ at point $M$. Then $DM = BC = p$ and $MB = CD = q$. Thus,$AM = AB - MB = x - q$.In $	riangle BCD$,$	an \alpha = \frac{p}{q}$.In $	riangle ADM$,$\angle DAM = \pi - (	heta + \alpha)$.Therefore,$	an(\pi - (	heta + \alpha)) = \frac{DM}{AM} = \frac{p}{x - q}$.This implies $-	an(	heta + \alpha) = \frac{p}{x - q}$,or $	an(	heta + \alpha) = \frac{p}{q - x}$.$q - x = p \cot(	heta + \alpha) = p \left( \frac{\cot 	heta \cot \alpha - 1}{\cot \alpha + \cot 	heta} ight)$.Since $\cot \alpha = \frac{q}{p}$,we have $q - x = p \left( \frac{\frac{q}{p} \cot 	heta - 1}{\frac{q}{p} + \cot 	heta} ight) = p \left( \frac{q \cot 	heta - p}{q + p \cot 	heta} ight) = p \left( \frac{q \cos 	heta - p \sin 	heta}{q \sin 	heta + p \cos 	heta} ight)$.$x = q - \frac{pq \cos 	heta - p^2 \sin 	heta}{q \sin 	heta + p \cos 	heta} = \frac{q^2 \sin 	heta + pq \cos 	heta - pq \cos 	heta + p^2 \sin 	heta}{p \cos 	heta + q \sin 	heta}$.$x = \frac{(p^2 + q^2) \sin 	heta}{p \cos 	heta + q \sin 	heta}$.

$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC \perp CD$. If $\angle ADB = \theta$,$BC = p$ and $CD = q$,then $AB$ is equal to:

Similar Questions

If ${c^2} = {a^2} + {b^2}$,then $4s(s - a)(s - b)(s - c) = $

If the two angles on the base of a triangle are $22.5^o$ and $112.5^o$,then the ratio of the height of the triangle to the length of the base is

The equation $\sin 3\theta = 4 \sin \theta \sin 2\theta \sin 4\theta$ in the interval $0 \le \theta \le \pi$ has:

$A$ triangle $ABC$ has area of $P$ square units and perimeter $2S$ units. If $h_1, h_2$ and $h_3$ are respectively the lengths of the altitudes of the triangle drawn from the vertices $A, B$ and $C$,then $P^2 \left[ \frac{(h_1 h_2 + h_2 h_3 + h_3 h_1)^2}{h_1^2 h_2^2 h_3^2} - 2 \right] =$

$ABC$ is a triangular park with $AB = AC = 100 \, m$. $A$ clock tower is situated at the mid-point $D$ of $BC$. The angles of elevation of the top of the tower at $A$ and $B$ are $\cot^{-1} 3.2$ and $\csc^{-1} 2.6$ respectively. The height of the tower is .... $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module