The two adjacent sides of a cyclic quadrilate | vedclass

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

Question

(A) Let the cyclic quadrilateral be $ABCD$ with sides $AB = 2$,$BC = 5$,$CD = 3$,and $DA = d$. The angle $\angle ABC = 60^o$.Since $ABCD$ is a cyclic

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let the cyclic quadrilateral be $ABCD$ with sides $AB = 2$,$BC = 5$,$CD = 3$,and $DA = d$. The angle $\angle ABC = 60^o$.Since $ABCD$ is a cyclic quadrilateral,the opposite angles are supplementary. Therefore,$\angle ADC = 180^o - 60^o = 120^o$.Using the law of cosines in $	riangle ABC$ for the diagonal $AC$:$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(60^o) = 2^2 + 5^2 - 2(2)(5)(0.5) = 4 + 25 - 10 = 19$.Using the law of cosines in $	riangle ADC$ for the diagonal $AC$:$AC^2 = AD^2 + CD^2 - 2(AD)(CD)\cos(120^o) = d^2 + 3^2 - 2(d)(3)(-0.5) = d^2 + 9 + 3d$.Equating the two expressions for $AC^2$:$d^2 + 3d + 9 = 19 \Rightarrow d^2 + 3d - 10 = 0$.Factoring the quadratic equation:$(d + 5)(d - 2) = 0$.Since the side length $d$ must be positive,$d = 2$.

The two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60^o$. If the third side is $3$,the remaining fourth side is

Similar Questions

In any $\triangle ABC$, the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is equal to:

In $\triangle ABC$ with usual notations, if $1 + \tan(\frac{A}{2}) \tan(\frac{B}{2}) = \frac{k}{s}$ (where $s$ is the semi-perimeter), then the value of $k$ is...

In a $\triangle ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then $\triangle ABC$ is

In a triangle $ABC$,$\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^{\circ}$. Then the measure of $\angle A$ is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module