For the triangle shown in the figure,side b = | vedclass

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

Question

(C) In $\Delta ABC$,the sum of angles is $180^{\circ}$. Given $\angle B = 120^{\circ}$ and $\angle C = \angle A = 	heta$,we have $120^{\circ} + 2	he

Vedclass · Accepted Answer

side $a = 	ext{side } c = \frac{10 \sqrt{3}}{3} \, m$

Vedclass · Answer

(C) In $\Delta ABC$,the sum of angles is $180^{\circ}$. Given $\angle B = 120^{\circ}$ and $\angle C = \angle A = 	heta$,we have $120^{\circ} + 2	heta = 180^{\circ}$,so $2	heta = 60^{\circ}$,which means $	heta = 30^{\circ}$.Draw a perpendicular $BM$ from $B$ to $AC$. Since $\Delta ABC$ is isosceles,$M$ is the midpoint of $AC$,so $CM = 5 \, m$.In right-angled $\Delta BMC$,$\cos 30^{\circ} = \frac{CM}{BC}$.$\frac{\sqrt{3}}{2} = \frac{5}{a}$.$a = \frac{10}{\sqrt{3}} = \frac{10 \sqrt{3}}{3} \, m$.Since the triangle is isosceles,$a = c = \frac{10 \sqrt{3}}{3} \, m$.

For the triangle shown in the figure,side $b = 10 \, m$,and angles $\angle C$ and $\angle A$ are equal. Find the lengths of sides $a$ and $c$.

Similar Questions

The sides of a triangle are in the ratio $1 : \sqrt{3} : 2$. Then the angles are in the ratio

If $a=2, b=3, c=4$ in a triangle $ABC$,then $\cos C=$

In a triangle $ABC$,$b = \sqrt{3}$,$c = 1$ and $\angle A = 30^\circ$,then the largest angle of the triangle is ....$^\circ$

In a $\triangle ABC$, $a^2 \sin 2C + c^2 \sin 2A$ is equal to (in $\Delta$)

In a $\triangle ABC$,$\frac{a-b}{a+b} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module