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

In $\triangle ABC$,$(\cot A+\cot B)(\cot B+\cot C)(\cot C+\cot A) =$

If in a $\triangle ABC$,$\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$,then $\angle C$ is equal to (in $^{\circ}$)

In a triangle $ABC$,with usual notations,$\cot \left(\frac{A+B}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right) = $

In a triangle $ABC$,with usual notations,if $\tan \left(\frac{A}{2}\right) = \frac{5}{6}$ and $\tan \left(\frac{C}{2}\right) = \frac{2}{5}$,then:

In $\triangle ABC$, with the usual notations, $\angle C = 90^\circ$, then $\sin(A - B)$ is equal to....

Vedclass Test Series

Exam Paper Generator

Online Exam Module