(A) Given $R = 4r$. In any triangle,$r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.Since $A = B$,$C = 180^{\circ} - 2A$. Thus,$r = 4R

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

$8 \cos^2 A - 8 \cos A + 1 = 0$

(A) Given $R = 4r$. In any triangle,$r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.Since $A = B$,$C = 180^{\circ} - 2A$. Thus,$r = 4R \sin^2(\frac{A}{2}) \sin(\frac{180^{\circ}-2A}{2}) = 4R \sin^2(\frac{A}{2}) \cos A$.Substituting $R = 4r$,we get $r = 4(4r) \sin^2(\frac{A}{2}) \cos A$,which simplifies to $1 = 16 \sin^2(\frac{A}{2}) \cos A$.Using the identity $2 \sin^2(\frac{A}{2}) = 1 - \cos A$,we have $1 = 8(1 - \cos A) \cos A$.$1 = 8 \cos A - 8 \cos^2 A$.Rearranging gives $8 \cos^2 A - 8 \cos A + 1 = 0$.

Class 11 Mathematics Trigonometrical Equations Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

Difficult

The circumradius $R$ of an isosceles triangle $ABC$ is four times its inradius $r$. If $A = B$,then which of the following is true?

A
$8 \cos^2 A - 8 \cos A + 1 = 0$
B
$4 \cos^2 A - 10 \cos A + 1 = 0$
C
$\cos^2 A - \cos A - 3 = 0$
D
$\cos^2 A - \cos A - 8 = 0$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Trigonometrical Equations

Subtopic

Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

If $\alpha, \beta$ and $\gamma$ are the lengths of the altitudes of a $\triangle ABC$ with area $\Delta$,then $\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ is equal to

MediumAP EAMCET 2012

View Solution

If in a triangle $ABC$,$a^2+2bc-(b^2+c^2)=ab \sin \frac{C}{2} \cos \frac{C}{2}$,then $\cot (B+C)=$

MediumTS EAMCET 2019

View Solution

If in a triangle $ABC$,$\cos A \cos B + \sin A \sin B \sin C = 1$,then $a : b : c =$

View Solution

The sides of a triangle inscribed in a given circle subtend angles $\alpha, \beta, \gamma$ at the center. The minimum value of the $A.M.$ of $\cos (\alpha + \frac{\pi}{2})$,$\cos (\beta + \frac{\pi}{2})$ and $\cos (\gamma + \frac{\pi}{2})$ is equal to

EasyAP EAMCET 2021

View Solution

If in a triangle $ABC$,$AB=5$ units,$\angle B=\cos ^{-1}\left(\frac{3}{5}\right)$ and the radius of the circumcircle of $\triangle ABC$ is $5$ units,then the area (in sq. units) of $\triangle ABC$ is:

DifficultJEE MAIN 2021

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The circumradius $R$ of an isosceles triangle $ABC$ is four times its inradius $r$. If $A = B$,then which of the following is true?

Similar Questions

If $\alpha, \beta$ and $\gamma$ are the lengths of the altitudes of a $\triangle ABC$ with area $\Delta$,then $\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ is equal to

If in a triangle $ABC$,$a^2+2bc-(b^2+c^2)=ab \sin \frac{C}{2} \cos \frac{C}{2}$,then $\cot (B+C)=$

If in a triangle $ABC$,$\cos A \cos B + \sin A \sin B \sin C = 1$,then $a : b : c =$

The sides of a triangle inscribed in a given circle subtend angles $\alpha, \beta, \gamma$ at the center. The minimum value of the $A.M.$ of $\cos (\alpha + \frac{\pi}{2})$,$\cos (\beta + \frac{\pi}{2})$ and $\cos (\gamma + \frac{\pi}{2})$ is equal to

If in a triangle $ABC$,$AB=5$ units,$\angle B=\cos ^{-1}\left(\frac{3}{5}\right)$ and the radius of the circumcircle of $\triangle ABC$ is $5$ units,then the area (in sq. units) of $\triangle ABC$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module