(A) Given,$a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$ Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we have: $a \le

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

(A) Given,$a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$ Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we have: $a \left( \frac{1 + \cos C}{2} \right) + c \left( \frac{1 + \cos A}{2} \right) = \frac{3b}{2}$ Multiplying by $2$: $a(1 + \cos C) + c(1 + \cos A) = 3b$ $a + a \cos C + c + c \cos A = 3b$ $(a + c) + (a \cos C + c \cos A) = 3b$ By the projection rule,$a \cos C + c \cos A = b$. Substituting this into the equation: $(a + c) + b = 3b$ $a + c = 2b$

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

Easy

In a $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

TS EAMCET 2021 All TS EAMCET

A
$2b = a + c$
B
$b^2 = ac$
C
$b^2 = \frac{2ac}{a+c}$
D
$a + b + c = 1$

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

Previous Year

TS EAMCET 2021 Paper All TS EAMCET years →

In a $\Delta ABC$,if $\sin A + \sin B + \sin C = 1 + \sqrt{2}$ and $\cos A + \cos B + \cos C = \sqrt{2}$,then the triangle is:

Medium

View Solution

In a triangle $ABC$,if $a < b < c$ and $\frac{a^3+b^3+c^3}{\sin^3 A+\sin^3 B+\sin^3 C}=8$,then the maximum value of $c$ is

MediumTS EAMCET 2020

View Solution

In a $\Delta ABC$,${a^2}\sin 2C + {c^2}\sin 2A = $

Medium

View Solution

In a triangle $ABC$,$\angle B = \frac{\pi}{3}$ and $\angle C = \frac{\pi}{4}$,and $D$ divides $BC$ internally in the ratio $1 : 3$. Then $\frac{\sin \angle BAD}{\sin \angle CAD}$ is equal to

MediumIIT 1995

View Solution

The number of solutions of $\sin ^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0$,where $-\pi \leq x \leq \pi$,is....................

DifficultJEE MAIN 2024

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

In a $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

Similar Questions

In a $\Delta ABC$,if $\sin A + \sin B + \sin C = 1 + \sqrt{2}$ and $\cos A + \cos B + \cos C = \sqrt{2}$,then the triangle is:

In a triangle $ABC$,if $a < b < c$ and $\frac{a^3+b^3+c^3}{\sin^3 A+\sin^3 B+\sin^3 C}=8$,then the maximum value of $c$ is

In a $\Delta ABC$,${a^2}\sin 2C + {c^2}\sin 2A = $

In a triangle $ABC$,$\angle B = \frac{\pi}{3}$ and $\angle C = \frac{\pi}{4}$,and $D$ divides $BC$ internally in the ratio $1 : 3$. Then $\frac{\sin \angle BAD}{\sin \angle CAD}$ is equal to

The number of solutions of $\sin ^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0$,where $-\pi \leq x \leq \pi$,is....................

Vedclass Test Series

Exam Paper Generator

Online Exam Module