In A B C, 1 2 (B-C) 1 2 A = - Vidyadip

MCQ

In $\triangle A B C, \frac{\cos \frac{1}{2}(B-C)}{\sin \frac{1}{2} A}=$

A
$\frac{ b - c }{ a }$
✓
$\frac{ b + c }{ a }$
C
$\frac{ a }{ b - c }$
D
$\frac{a}{b+c}$

✓

Answer

Correct option: B.

$\frac{ b + c }{ a }$

(B) $\frac{\cos \frac{ B - C }{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{ B + C }{2} \cos \frac{B- C }{2}}{\sin \frac{B+ C }{2} \sin \frac{A}{2}}$
$=\frac{2 \sin \frac{B+ C }{2} \cos \frac{B- C }{2}}{2 \sin \left(\frac{\pi}{2}-\frac{ A }{2}\right) \sin \frac{ A }{2}}$
$=\frac{\sin B+\sin C}{\sin A}=\frac{b+c}{a}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Trigonometric Functions→Trigonometric Functions — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

The integral value of $\int_{-2}^0\left[x^3+3 x^2+3 x+3+(x+1) \cos (x+1)\right] d x$ is

The joint equation of lines passing through the origin and trisecting the first quadrant is

In $\triangle A B C$, if $\cos A=\frac{\sin B}{2 \sin C}$, then $\triangle A B C$ is

The probability distribution of a discrete random variable $X$ is given below:

$\text{X}:$	$2$	$3$	$4$	$5$
$\text{P}(\text{X}):$	$\frac{5}{\text{k}}$	$\frac{7}{\text{k}}$	$\frac{9}{\text{k}}$	$\frac{11}{\text{k}}$

The value of $k$ is:

If the equation $p x^2-8 x y+3 y^2+14 x+2 y+ q =0$ represents a pair of lines perpendicular to each other, then the values of $p$ and $q$ are

If the vector $8 \hat{ i }+a \hat{ j }$ of magnitude 10 is in the direction of the vector $4 \hat{ i }-3 \hat{ j }$, then the value of a is equal to

The area bounded by the parabola $x^2=2 y$ and the line $y=3 x$ is

If $\sin 3 \alpha=4 \sin \alpha \sin (x+\alpha) \sin (x-\alpha)$, then $x=$

If k is a scalar and I is a unit matrix of order 3 , then $\operatorname{adj}( kI )=$

A man of height $6ft$ walks at a uniform speed of $9ft/\sec.$ from a lamp fixed at $15ft$ height. The length of his shadow is increasing at the rate of: