In a triangle ABC with a fixed base BC,the ve | vedclass

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

Question

(C) In $	riangle ABC$,we have $A + B + C = 180^{\circ}$,so $\frac{B+C}{2} = 90^{\circ} - \frac{A}{2}$.Given $\cos B + \cos C = 4 \sin^2 \frac{A}{2}$.

Vedclass · Accepted Answer

$(A, C)$

Vedclass · Answer

(C) In $	riangle ABC$,we have $A + B + C = 180^{\circ}$,so $\frac{B+C}{2} = 90^{\circ} - \frac{A}{2}$.Given $\cos B + \cos C = 4 \sin^2 \frac{A}{2}$.Using the sum-to-product formula: $2 \cos \frac{B+C}{2} \cos \frac{B-C}{2} = 4 \sin^2 \frac{A}{2}$.Since $\cos \frac{B+C}{2} = \sin \frac{A}{2}$,we have $2 \sin \frac{A}{2} \cos \frac{B-C}{2} = 4 \sin^2 \frac{A}{2}$.Dividing by $2 \sin \frac{A}{2}$ (assuming $A 
eq 0$),we get $\cos \frac{B-C}{2} = 2 \sin \frac{A}{2}$.Multiplying both sides by $2 \cos \frac{A}{2}$,we get $2 \cos \frac{A}{2} \cos \frac{B-C}{2} = 4 \sin \frac{A}{2} \cos \frac{A}{2}$.Using $2 \cos \frac{A}{2} = 2 \sin \frac{B+C}{2}$,we get $2 \sin \frac{B+C}{2} \cos \frac{B-C}{2} = 2 \sin A$.This simplifies to $\sin B + \sin C = 2 \sin A$.By the Sine Rule,$\frac{b}{2R} + \frac{c}{2R} = 2 \frac{a}{2R}$,which implies $b + c = 2a$.Since $b + c = AB + AC = 2BC$,and $BC$ is a fixed base,the sum of the distances of $A$ from two fixed points $B$ and $C$ is constant $(2BC > BC)$.Therefore,the locus of point $A$ is an ellipse with foci at $B$ and $C$.

Similar Questions

In a triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C = $

In $\triangle ABC$,if $a, b, c$ are $5, 12$ and $13$ respectively,then $b^2 \sin 2C + c^2 \sin 2B =$

The number of roots of the equation $(81)^{\sin ^{2} x} + (81)^{\cos ^{2} x} = 30$ in the interval $[0, \pi]$ is equal to

In $\Delta ABC$,if $a, b, c$ are in $A.P.$ (with usual notations),identify the incorrect statement -

If $\tan (\pi \cos \theta)=\cot (\pi \sin \theta)$,then a value of $\cos \left(\theta-\frac{\pi}{4}\right)$ among the following is

Vedclass Test Series

Exam Paper Generator

Online Exam Module