(A) We know the formula for the sine of half-angles in a triangle: $\sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}}$.Squaring both sides,we get $\

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(A) We know the formula for the sine of half-angles in a triangle: $\sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}}$.Squaring both sides,we get $\sin^2 \frac{A}{2} = \frac{(s - b)(s - c)}{bc}$.Rearranging the terms,we have $(s - b)(s - c) = bc \sin^2 \frac{A}{2}$.Comparing this with the given equation $(s - b)(s - c) = x \sin^2 \frac{A}{2}$,we find that $x = bc$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

Medium

In a $\Delta ABC$,if $2s = a + b + c$ and $(s - b)(s - c) = x \sin^2 \frac{A}{2}$,then $x =$

A
$bc$
B
$ca$
C
$ab$
D
$abc$

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

In any $\Delta ABC$,if $a \cos B = b \cos A$,then the triangle is

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Difficult

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In a $\triangle ABC$,$b:c = \sqrt{3}:\sqrt{2}$ and the angles $A, B, C$ are in $AP$,then $\angle A = $ (in $^{\circ}$)

EasyAP EAMCET 2020

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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:

MediumMHT CET 2025

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Let $ABC$ be an acute-angled triangle and let $D$ be the mid-point of $BC$. If $AB = AD$,then $\tan(B) / \tan(C)$ equals

KVPY 2013

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In a $\Delta ABC$,if $2s = a + b + c$ and $(s - b)(s - c) = x \sin^2 \frac{A}{2}$,then $x =$

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