(B) Given: $b = 20, c = 21, \sin A = 3/5$.Since $\sin^2 A + \cos^2 A = 1$,we have $\cos A = \pm \sqrt{1 - (3/5)^2} = \pm 4/5$.Assuming $A$ is an acute

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(B) Given: $b = 20, c = 21, \sin A = 3/5$.Since $\sin^2 A + \cos^2 A = 1$,we have $\cos A = \pm \sqrt{1 - (3/5)^2} = \pm 4/5$.Assuming $A$ is an acute angle,$\cos A = 4/5$.Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$.$a^2 = 20^2 + 21^2 - 2(20)(21)(4/5)$.$a^2 = 400 + 441 - 840(4/5) = 841 - 672 = 169$.Therefore,$a = \sqrt{169} = 13$.

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

Easy

In a $\Delta ABC,$ if $b = 20, c = 21$ and $\sin A = 3/5$,then $a = $

A
$12$
B
$13$
C
$14$
D
$15$

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Relation between sides and angles, Solutions of triangles

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