(D) Given $b = 3, c = 4$ and $B = \frac{\pi}{3}$.We calculate $c \sin B = 4 \sin \frac{\pi}{3} = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}$.Since $\sqrt

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(D) Given $b = 3, c = 4$ and $B = \frac{\pi}{3}$.We calculate $c \sin B = 4 \sin \frac{\pi}{3} = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}$.Since $\sqrt{3} \approx 1.732$,we have $2\sqrt{3} \approx 3.464$.Comparing this with $b$,we find $c \sin B = 2\sqrt{3} > 3 = b$.Since $b Therefore,no triangle can be constructed.

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

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If $b = 3, c = 4$ and $B = \frac{\pi}{3}$,then the number of triangles that can be constructed is

A
Infinite
B
Two
C
One
D
Nil

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Class 11

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

In $\Delta ABC,$ if $(a + b + c)(a - b + c) = 3ac,$ then

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In a triangle $ABC$,$\frac{2\cos A}{a} + \frac{\cos B}{b} + \frac{2\cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$,then the value of angle $A$ is .....$^o$

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If the angles of a triangle $ABC$ are in the ratio $1: 2: 3$,then the corresponding sides are in the ratio

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We are given $b, c$ and $\sin B$ such that $B$ is acute and $b < c \sin B$. Then

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In triangle $ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then the triangle is:

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If $b = 3, c = 4$ and $B = \frac{\pi}{3}$,then the number of triangles that can be constructed is

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