(B) Using the Law of Sines,we have $\frac{a}{\sin A} = \frac{b}{\sin B}$.Substituting the given values: $\frac{5}{3/4} = \frac{7}{\sin B}$.$\frac{20}{

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(B) Using the Law of Sines,we have $\frac{a}{\sin A} = \frac{b}{\sin B}$.Substituting the given values: $\frac{5}{3/4} = \frac{7}{\sin B}$.$\frac{20}{3} = \frac{7}{\sin B} \implies \sin B = \frac{21}{20}$.Since the value of $\sin B$ must be $\le 1$ and $\frac{21}{20} > 1$,no such angle $B$ exists.Therefore,no triangle $ABC$ can be formed with the given parameters.

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

Medium

In a triangle $ABC$,with usual notations,if $a=5$,$b=7$,and $\sin A=\frac{3}{4}$,then the total number of triangles possible is:

MHT CET 2025 All MHT CET

A
$1$
B
$0$
C
$2$
D
$5$

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

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In a triangle $ABC$,with usual notations,if $a=5$,$b=7$,and $\sin A=\frac{3}{4}$,then the total number of triangles possible is:

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