(A) Using the Law of Sines,we have $\frac{\sin A}{a} = \frac{\sin B}{b}$,which implies $a \sin B = b \sin A$.From option $(A)$,$b \sin A = a$,so $a \s

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$b \sin A = a, A < \frac{\pi}{2}$

(A) Using the Law of Sines,we have $\frac{\sin A}{a} = \frac{\sin B}{b}$,which implies $a \sin B = b \sin A$.From option $(A)$,$b \sin A = a$,so $a \sin B = a$,which gives $\sin B = 1$,meaning $B = \frac{\pi}{2}$.Since $A From option $(B)$,$b \sin A > a$,which implies $a \sin B > a$,or $\sin B > 1$,which is impossible.Similarly,for option $(C)$,$b \sin A > a$ leads to $\sin B > 1$,which is also impossible.

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

Medium

Which of the following conditions allows for the existence of a triangle $ABC$?

IIT 1986 All IIT

A
$b \sin A = a, A < \frac{\pi}{2}$
B
$b \sin A > a, A > \frac{\pi}{2}$
C
$b \sin A > a, A < \frac{\pi}{2}$
D
None of these

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Relation between sides and angles, Solutions of triangles

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Which of the following conditions allows for the existence of a triangle $ABC$?

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