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Question
In ∆ABC, if sin2 A + sin2 B = sin2 C. show that the triangle is right-angled.
Solution
In ∆ ABC,
Given, \[\sin^2 A + \sin^2 B = \sin^2 C . . . . . . \left( 1 \right)\]
Suppose
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]
⇒ \[\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}\]
On putting these values in equation (1), we get:
\[\frac{a^2}{k^2} + \frac{b^2}{k^2} = \frac{c^2}{k^2} \Rightarrow a^2 + b^2 = c^2\]
Thus, ∆ ABC is right-angled.
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