TS EAMCET · Maths · Properties of Triangles
If \(\triangle A B C\) is a non isosceles triangle and \(\angle C=90^{\circ}\), then \(\frac{a^2+b^2}{a^2-b^2} \sin (A-B)=\)
- A \(1\)
- B \(2\)
- C \(0\)
- D \(-1\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
Given that, in \(\triangle A B C\), \(\begin{aligned} \angle C & =90^{\circ} \\ \angle A+\angle B+\angle C & =180^{\circ} \\ \therefore \quad \angle A+\angle B & =90^{\circ}\end{aligned}\) Then, \(\frac{a^2+b^2}{a^2-b^2} \sin (A-B)\) Using sine rule,we get…
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