Question : In $\triangle$ABC, AB = $a - b$, AC = $\sqrt{a^2+b^2}$, and BC = $\sqrt{2ab}$, then find angle B.
Option 1: 60°
Option 2: 30°
Option 3: 90°
Option 4: 45°
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Correct Answer: 90°
Solution :
Given:
In $\triangle$ABC, AB = $a - b$, AC = $\sqrt{a^2+b^2}$ and BC = $\sqrt{2ab}$
Now, AB
2
+ BC
2
= $(a-b)^2+(\sqrt{2ab})^2$ = $a^2+b^2$ = $(\sqrt{a^2+b^2})^2$ = AC
2
So, ABC is a right-angled triangle.
$\therefore \angle $B = 90°
Hence, the correct answer is 90°.
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