Q.In \(\triangle\)ABC, if AB = \((2a-1)\) cm, AC = \(2\sqrt{2}a\) cm, and BC = \((2a+1)\) cm, then write the value of \(\angle\)BAC.

Here, \(AB^2 + AC^2 = (2a - 1)^2 + (2\sqrt{2}a)^2\)
\(= 4a^2 - 4a + 1 + 8a\)
\(= 4a^2 + 4a + 1\)
\(= (2a + 1)^2\)
\(= BC^2\)
\(\therefore \triangle ABC\) is a right-angled triangle, where BC is the hypotenuse.
\(\therefore \angle BAC = 90^\circ\)
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