In a right angled triangle, the square of the hypotenuse is twice the product of the other two sides. Then the angle of the triangle is:

15^{\circ}

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30^{\circ}

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45^{\circ}

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60^{\circ}

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Solution

The correct option is C $45^{\circ}$
$b^{2} s e c^{2} θ = b^{2} t a n θ$
$\frac{1}{c o s^{2} θ} = \frac{s i n θ}{c o s θ}$
$s i n θ = c o s θ$
$θ = 45^{\circ}$

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In a right angled triangle, the square of the hypotenuse is twice the product of the other two sides. Then the angle of the triangle is:

The correct option is C 45∘b2sec2θ=b2tanθ1cos2θ=sinθcosθsinθ=cosθθ=45∘

The correct option is C $45^{\circ}$
$b^{2} s e c^{2} θ = b^{2} t a n θ$
$\frac{1}{c o s^{2} θ} = \frac{s i n θ}{c o s θ}$
$s i n θ = c o s θ$
$θ = 45^{\circ}$