Advanced Maths collection
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Let sides a/r, a, ar

(ar)² = (a)² + (a/r)²

r⁴ - r² - 1 = 0

r² = t

t² - t - 1 = 0

t = r² = (1 ± √5)/2

r = ±√(1 ± √5)/2

B
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@xDAEMONx upar negative remainder aa raha iska koi upay ha to batana
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35^23^23 % 16

3^23^23 % 16

(81)^4k * (3)³ % 16

1 * 27 % 16

11
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Both terms are in square root so values are real and positive.

So for maximum right hand term should be minimum i.e. zero

2a² - 1 - cos²x = 0

2a² = 1 + cos²x

Put

√sin²x + cos²x + 1

√2

B
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