A geometry problem by A Former Brilliant Member

Geometry Level 3

If cos ⁡ θ − sin ⁡ θ cos ⁡ θ + sin ⁡ θ = 1 − 3 1 + 3 \dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\dfrac{1-\sqrt{3}}{1+\sqrt{3}} , then find acute angle θ \theta .

0 ∘ 0^\circ 3 0 ∘ 30^\circ 6 0 ∘ 60^\circ 4 5 ∘ 45^\circ

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Sabhrant Sachan
Jun 20, 2016

We have cos ⁡ θ − sin ⁡ θ cos ⁡ θ + sin ⁡ θ = 1 − 3 1 + 3 , Apply C and D cos ⁡ θ − sin ⁡ θ + cos ⁡ θ + sin ⁡ θ cos ⁡ θ − sin ⁡ θ − cos ⁡ θ − sin ⁡ θ = 1 − 3 + 1 + 3 1 − 3 − 1 − 3 ⟹ − 2 cos ⁡ θ 2 sin ⁡ θ = − 2 2 3 tan ⁡ θ = 3 ⟹ θ = 6 0 ∘ \text{We have } \quad \dfrac{\cos{\theta}-\sin{\theta}}{\cos{\theta}+\sin{\theta}} = \dfrac{1-\sqrt{3}}{1+\sqrt{3}} \quad , \quad \text{Apply C and D } \\ \quad \dfrac{\cos{\theta}-\sin{\theta}+\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}-\cos{\theta}-\sin{\theta}} = \dfrac{1-\sqrt{3}+1+\sqrt{3}}{1-\sqrt{3}-1-\sqrt{3}} \implies -\dfrac{2\cos{\theta}}{2\sin{\theta}} = -\dfrac{2}{2\sqrt3} \\ \quad \tan{\theta} = \sqrt3 \\\quad \implies \boxed{\theta = 60^\circ}

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Nice solution(+1) ! Did the same way.

Rishabh Tiwari - 4 years, 12 months ago

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