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Geometry Level 3

3 + cot ⁡ ( 7 6 ∘ ) cot ⁡ ( 1 6 ∘ ) cot ⁡ ( 7 6 ∘ ) + cot ⁡ ( 1 6 ∘ ) = cot ⁡ ( x ∘ ) \large \frac{3+\cot(76^\circ)\cot(16^\circ)}{\cot(76^\circ)+\cot(16^\circ)}=\cot(x^\circ)

Given the above, find x x .

88 46 92 44

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1 solution

3 + cot ⁡ 7 6 ∘ cot ⁡ 1 6 ∘ cot ⁡ 7 6 ∘ + cot ⁡ 1 6 ∘ = 3 + cot ⁡ ( 60 + 16 ) ∘ cot ⁡ 1 6 ∘ cot ⁡ ( 60 + 16 ) ∘ + cot ⁡ 1 6 ∘ Let t = tan ⁡ 1 6 ∘ = 3 + 1 − 3 t 3 + t × 1 t 1 − 3 t 3 + t + 1 t = 3 t 2 + 2 3 t + 1 3 + 2 t − 3 t 2 = ( 3 t + 1 ) 2 ( 3 − t ) ( 3 t + 1 ) = 3 t + 1 3 − t = 1 + tan ⁡ 6 0 ∘ tan ⁡ 1 6 ∘ tan ⁡ 6 0 ∘ − tan ⁡ 1 6 ∘ = cot ⁡ ( 60 − 16 ) ∘ = cot ⁡ 4 4 ∘ \begin{aligned} \frac {3+\cot 76^\circ \cot 16^\circ}{\cot 76^\circ + \cot 16^\circ} & = \frac {3+\cot (60+16)^\circ \cot 16^\circ}{\cot (60+16)^\circ + \cot 16^\circ} & \small \color{#3D99F6} \text{Let }t = \tan 16^\circ \\ & = \frac {3+\frac {1-\sqrt 3t}{\sqrt 3+t}\times \frac 1t}{\frac {1-\sqrt 3t}{\sqrt 3+t}+\frac 1t} \\ & = \frac {3t^2+2\sqrt 3t+1}{\sqrt 3 +2t-\sqrt 3 t^2} \\ & = \frac {(\sqrt 3t+1)^2}{(\sqrt 3-t)(\sqrt 3t+1)} \\ & = \frac {\sqrt 3t+1}{\sqrt 3-t} \\ & = \frac {1+\tan 60^\circ\tan 16^\circ}{\tan 60^\circ - \tan 16^\circ} \\ & = \cot (60-16)^\circ \\ & = \cot 44^\circ \end{aligned}

Therefore, x = 44 x= \boxed{44} .

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