Trig - Oh no - metry

Geometry Level pending

If 3 + cot ⁡ ( 6 0 ∘ + A ∘ ) cot ⁡ A ∘ cot ⁡ ( 6 0 ∘ + A ∘ ) + cot ⁡ A ∘ = tan ⁡ ( x ∘ + A ∘ ) \dfrac{3+\cot (60^\circ+A^\circ) \cot A^\circ}{\cot(60^\circ+A^\circ) + \cot A^\circ} = \tan (x^\circ + A^\circ) , find the value of x x from the following list.

30 0 60 65 45

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1 solution

Chew-Seong Cheong
May 12, 2016

Let t = tan ⁡ A ∘ t = \tan A^\circ , then we have:

tan ⁡ ( x ∘ + A ∘ ) = 3 + cot ⁡ ( 6 0 ∘ + A ∘ ) cot ⁡ A ∘ cot ⁡ ( 6 0 ∘ + A ∘ ) + cot ⁡ A ∘ = 3 + 1 − 3 t 3 + t ⋅ 1 t 1 − 3 t 3 + t + 1 t = 3 3 t + 3 t 2 + 1 − 3 t t − 3 t 2 + 3 + t = 3 t 2 + 2 3 t + 1 3 + 2 t − 3 t 2 = ( 3 t + 1 ) 2 ( 1 + 3 t ) ( 3 − t ) = 1 + 3 t 3 − t = cot ⁡ ( 6 0 ∘ − A ∘ ) = tan ⁡ ( 3 0 ∘ + A ∘ ) \begin{aligned} \tan (x^\circ + A^\circ) & = \frac{3 + \cot (60^\circ + A^\circ)\cot A^\circ}{\cot (60^\circ + A^\circ) + \cot A^\circ} \\ & = \frac{3+\dfrac{1-\sqrt{3}t}{\sqrt{3}+t}\cdot \dfrac{1}{t}}{\dfrac{1-\sqrt{3}t}{\sqrt{3}+t} + \dfrac{1}{t}} \\ & = \frac{3\sqrt{3}t + 3t^2 +1- \sqrt{3}t}{t - \sqrt{3}t^2+\sqrt{3}+t} \\ & = \frac{3t^2+2\sqrt{3}t +1}{\sqrt{3}+2t - \sqrt{3}t^2} \\ & = \frac{(\sqrt{3}t+1)^2}{(1+\sqrt{3}t)(\sqrt{3}-t)} \\ & = \frac{1+\sqrt{3}t}{\sqrt{3}-t} \\ & = \cot (60^\circ - A^\circ) \\ & = \tan (30^\circ + A^\circ) \end{aligned}

⟹ x = 30 \implies x = \boxed{30}

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