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I have the following question with me:

"Consider a triangle $ABC$ and let $M$ be the midpoint of the side $BC$. Suppose $\angle MAC$ = $\angle ABC$ and $\angle BAM = 105^{\circ}$ . Find the measure of $∠ABC$."

Upon taking $\angle ABC$ as $x$ and upon simplifying, I get the following equation, if my calculations are correct,

$$(4+\sqrt{3})\cos2x = 2 + \sin2x$$

How do I solve this equation?

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Alternative solution.

Let $BC = 2a$. By the power of the point of $C$ with respect to circle $(ABM)$, which is tangent to a line $AC$, we have $$CA^2= CM\cdot CB =2a^2$$

By rule of sine for $ABC$ we have $${2a\over \sin (\alpha +105^{\circ})} ={a\sqrt{2}\over \sin \alpha}$$

so $$\cot \alpha = {\sqrt{2}+\sin 15^{\circ}\over \cos 15^{\circ}} =\sqrt{3}\implies \alpha = 30^{\circ}$$

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  • $\begingroup$ How is $AC$ tangent to circumcircle of $ABM$? $\endgroup$ – saisanjeev Oct 2 '18 at 7:29
  • $\begingroup$ Because $\angle MAC = \angle ABC$ and tangent chord property. $\endgroup$ – Aqua Oct 2 '18 at 7:30
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You can either use t-method, letting $t=\tan x$ so that $\sin 2x=\frac {2t}{1+t^2}$ and $cos 2x=\frac {1-t^2}{1+t^2}$. Plugging this in and solving the quadratic gives you values of $t$ which you can these use to solve for $x$. Alternatively, you can use auxiliary angles where $A \cos 2x-B \sin 2x=\sqrt{(A^2+B^2)\cos (2x+\arctan \frac {B}{A})}$

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