# Quadrilateral ABCD ​ is inscribed in circle O. What is ​ m∠A ​ ?

May 2, 2017

$\angle C = {138}^{o}$

$\angle A = {180}^{o} - {138}^{o} = {42}^{o}$

#### Explanation:

ABCD can be considered as two triangles back to back. The sum of the internal angles of a triangle is ${180}^{o}$. So the sum of the internal angles of ABCD $= 2 \times {180}^{o} = {360}^{0}$

Also $\angle A + \angle C = {180}^{o} = \angle B + \angle D$

$\implies \textcolor{g r e e n}{2 x + \left(3 x - 5\right)} = {180}^{0} = \textcolor{b r o w n}{\left(x + 5\right) + \angle C}$

$\textcolor{g r e e n}{5 x - 5} = 180$

$5 x = 185$

Divide both sides by 5

$x = \frac{185}{5} = 37$
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$\textcolor{b l u e}{\text{Consider } \angle A + \angle C}$

$\left(x + 5\right) + \angle C = {180}^{o}$

But $x = 37$ giving:

$37 + 5 + \angle C = {180}^{o}$

Subtract 42 from both sides (37+5=42)

$\angle C = {138}^{o}$
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Check

$\left(x + 5\right) + 2 x + \left(3 x - 5\right) + \angle C = 360$

(37+5)+2(37)+[3(37)-5]+138=?

$42 + 74 + 106 + 138 = 360 \textcolor{red}{\leftarrow \text{ works!}}$