Find x

Geometry Level 2

The above figure is quadrilateral $ABCD$ . Given that $\angle BAD + \angle ACD=145^\circ$ , find the measure of $\angle BAC$ in degrees.

1 solution

A Former Brilliant Member
Jul 10, 2017

Let $\angle BAC=x$ . From, $\angle BAD+\angle ACD=145$ $\color{#D61F06}\implies$ $\angle BAD = 145 - \angle ACD$

Consider $\triangle ACD:$

$\angle CAD+\angle ACD=180-50=130$ $\color{#D61F06}\implies$ $\angle CAD = 130-\angle ACD$

From the figure,

$x=\angle BAD-\angle ACD$

$x=145-\angle ACD- (130-\angle ACD)$

$x=145-\angle ACD - 130 + \angle ACD$

$x=145-130=$ $\boxed{15^\circ}$