Question

In the given figure, $\angle$BCA = $\angle$RPQ = ${15}^{\circ }$. Find $\angle$BAC is

• A. 15${}^{\circ }$
• B. 30${}^{\circ }$
• C. 45${}^{\circ }$
• D. 60${}^{\circ }$

Answer 1

The correct option is C 45${}^{\circ }$

Along the horizontal line through point A, we have,
90${}^{\circ }$ + $\angle$BAC + 45${}^{\circ }$ = 180${}^{\circ }$
or, $\angle$BAC = 180${}^{\circ }$ - 45${}^{\circ }$ - 90${}^{\circ }$
$\angle$BAC = 45${}^{\circ }$