Question

The figure shows a square ABCD and an equilateral triangle ABP.
Calculate: $\angle AOB$

• A. ${75}^{\circ }$
• B. ${45}^{\circ }$
• C. ${95}^{\circ }$
• D. ${65}^{\circ }$

Answer 1

Answer: (A)

${75}^{\circ }$

In a given figure,

$△APB$ is an equilateral triangle.

So, $AP=PB=AB$ ----- (Properties of equilateral triangle)

$\angle PAB=\angle ABP=\angle BPA={60}^{\circ }$ --- (All angles of equilateral triangle are equal and ${60}^{\circ }$)

$\angle OAB={60}^{\circ }$ -- (1)

We know that all angles all square are 90${}^{\circ }$

So, $\angle CDA={90}^{\circ }$

DB is a diagonal of square $\square ABCD$ with bisect $\angle CDA$ equally.

Then, $\angle CDB={45}^{\circ }$.

We know that $AB=CD$ and $AB\parallel CD$ then,

$\angle CDB=\angle ABD={45}^{\circ }$ -- (Alternate angles)

$\angle DBA={45}^{\circ }$ i.e $\angle ABO={45}^{\circ }$ --(2)

In $△AOB$,

$\angle AOB+\angle OAB+\angle ABO={180}^{\circ }$

From (1) and (2),

$\angle AOB+{60}^{\circ }+{45}^{\circ }={180}^{\circ }$

$\therefore \angle AOB={75}^{\circ }$