Question

The figure below shows two right angled triangles $\Delta OAB$ and $\Delta OQP$ with right angles at vertex A and P, respectively, having the common vertex O. The lengths of some of the sides are indicated in the figure: $OP = 2$ , $PQ = 1$ , $OA = 3$ , $AB = 1$ . $AB$ and $OP$ are parallel.

What is $\angle QOB$ ?

Accepted Answer

B. Using coordinates with $O$ at origin $(0,0)$: Since $AB \parallel OP$, let $OP$ lie along the y-axis or a horizontal line. $OQ = \sqrt{2^2 + 1^2} = \sqrt{5}$. $OB = \sqrt{3^2 + 1^2} = \sqrt{10}$. Calculating vectors or angles shows that $\Delta QOB$ is a right isosceles triangle with $\angle QOB = 45^\circ$.

Angle Calculation in Right Triangles

Explanation

Practise this under exam conditions