Question

Triangle $PQR$ is right angled at $Q$ and has side lengths $PQ=14$ and $QR=48.$ If $M$ is the mid point of $PR$ and $\angle MQP=\theta ,$ then $\cos \theta$ is equal to:

• A. $\frac{7}{50}$
• B. $\frac{7}{25}$
• C. $\frac{7}{24}$
• D. $\frac{24}{25}$

Answer 1

The correct option is B $\frac{7}{25}$

Given $PQR$ is a right triangle with $Q$ as right angle.

$M$ is the mid point of $PR$

$PQ=14,QR=48$
$\angle MQP=\theta$

To find $\cos \theta =?$

Draw $MN\parallel QR$

$MN$ meets $PQ$ at $N$

$\therefore$ By mid point theorem for triangles, we get

$N$ is the mid point of $PQ$

$\therefore PN=NQ=\frac{1}{2}\times PQ=\frac{1}{2}\times 14$ units $=7$ units and

$MN=\frac{1}{2}\times QR=\frac{1}{2}\times 48$ units $=24$ units.

Again $MN\parallel QR$

$\Rightarrow \angle MNQ=\angle NQR={90}^o.$ ..... (corresponding angles)

So $\Delta MNQ$ is a right angle triangle with $MQ$ as hypotenuse.

$\therefore MQ=\sqrt{{NQ}^2+{MN}^2}=\sqrt{7^2+{24}^2}=25$ units

$\therefore \cos \theta =\frac{MQ}{NQ}=\frac{7}{25}$