Question

In a $\Delta PQR,PS\perp QR.$ If $QS:SP:SR=9:12:16,$ then the value of $\angle PQR$ is always equal to

• A. ${90}^0–\angle PRQ$
• B. ${90}^0+\angle PRQ$
• C. ${180}^0–\angle PRQ$
• D. ${180}^0–\angle QPR$

Answer 1

The correct option is A ${90}^0–\angle PRQ$

Given: $PS\perp QR,QS:SP:SR=9:12:16$
Let QS = 9x, SP = 12x and SR = 16x
Now, in $\Delta PQS$, by Pythagoras Theorem,



$P{Q}^2=P{S}^2+Q{S}^2$
$P{Q}^2=(12x{)}^2+(9x{)}^2$
$P{Q}^2=144x^2+81x^2$
$P{Q}^2=225x^2$
$PQ=15x$
Similarly, in ΔPSR, by Pythagoras Theorem,
$P{R}^2=P{S}^2+S{R}^2$
$P{R}^2=(12x{)}^2+(16x{)}^2$
$P{R}^2=144x^2+256x^2$
$P{R}^2=400x^2$
$PR=20x$

Now, $P{Q}^2+P{R}^2=(15x{)}^2+(20)x^2$
$=225x^2+400x^2$
$=625x^2$
$P{Q}^2+P{R}^2=(25x{)}^2$ .........(i)
and, $Q{R}^2=(QS+SR{)}^2$
$=(9x+16x{)}^2$
$Q{R}^2=(25x{)}^2$ ............(ii)
From (i) and (ii),
$P{Q}^2+P{R}^2=Q{R}^2$
Thus, by converse of Pythagoras Theorem,
$\Delta PQR$ is a right angle triangle, right-angled at P.
$\therefore \angle QPR={90}^0$
Now, in $\Delta PQR$, by Angle Sum Property,
$\angle PQR+\angle QPR+\angle PRQ={180}^0$
$\Rightarrow \angle PQR+{90}^0+\angle PRQ={180}^0$
$\Rightarrow \angle PQR={90}^0–\angle PRQ$
Hence, the correct answer is option (1).