Question : In $\Delta$PQR, $\angle$P : $\angle$Q : $\angle$R = 2 : 2 : 5. A line parallel to QR is drawn which touches PQ and PR at A and B respectively. What is the value of $\angle$PBA – $\angle$PAB?
Option 1: 60º
Option 2: 30º
Option 3: 20º
Option 4: 50º
Correct Answer: 60º
Solution :
In a triangle, the angles are in the ratio 2 : 2 : 5.
Let the angles are $2x$, $2x$, and $5x$ respectively, where $x$ is a constant.
Since the sum of the angles in a triangle is \(180^\circ\).
$⇒2x + 2x + 5x = 180^\circ$
$⇒x = 20^\circ$
$\angle P = \angle Q = 40^\circ$ and $\angle R = 100^\circ$.
Since a line AB parallel to QR is drawn which touches PQ and PR at A and B, by the corresponding angles theorem.
$⇒\angle PAB = \angle Q = 40^\circ$ and $\angle PBA = \angle R = 100^\circ$.
$⇒\angle PBA - \angle PAB = 100^\circ - 40^\circ = 60^\circ$
Hence, the correct answer is 60º.
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