In figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to ______.
Options
50°
30°
60°
100°
Solution
In figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to 100°.
Explanation:
From ∆APB and ∆CPD,
∠APB = ∠CPD = 50° ......(Since they are vertically opposite angles)
`(AP)/(PD) = 6/5` ......(i)
Also, `(BP)/(CP) = 3/2.5`
Or `(BP)/(CP) = 6/5` ......(ii)
From equations (i) and (ii),
We get,
`(AP)/(PD) = (BP)/(CP)`
So, ∆APB ∼ ∆DPC ......[Using SAS similarity criterion]
∴ ∠A = ∠D = 30° ......[Since, corresponding angles of similar triangles]
Since, Sum of angles of a triangle = 180°,
In ∆APB,
∠A + ∠B + ∠APB = 180°
So, 30° + ∠B + 50° = 180°
Then, ∠B = 180° – (50° + 30°)
∠B = 180 – 80° = 100°
Therefore, ∠PBA = 100°
