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In Fig.6.3, 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

Solution:

Given, two line segments AC and BD intersect each other at the point P.

The length of the line segments

PA = 6 cm

PB = 3 cm

PC = 2.5 cm

PD = 5 cm

Given, ∠APB = 50° and ∠CDP = 30°

We have to find the value of ∠PBA.

SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.

Checking for the ratio of the corresponding sides,

PA/PD = 6/5

PB/PC = 3/2.5 = 30/25 = 6/5

Thus, PA/PD = PB/PC

As the ratio of the corresponding sides are in the same proportion, the corresponding angles

∠APB = ∠DPC (Vertically opposite angles) are also equal.

Therefore, the triangles APB and DPC are similar.

So, ∠A = ∠D = 30°

In the triangle APB,

We know that the sum of all the three interior angles of a triangle is always equal to 180°.

∠A + ∠B + ∠APB = 180°

30° + ∠B + 50° = 180°

80° + ∠B = 180°

∠B = 180° - 80°

∠B = 100°

Therefore, ∠PBA = 100°

✦ Try This: In figure, BD and CE intersect each other at the point P. Is ΔPBC ~ ΔPDE ? Why?

In figure, BD and CE intersect each other at the point P. Is ΔPBC ~ ΔPDE ? Why?

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6

NCERT Exemplar Class 10 Maths Exercise 6.1 Problem 5

In Fig.6.3, 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

Summary:

In Fig.6.3, 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°

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