NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.1

Question 5

Step 1: Look at the lengths of the line segments.

We are given:

• \(PA = 6\,\text{cm}\)
• \(PB = 3\,\text{cm}\)
• \(PC = 2.5\,\text{cm}\)
• \(PD = 5\,\text{cm}\)

So, \(\dfrac{PA}{PB} = \dfrac{6}{3} = 2\).

And, \(\dfrac{PD}{PC} = \dfrac{5}{2.5} = 2\).

Step 2: Since the ratios are equal (both are 2), triangles \(\triangle APB\) and \(\triangle DPC\) are similar.

This means their angles are equal in pairs: \(\angle A = \angle D\), \(\angle B = \angle C\), etc.

Step 3: From the question, we know \(\angle CDP = 30^\circ\).

Because of similarity, \(\angle A = \angle D = 30^\circ\).

Step 4: In triangle \(APB\):

• \(\angle APB = 50^\circ\) (given)
• \(\angle PAB = 30^\circ\) (from similarity)
• \(\angle PBA = ?\)

Step 5: The sum of angles in any triangle is \(180^\circ\).

So,

\(\angle PBA = 180^\circ - (\angle APB + \angle PAB)\)

\(= 180^\circ - (50^\circ + 30^\circ)\)

\(= 180^\circ - 80^\circ\)

\(= 100^\circ\).

Final Answer: \(\angle PBA = 100^\circ\).