# In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2 (∠QOS - ∠POS).

**Solution:**

Given: OR is perpendicular to PQ. ∠ROQ = ∠ROP = 90°.

To prove: ∠ROS = 1/2(∠QOS - ∠POS).

When a ray intersects a line, the sum of adjacent angles so formed is 180°.

Let ∠ROS = a, ∠POS = b and ∠SOQ = c.

To prove that: a = 1/2(c - b).

Since ∠ROQ = ∠ROP = 90°,

We can say, ∠POS + ∠SOR = ∠POR

b + a = 90°.….(1)

Line PQ is intersected by ray OS.

Hence ∠POS + ∠SOQ = b + c = 180°

b + c = 180°….. (2)

From equation (1), we get a + b = 90°

Multiplying by 2 on both sides we get,

2(a + b) = 2 × 90°

2(a + b) = 180°…..(3)

Comparing equations (3) and (2),

2(a + b) = b + c

2a + 2b = b + c

2a = b + c - 2b

2a = c - b

a = 1/2 (c - b)

∴ ∠ROS = 1/2 (∠QOS - ∠POS)

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 6

**Video Solution:**

## In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2 (∠QOS - ∠POS).

NCERT Solutions Class 9 Maths Chapter 6 Exercise 6.1 Question 5

**Summary:**

In Fig. 6.17, if POQ is a line OR is perpendicular to PQ, and OS is another ray lying between rays OP and OR, then ∠ROS = 1/2 (∠QOS − ∠POS).

**☛ Related Questions:**

- In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.
- In Fig. 6.30, if AB||CD, EF ⊥ CD and ∠GED =126°, find ∠AGE, ∠GEF and ∠FGE.
- In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.[Hint: Draw a line parallel to ST through point R.]
- In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

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