## RS Aggarwal Class 7 Solutions Chapter 13 Lines and Angles Ex 13

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 13 Lines and Angles Ex 13.

**Question 1.**

**Solution:**

(i) The given angle = 35°

Let x be its complementary, then

x + 35° = 90°

⇒ x = 90° – 35° = 55°

Complement angle = 55°

(ii) The given angle = 47°

Let x be its complement, then

x + 47° = 90 ⇒ x = 90° – 47° = 43°

Complement angle = 43°

(iii) The given angles = 60°

Let x be its complement angle

x + 60° = 90° ⇒ x = 90° – 60° = 30°

Complement angle = 30°

(iv) The given angle = 73°

Let x be its complement angle

x + 73° = 90°

⇒ x = 90° – 73° = 17°

Complement angle = 17°

**Question 2.**

**Solution:**

(i) Given angle = 80°

Let x be its supplement angle, then

x + 80° = 180°

⇒ x = 180° – 80° = 100°

Supplement angle = 100°

(ii) Given angle = 54°

Let x be its supplement angle, then

x + 54° = 180°

⇒ x = 180° – 54° = 126°

Supplement angle = 126°

(iii) Given angle = 105°

let x be its supplement angle, then

x + 105° = 180°

⇒ x = 180° – 105° = 75°

Supplement angle = 75°

(iv) Given angle = 123°

Let x be its supplement angle, then

x + 123° = 180°

⇒ x = 180° – 123° = 57°

⇒ Supplement angle = 57°

**Question 3.**

**Solution:**

Let smaller angle =x

Then larger angle = x + 36°

But x + x + 36° = 180° (Angles are supplementary)

2x = 180° – 36°= 144°

x = 72°

Smaller angle = 72°

and larger angle = 72° + 36° = 108°

**Question 4.**

**Solution:**

Let angle be = x

Then other supplement angle = 180°- x

x = 180° – x

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = 90°

Hence angles are 90°, and 90°

**Question 5.**

**Solution:**

Sum of two supplementary angles is 180°

If one is acute, then second will be obtuse or both angles will be equal

Hence both angles can not be acute or obtuse

Both can be right angles only

**Question 6.**

**Solution:**

In the given figure,

AOB is a straight line and the ray OC stands on it.

∠AOC = 64° and ∠BOC = x°

∠AOC + ∠BOC = 180° (Linear pair)

⇒ 64° + x = 180°

⇒ x = 180° – 64° = 116°

Hence x = 116°

**Question 7.**

**Solution:**

AOB is a straight line and ray OC stands on it ∠AOC = (2x – 10)°, ∠BOC = (3x + 20)°

∠AOC + ∠BOC = 180° (Linear pair)

⇒ 2x – 10° + 3x + 20° = 180°

⇒ 5x + 10° = 180°

⇒ 5x = 170°

⇒ x = 34°

∠AOC = (2x – 10)° = 2 x 34° – 10 = 68° – 10° = 58°

∠BOC = (3x + 20)° = 3 x 34° + 20° – 102° + 20° = 122°

**Question 8.**

**Solution:**

AOB is a straight line and rays OC and OD stands on it ∠AOC = 65°, ∠BOD = 70° and ∠COD = x

But ∠AOC + ∠COD + ∠BOD = 180° (Angles on one side of the straight line)

⇒ 65° + x + 70° = 180°

⇒ 135° + x = 180°

⇒ x = 180° – 135°

⇒ x = 45°

Hence x = 45°

**Question 9.**

**Solution:**

Two straight lines AB and CD intersect each other at O.

∠AOC = 42°

AB and CD intersect each other at O.

∠AOC = ∠BOD (Vertically opposite angles)

and ∠AOD = ∠BOC

But ∠AOC = 42°

∠BOD = 42°

AOB is a straight line and OC stands on it

∠AOC + ∠BOC = 180°

⇒ 42° = ∠BOC = 180°

⇒ ∠BOC = 180° – 42° = 138°

But ∠AOD = ∠BOC (vertically opposite angles)

∠AOD = 138°

Hence ∠AOD = 138°, ∠BOD = 42° and ∠COB =138°

**Question 10.**

**Solution:**

Two straight lines PQ and RS intersect at O.

∠POS = 114°

Straight lines,

PQ and RS intersect each other at O

∠POS = ∠QOR (Vertically opposite angles)

But ∠POS = 114°

∠QOR = 114° or ∠ROQ = 114°

But ∠POS + ∠POR = 180° (Linear pair)

⇒ 114° + ∠POR = 180°

⇒ ∠POR = 180° – 114° = 66°

But ∠QOS = ∠POR (vertically opposite angles)

∠QOS = 66°

Hence ∠POR = 66°, ∠ROQ =114° and ∠QOS = 66°

**Question 11.**

**Solution:**

In the given figure, rays OA, OB, OC and OD meet at O and ZAOB – 56°,

∠BOC = 100°, ∠COD = x and ∠DOA = 74°

But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (Angles at a point)

56° + 100° + x° + 74° = 360°

⇒ 230° + x° = 360°

⇒ x° = 360° – 230° = 130°

⇒ x = 130°

Hope given RS Aggarwal Solutions Class 7 Chapter 13 Lines and Angles Ex 13 are helpful to complete your math homework.

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