Selina Solutions Concise Mathematics Class 6 Chapter 25 Properties of Angles and Lines Exercise 25(A) are designed by experts after conducting research on each concept. The sum of two adjacent angles in a straight line is 180^{0}, which is the main concept discussed under this exercise. Students can improve problem solving and time management skills by using these solutions. This also helps them to appear for the examination fearlessly. To get their doubts cleared about the concepts, students can refer to Selina Solutions Concise Mathematics Class 6 Chapter 25 Properties of Angles and Lines Exercise 25(A) PDF, from the links provided here for free download.

## Selina Solutions Concise Mathematics Class 6 Chapter 25: Properties of Angles and Lines Exercise 25(A) Download PDF

### Access other exercises of Selina Solutions Concise Mathematics Class 6 Chapter 25: Properties of Angles and Lines

### Access Selina Solutions Concise Mathematics Class 6 Chapter 25: Properties of Angles and Lines Exercise 25(A)

Exercise 25(A)

**1. Two straight lines AB and CD intersect each other at a point O and angle AOC = 50 ^{0}; find: **

**(i) angle BOD**

**(ii) âˆ AOD **

**(iii) âˆ BOC**

**Solution:**

(i) âˆ BOD

Given âˆ AOC = 50^{0}

We know that,

Vertically opposite angles are equal

So, âˆ BOD = âˆ AOC

Therefore, âˆ BOD = 50^{0}

(ii) âˆ AOD

âˆ AOD + âˆ BOD = 180^{0}

âˆ AOD + 50^{0} = 180^{0} {From (i)}

âˆ AOD = 180^{0} â€“ 50^{0}

We get,

âˆ AOD = 130^{0}

(iii) âˆ BOC

We know that,

Vertically opposite angles are equal

So, âˆ BOC = âˆ AOD

Therefore, âˆ BOC = 130^{0}

**2. The adjoining figure, shows two straight lines AB and CD intersecting at P. If âˆ BPC = 4x â€“ 5 ^{0} and âˆ APD = 3x + 15^{0}; find:**

**(i) the value of x**

**(ii) âˆ APD**

**(iii) âˆ BPD**

**(iv) âˆ BPC**

**Solution:**

(i) The value of x is calculated as,

3x + 15^{0} = 4x â€“ 5^{0}

3x â€“ 4x = – 5^{0} – 15^{0}

– x =- 20^{0}

x = 20^{0}

(ii) the value of âˆ APD is calculated as,

âˆ APD = 3x + 15^{0}

= 3 Ã— 20^{0} + 15^{0}

We get,

= 60^{0} + 15^{0}

= 75^{0}

(iii) The value of âˆ BPD is calculated as,

âˆ BPD = 180 â€“ âˆ BPC

= 180^{0} â€“ (4x – 5^{0})

= 180^{0} â€“ (4 Ã— 20^{0} â€“ 5^{0})

= 180^{0} â€“ 80^{0} + 5^{0}

We get,

= 105^{0}

(iv) The value of âˆ BPC is calculated as,

âˆ BPC = (4x â€“ 5^{0})

= (4 Ã— 20^{0} â€“ 5^{0})

We get,

= 80^{0} – 5^{0}

= 75^{0}

**3. The given diagram, shows two adjacent angles AOB and AOC, whose exterior sides are along the same straight line. Find the value of x.**

**Solution:**

Here, the exterior arms of the adjacent angles are in a straight line,

Hence, the adjacent angles are supplementary

âˆ AOB + âˆ AOC =180^{0}

68^{0} + (3x – 20^{0}) = 180^{0}

68^{0} + 3x â€“ 20^{0} = 180^{0}

3x = 180^{0} + 20^{0} – 68^{0}

3x = 200^{0} – 68^{0}

We get,

3x = 132^{0}

x = 132^{0} / 3

x = 44^{0}

**4. Each figure given below shows a pair of adjacent angles AOB and BOC. Find whether or not the exterior arms OA and OC are in the same straight line.**

**(i)**

** **

**(ii)**

** **

**(iii) **

**Solution:**

(i) We know that,

The sum of adjacent angles AOB and COB = 180^{0}

Hence,

âˆ AOB + âˆ COB = 180^{0}

(90^{0} â€“ x) + (90^{0} + x) = 180^{0}

90^{0} â€“ x + 90^{0} + x = 180^{0}

We get,

180^{0} = 180^{0}

The exterior arms OA and OC are in the same straight line

(ii) âˆ AOB + âˆ BOC = 97^{0} + 83^{0}

= 180^{0}

The sum of adjacent angles AOB and BOC is 180^{0}

Hence, the exterior arms OA and OC are in the same straight line

(iii) âˆ COB + âˆ AOB = 88^{0} + 112^{0}

We get,

= 200^{0} [which is not equal to 180^{0}]

Hence, the exterior arms OA and OC are not in the same straight line

**5. A line segment AP stands at point P of a straight line BC such that âˆ APB = 5x â€“ 40 ^{0} and âˆ APC = x + 10^{0}; find the value of x and angle APB.**

**Solution:**

Given

A line segment AP stands at P and

âˆ APB = 5x â€“ 40^{0}

âˆ APC = x + 10^{0}

(i) BPC is a straight line

âˆ APB + âˆ APC = 180^{0}

5x â€“ 40^{0} + x + 10^{0}_{ }= 180^{0}

6x â€“ 30^{0} = 180^{0}

6x = 180^{0} + 30^{0}

We get,

6x = 210^{0}

x = 210^{0} / 6

x = 35^{0}

(ii) âˆ APB = 5x – 40^{0}

= 5 Ã— 35^{0} – 40^{0}

We get,

= 175^{0} â€“ 40^{0}

= 135^{0}