RD Sharma Solutions For Class 7 Maths Chapter 15 Properties of Triangles Exercise 15.3

Students can view and download the PDF of RD Sharma Solutions for Class 7 Maths Exercise 15.3 of Chapter 15 Properties of Triangles available here. The BYJU’S experts in Maths formulate the solutions for questions present in the textbook based on the understanding abilities of students. This exercise explains exterior and interior angles of a triangle. RD Sharma Solutions for Class 7 includes answers to all questions present in this exercise. Several topics of this exercise are listed below:

• Exterior angle
• Interior opposite angle
• Interior adjacent angle
• Theorem based on the adjacent angle

RD Sharma Solutions for Class 7 Maths Chapter 15 – Properties of Triangles Exercise 15.3

Download PDF

carouselExampleControls112

Previous Next

Access answers to Maths RD Sharma Solutions For Class 7 Chapter 15 – Properties of Triangles Exercise 15.3

Exercise 15.3 Page No: 15.19

1. In Fig. 35, ∠CBX is an exterior angle of ∆ABC at B. Name

(i) The interior adjacent angle

(ii) The interior opposite angles to exterior ∠CBX

Also, name the interior opposite angles to an exterior angle at A.

Solution:

(i) The interior adjacent angle is ∠ABC

(ii) The interior opposite angles to exterior ∠CBX is ∠BAC and ∠ACB

Also the interior angles opposite to exterior are ∠ABC and ∠ACB

2. In the fig. 36, two of the angles are indicated. What are the measures of ∠ACX and ∠ACB?

Solution:

Given that in △ABC, ∠A = 50^o and ∠B = 55^o

We know that the sum of angles in a triangle is 180^o

Therefore we have

∠A + ∠B + ∠C = 180^o

50^o+ 55^o+ ∠C = 180^o

∠C = 75^o

∠ACB = 75^o

∠ACX = 180^o− ∠ACB = 180^o − 75^o = 105^o

3. In a triangle, an exterior angle at a vertex is 95^o and its one of the interior opposite angles is 55^o. Find all the angles of the triangle.

Solution:

We know that the sum of interior opposite angles is equal to the exterior angle.

Hence, for the given triangle, we can say that:

∠ABC+ ∠BAC = ∠BCO

55^o + ∠BAC = 95^o

∠BAC= 95^o– 55^o

∠BAC = 40^o

We also know that the sum of all angles of a triangle is 180^o.

Hence, for the given △ABC, we can say that:

∠ABC + ∠BAC + ∠BCA = 180^o

55^o + 40^o + ∠BCA = 180^o

∠BCA = 180^o –95^o

∠BCA = 85^o

4. One of the exterior angles of a triangle is 80^o, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?

Solution:

Let us assume that A and B are the two interior opposite angles.

We know that ∠A is equal to ∠B.

We also know that the sum of interior opposite angles is equal to the exterior angle.

Therefore from the figure we have,

∠A + ∠B = 80^o

∠A +∠A = 80^o (because ∠A = ∠B)

2∠A = 80^o

∠A = 80/2 =40^o

∠A= ∠B = 40^o

Thus, each of the required angles is of 40^o.

5. The exterior angles, obtained on producing the base of a triangle both ways are 104^o and 136^o. Find all the angles of the triangle.

Solution:

In the given figure, ∠ABE and ∠ABC form a linear pair.

∠ABE + ∠ABC =180^o

∠ABC = 180^o– 136^o

∠ABC = 44^o

We can also see that ∠ACD and ∠ACB form a linear pair.

∠ACD + ∠ACB = 180^o

∠ACB = 180^o– 104^o

∠ACB = 76^o

We know that the sum of interior opposite angles is equal to the exterior angle.

Therefore, we can write as

∠BAC + ∠ABC = 104^o

∠BAC = 104^o – 44^o = 60^o

Thus,

∠ACE = 76^o and ∠BAC = 60^o

6. In Fig. 37, the sides BC, CA and BA of a △ABC have been produced to D, E and F respectively. If ∠ACD = 105^o and ∠EAF = 45^o; find all the angles of the △ABC.

Solution:

In a △ABC, ∠BAC and ∠EAF are vertically opposite angles.

Hence, we can write as

∠BAC = ∠EAF = 45^o

Considering the exterior angle property, we have

∠BAC + ∠ABC = ∠ACD = 105^o

On rearranging we get

∠ABC = 105^o– 45^o = 60^o

We know that the sum of angles in a triangle is 180^o

∠ABC + ∠ACB +∠BAC = 180°

∠ACB = 75^o

Therefore, the angles are 45^o, 60^o and 75^o.

7. In Fig. 38, AC perpendicular to CE and C ∠A: ∠B: ∠C= 3: 2: 1. Find the value of ∠ECD.

Solution:

In the given triangle, the angles are in the ratio 3: 2: 1.

Let the angles of the triangle be 3x, 2x and x.

We know that sum of angles in a triangle is 180^o

3x + 2x + x = 180^o

6x = 180^o

x = 30^o

Also, ∠ACB + ∠ACE + ∠ECD = 180^o

x + 90^o + ∠ECD = 180^o (∠ACE = 90^o)

We know that x = 30^o

Therefore

∠ECD = 60^o

8. A student when asked to measure two exterior angles of △ABC observed that the exterior angles at A and B are of 103^o and 74^o respectively. Is this possible? Why or why not?

Solution:

We know that sum of internal and external angle is equal to 180^o

Internal angle at A + External angle at A = 180^o

Internal angle at A + 103^o =180^o

Internal angle at A = 77^o

Internal angle at B + External angle at B = 180^o

Internal angle at B + 74^o = 180^o

Internal angle at B = 106^o

Sum of internal angles at A and B = 77^o + 106^o = 183^o

It means that the sum of internal angles at A and B is greater than 180^o, which cannot be possible.

9. In Fig.39, AD and CF are respectively perpendiculars to sides BC and AB of △ABC. If ∠FCD = 50^o, find ∠BAD

Solution:

We know that the sum of all angles of a triangle is 180^o

Therefore, for the given △FCB, we have

∠FCB + ∠CBF + ∠BFC = 180^o

50^o + ∠CBF + 90^o = 180^o

∠CBF = 180^o – 50^o – 90^o = 40^o

Using the above steps for △ABD, we can say that:

∠ABD + ∠BDA + ∠BAD = 180^o

∠BAD = 180^o – 90^o – 40^o = 50^o

10. In Fig.40, measures of some angles are indicated. Find the value of x.

Solution:

We know that the sum of the angles of a triangle is 180^o

From the figure we have,

∠AED + 120^o = 180^o (Linear pair)

∠AED = 180^o – 120^o = 60^o

We know that the sum of all angles of a triangle is 180^o.

Therefore, for △ADE, we have

∠ADE + ∠AED + ∠DAE = 180^o

60^o+ ∠ADE + 30^o =180^o

∠ADE = 180^o – 60^o – 30^o = 90^o

From the given figure, we have

∠FDC + 90^o = 180^o (Linear pair)

∠FDC = 180^o – 90^o = 90^o

Using the same steps for △CDF, we get

∠CDF + ∠DCF + ∠DFC = 180^o

90^o + ∠DCF + 60^o = 180^o

∠DCF = 180^o – 60^o – 90^o = 30^o

Again from the figure we have

∠DCF + x = 180^o (Linear pair)

30^o + x = 180^o

x = 180^o – 30^o = 150^o

11. In Fig. 41, ABC is a right triangle right angled at A. D lies on BA produced and DE perpendicular to BC intersecting AC at F. If ∠AFE = 130^o, find

(i) ∠BDE

(ii) ∠BCA

(iii) ∠ABC

Solution:

(i) Here,

∠BAF + ∠FAD = 180^o (Linear pair)

∠FAD = 180^o – ∠BAF = 180^o – 90^o = 90^o

Also from the figure,

∠AFE = ∠ADF + ∠FAD (Exterior angle property)

∠ADF + 90^o = 130^o

∠ADF = 130^o − 90^o = 40^o

∠BDE = 40^o

(ii) We know that the sum of all the angles of a triangle is 180^o.

Therefore, for △BDE, we have

∠BDE + ∠BED + ∠DBE = 180^o

∠DBE = 180^o – ∠BDE – ∠BED

∠DBE = 180^o – 40^o – 90^o = 50^o …. Equation (i)

Again from the figure we have,

∠FAD = ∠ABC + ∠ACB (Exterior angle property)

90^o = 50^o + ∠ACB

∠ACB = 90^o – 50^o = 40^o

(iii) From equation we have

∠ABC = ∠DBE = 50^o

12. ABC is a triangle in which ∠B = ∠C and ray AX bisects the exterior angle DAC. If ∠DAX = 70^o. Find ∠ACB.

Solution:

Given that ABC is a triangle in which ∠B = ∠C

Also given that AX bisects the exterior angle DAC

∠CAX = ∠DAX (AX bisects ∠CAD)

∠CAX =70^o [given]

∠CAX +∠DAX + ∠CAB =180^o

70^o+ 70^o + ∠CAB =180^o

∠CAB =180^o –140^o

∠CAB =40^o

∠ACB + ∠CBA + ∠CAB = 180^o (Sum of the angles of ∆ABC)

∠ACB + ∠ACB+ 40^o = 180^o (∠C = ∠B)

2∠ACB = 180^o – 40^o

∠ACB = 140/2

∠ACB = 70^o

13. The side BC of △ABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC= 30^o and ∠ACD = 115^o, find ∠ALC

Solution:

Given that ∠ABC= 30^o and ∠ACD = 115^o

From the figure, we have

∠ACD and ∠ACL make a linear pair.

∠ACD+ ∠ACB = 180^o

115^o + ∠ACB =180^o

∠ACB = 180^o – 115^o

∠ACB = 65^o

We know that the sum of all angles of a triangle is 180^o.

Therefore, for△ ABC, we have

∠ABC + ∠BAC + ∠ACB = 180^o

30^o + ∠BAC + 65^o = 180^o

∠BAC = 85^o

∠LAC = ∠BAC/2 = 85/2

Using the same steps for △ALC, we get

∠ALC + ∠LAC + ∠ACL = 180^o

∠ALC + 85/2 + 65^o = 180^o

We know that ∠ACL = ∠ACB

∠ALC = 180^o – 85/2 – 65^o

∠ALC = 72 ½^o

14. D is a point on the side BC of △ABC. A line PDQ through D, meets side AC in P and AB produced at Q. If ∠A = 80^o, ∠ABC = 60^o and ∠PDC = 15^o, find

(i) ∠AQD

(ii) ∠APD

Solution:

From the figure we have

∠ABD and ∠QBD form a linear pair.

∠ABC + ∠QBC =180^o

60^o + ∠QBC = 180^o

∠QBC = 120^o

∠PDC = ∠BDQ (Vertically opposite angles)

∠BDQ = 15^o

(i) In △QBD:

∠QBD + ∠QDB + ∠BQD = 180^o (Sum of angles of △QBD)

120^o+ 15^o + ∠BQD = 180^o

∠BQD = 180^o – 135^o

∠BQD = 45^o

∠AQD = ∠BQD = 45^o

(ii) In △AQP:

∠QAP + ∠AQP + ∠APQ = 180^o (Sum of angles of △AQP)

80^o + 45^o + ∠APQ = 180^o

∠APQ = 55^o

∠APD = ∠APQ = 55^o

15. Explain the concept of interior and exterior angles and in each of the figures given below. Find x and y (Fig. 42)

Solution:

The interior angles of a triangle are the three angle elements inside the triangle.

The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Using these definitions, we will obtain the values of x and y.

(i) From the given figure, we have

∠ACB + x = 180^o (Linear pair)

75^o+ x = 180^o

x = 105^o

We know that the sum of all angles of a triangle is 180^o

Therefore, for △ABC, we can say that:

∠BAC+ ∠ABC +∠ACB = 180^o

40^o+ y +75^o = 180^o

y = 65^o

(ii) From the figure, we have

x + 80^o = 180^o (Linear pair)

x = 100^o

In △ABC, we have

We also know that the sum of angles of a triangle is 180^o

x + y + 30^o = 180^o

100^o + 30^o + y = 180^o

y = 50^o

(iii) We know that the sum of all angles of a triangle is 180°.

Therefore, for △ACD, we have

30^o + 100^o + y = 180^o

y = 50^o

Again from the figure we can write as

∠ACB + 100° = 180°

∠ACB = 80^o

Using the above rule for △ACB, we can say that:

x + 45^o + 80^o = 180^o

x = 55^o

(iv) We know that the sum of all angles of a triangle is 180^o.

Therefore, for △DBC, we have

30^o + 50^o + ∠DBC = 180^o

∠DBC = 100^o

From the figure we can say that

x + ∠DBC = 180^o is a Linear pair

x = 80^o

From the exterior angle property we have

y = 30^o + 80^o = 110^o

16. Compute the value of x in each of the following figures:

Solution:

(i) From the given figure, we can write as

∠ACD + ∠ACB = 180^o is a linear pair

On rearranging we get

∠ACB = 180^o – 112^o = 68^o

Again from the figure we have,

∠BAE + ∠BAC = 180^o is a linear pair

On rearranging we get,

∠BAC = 180^o– 120^o = 60^o

We know that the sum of all angles of a triangle is 180^o.

Therefore, for △ABC:

x + ∠BAC + ∠ACB = 180^o

x = 180^o – 60^o – 68^o = 52^o

x = 52^o

(ii) From the given figure, we can write as

∠ABC + 120^o = 180^o is a linear pair

∠ABC = 60^o

Again from the figure we can write as

∠ACB+ 110^o = 180^o is a linear pair

∠ACB = 70^o

We know that the sum of all angles of a triangle is 180^o.

Therefore, consider △ABC, we get

x + ∠ABC + ∠ACB = 180^o

x = 50^o

(iii) From the given figure, we can write as

∠BAD = ∠ADC = 52^o are alternate angles

We know that the sum of all the angles of a triangle is 180^o.

Therefore, consider △DEC, we have

x + 40^o + 52^o = 180^o

x = 88^o

(iv) In the given figure, we have a quadrilateral and also we know that sum of all angles in a quadrilateral is 360^o.

Thus,

35^o + 45^o + 50^o + reflex ∠ADC = 360^o

On rearranging we get,

Reflex ∠ADC = 230^o

230^o + x = 360^o (A complete angle)

x = 130^o