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 BYJUâ€™S experts in Maths formulate the solutions for questions present in this exercise. 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

## Download the PDF of RD Sharma Solutions For Class 7 Chapter 15 – Properties of triangles Exercise 15.3

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

**1. In Fig. 35, âˆ CBX is an exterior angle of âˆ†ABC at B. Name**

**(i) The interior adjacent angle**

**(ii) The interior opposite angles to exterior âˆ CBX**

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

**Solution:**

(i) The interior adjacent angle is âˆ ABC

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

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

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

**Solution:**

Given that in â–³ABC, âˆ A = 50^{o} and âˆ B = 55^{o}

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

Therefore we have

âˆ A + âˆ B + âˆ C = 180^{o}

50^{o}+ 55^{o}+ âˆ C = 180^{o}

âˆ C = 75^{o}

âˆ ACB = 75^{o}

âˆ ACX = 180^{o}âˆ’ âˆ ACB = 180^{o }âˆ’ 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:

âˆ ABC+ âˆ BAC = âˆ BCO

55^{o} + âˆ BAC = 95^{o}

âˆ BAC= 95^{o}â€“ 95^{o}

âˆ BAC = 40^{o}

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

Hence, for the given â–³ABC, we can say that:

âˆ ABC + âˆ BAC + âˆ BCA = 180^{o}

55^{o} + 40^{o} + âˆ BCA = 180^{o}

âˆ BCA = 180^{o} â€“95^{o}

âˆ 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 âˆ A is equal to âˆ B.

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

Therefore from the figure we have,

âˆ A + âˆ B = 80^{o}

âˆ A +âˆ A = 80^{o} (because âˆ A = âˆ B)

2âˆ A = 80^{o}

âˆ A = 40/2 =40^{o}

âˆ A= âˆ 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, âˆ ABE and âˆ ABC form a linear pair.

âˆ ABE + âˆ ABC =180^{o}

âˆ ABC = 180^{o}â€“ 136^{o}

âˆ ABC = 44^{o}

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

âˆ ACD + âˆ ACB = 180^{o}

âˆ AUB = 180^{o}â€“ 104^{o}

âˆ ACB = 76^{o}

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

Therefore, we can write as

âˆ BAC + âˆ ABC = 104^{o}

âˆ BAC = 104^{o} â€“ 44^{o} = 60^{o}

Thus,

âˆ ACE = 76^{o} and âˆ BAC = 60^{o}

**6. In Fig. 37, the sides BC, CA and BA of a â–³ABC have been produced to D, E and F respectively. If âˆ ACD = 105 ^{o} and âˆ EAF = 45^{o}; find all the angles of the â–³ABC.**

**Solution:**

In a â–³ABC, âˆ BAC and âˆ EAF are vertically opposite angles.

Hence, we can write as

âˆ BAC = âˆ EAF = 45^{o}

Considering the exterior angle property, we have

âˆ BAC + âˆ ABC = âˆ ACD = 105^{o}

On rearranging we get

âˆ ABC = 105^{o}â€“ 45^{o} = 60^{o}

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

âˆ ABC + âˆ ACS +âˆ BAC = 180Â°

âˆ ACB = 75^{o}

Therefore, the angles are 45^{o}, 65^{o} and 75^{o}.

**7. In Fig. 38, AC perpendicular to CE and C âˆ A: âˆ B: âˆ C= 3: 2: 1. Find the value of âˆ 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, âˆ ACB + âˆ ACE + âˆ ECD = 180^{o}

x + 90^{o} + âˆ ECD = 180^{o} (âˆ ACE = 90^{o})

We know that x = 30^{o}

Therefore

âˆ ECD = 60^{o}

**8. A student when asked to measure two exterior angles of â–³ABC 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 â–³ABC. If âˆ FCD = 50 ^{o}, find âˆ BAD**

**Solution:**

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

Therefore, for the given â–³FCB, we have

âˆ FCB + âˆ CBF + âˆ BFC = 180^{o}

50^{o} + âˆ CBF + 90^{o }= 180^{o}

âˆ CBF = 180^{o} â€“ 50^{o} â€“ 90^{o} = 40^{o}

Using the above steps for â–³ABD, we can say that:

âˆ ABD + âˆ BDA + âˆ BAD = 180^{o}

âˆ BAD = 180^{o} â€“ 90^{o} â€“ 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,

âˆ AED + 120^{o} = 180^{o} (Linear pair)

âˆ AED = 180^{o} â€“ 120^{o} = 60^{o}

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

Therefore, for â–³ADE, we have

âˆ ADE + âˆ AED + âˆ DAE = 180^{o}

60^{o}+ âˆ ADE + 30^{o} =180^{o}

âˆ ADE = 180^{o }â€“ 60^{o }â€“ 30^{o} = 90^{o}

From the given figure, we have

âˆ FDC + 90^{o} = 180^{o} (Linear pair)

âˆ FDC = 180^{o }â€“ 90^{o} = 90^{o}

Using the same steps for â–³CDF, we get

âˆ CDF + âˆ DCF + âˆ DFC = 180^{o}

90^{o} + âˆ DCF + 60^{o} = 180^{o}

âˆ DCF = 180^{o }â€“ 60^{o }â€“ 90^{o }= 30^{o}

Again from the figure we have

âˆ DCF + x = 180^{o} (Linear pair)

30^{o} + x = 180^{o}

x = 180^{o} â€“ 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 âˆ AFE = 130 ^{o}, find**

**(i) âˆ BDE**

**(ii) âˆ BCA**

**(iii) âˆ ABC**

**Solution:**

(i) Here,

âˆ BAF + âˆ FAD = 180^{o} (Linear pair)

âˆ FAD = 180^{o }– âˆ BAF = 180^{o }â€“ 90^{o} = 90^{o}

Also from the figure,

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

âˆ ADF + 90^{o} = 130^{o}

âˆ ADF = 130^{o} âˆ’ 90^{o} = 40^{o}

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

Therefore, for â–³BDE, we have

âˆ BDE + âˆ BED + âˆ DBE = 180^{o}

âˆ DBE = 180^{o }â€“ âˆ BDE

âˆ BED = 180^{o }â€“ 90^{o }â€“ 40^{o }= 50^{o} â€¦. Equation (i)

Again from the figure we have,

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

90^{o} = 50^{o} + âˆ ACB

âˆ ACB = 90^{o }â€“ 50^{o} = 40^{o}

(iii) From equation we have

âˆ ABC = âˆ DBE = 50^{o}

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

**Solution:**

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

Also given that AX bisects the exterior angle DAC

âˆ CAX = âˆ DAX (AX bisects âˆ CAD)

âˆ CAX =70^{o }[given]

âˆ CAX +âˆ DAX + âˆ CAB =180^{o}

70^{o}+ 70^{o} + âˆ CAB =180^{o}

âˆ CAB =180^{o} â€“140^{o}

âˆ CAB =40^{o}

âˆ ACB + âˆ CBA + âˆ CAB = 180^{o} (Sum of the angles of âˆ†ABC)

âˆ ACB + âˆ ACB+ 40^{o} = 180^{o} (âˆ C = âˆ B)

2âˆ ACB = 180^{o }â€“ 40^{o}

âˆ ACB = 140/2

âˆ ACB = 70^{o}

**13. The side BC of â–³ABC is produced to a point D. The bisector of âˆ A meets side BC in L. If âˆ ABC= 30 ^{o} and âˆ ACD = 115^{o}, find âˆ ALC**

**Solution:**

Given that âˆ ABC= 30^{o} and âˆ ACD = 115^{o}

From the figure, we have

âˆ ACD and âˆ ACL make a linear pair.

âˆ ACD+ âˆ ACB = 180^{o}

115^{o} + âˆ ACB =180^{o}

âˆ ACB = 180^{o }â€“ 115^{o}

âˆ ACB = 65^{o}

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

Therefore, forâ–³ ABC, we have

âˆ ABC + âˆ BAC + âˆ ACB = 180^{o}

30^{o} + âˆ BAC + 65^{o} = 180^{o}

âˆ BAC = 85^{o}

âˆ LAC = âˆ BAC/2 = 85/2

Using the same steps for â–³ALC, we get

âˆ ALC + âˆ LAC + âˆ ACL = 180^{o}

âˆ ALC + 82/2 + 65^{o} = 180^{o}

We know that âˆ ALC = âˆ ACB

âˆ ALC = 180^{o} – 82/2 – 65^{o}

âˆ ALC = 72 Â½^{o}

**14. D is a point on the side BC of â–³ABC. A line PDQ through D, meets side AC in P and AB produced at Q. If âˆ A = 80 ^{o}, âˆ ABC = 60^{o} and âˆ PDC = 15^{o}, find**

**(i) âˆ AQD**

**(ii) âˆ APD**

**Solution:**

From the figure we have

âˆ ABD and âˆ QBD form a linear pair.

âˆ ABC + âˆ QBC =180^{o}

60^{o} + âˆ QBC = 180^{o}

âˆ QBC = 120^{o}

âˆ PDC = âˆ BDQ (Vertically opposite angles)

âˆ BDQ = 75^{o}

In â–³QBD:

âˆ QBD + âˆ QDB + âˆ BDQ = 180^{o} (Sum of angles of â–³QBD)

120^{o}+ 15^{o} + âˆ BQD = 180^{o}

âˆ BQD = 180^{o }â€“ 135^{o}

âˆ BQD = 45^{o}

âˆ AQD = âˆ BQD = 45^{o}

In â–³AQP:

âˆ QAP + âˆ AQP + âˆ APQ = 180^{o} (Sum of angles of â–³AQP)

80^{o} + 45^{o} + âˆ APQ = 180^{o}

âˆ APQ = 55^{o}

âˆ APD = âˆ APQ

**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

âˆ 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 â–³ABC, we can say that:

âˆ BAC+ âˆ ABC +âˆ 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 â–³ABC, 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 180Â°.

Therefore, for â–³ACD, we have

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

y = 50^{o}

Again from the figure we can write as

âˆ ACB + 100Â° = 180Â°

âˆ ACB = 80^{o}

Using the above rule for â–³ACD, 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 â–³DBC, we have

30^{o} + 50^{o} + âˆ DBC = 180^{o}

âˆ DBC = 100^{o}

From the figure we can say that

x + âˆ 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

âˆ ACD + âˆ ACB = 180^{o} is a linear pair

On rearranging we get

âˆ ACB = 180^{o }â€“ 112^{o} = 68^{o}

Again from the figure we have,

âˆ BAE + âˆ BAC = 180^{o} is a linear pair

On rearranging we get,

âˆ BAC = 180^{o}â€“ 120^{o} = 60^{o}

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

Therefore, for â–³ABC:

x + âˆ BAC + âˆ ACB = 180^{o}

x = 180^{o }â€“ 60^{o }â€“ 68^{o} = 52^{o}

x = 52^{o}

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

âˆ ABC + 120^{o} = 180^{o} is a linear pair

âˆ ABC = 60^{o}

Again from the figure we can write as

âˆ ACB+ 110^{o} = 180^{o} is a linear pair

âˆ ACB = 70^{o}

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

Therefore, consider â–³ABC, we get

x + âˆ ABC + âˆ ACB = 180^{o}

x = 50^{o}

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

âˆ BAD = âˆ ADC = 52^{o} are alternate angles

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

Therefore, consider â–³DEC, 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 is quadrilateral is 360^{o}.

Thus,

35^{o} + 45^{o} + 50^{o} + reflex âˆ ADC = 360^{o}

On rearranging we get,

Reflex âˆ ADC = 230^{o}

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

x = 130^{o}