Get free PDF of RD Sharma Solutions for Class 7 Maths Exercise 15.2 of Chapter 15 Properties of Triangles from the provided links. These PDF’s can be easily downloaded by the students. Our experts have uniquely formulated these questions to help the students gain competence. RD Sharma Solutions for Class 7 is one of the best study material designed for CBSE students. This exercise deals with the angle sum property of a triangle and some more results on triangular properties.

## Download the PDF of RD Sharma Solutions For Class 7 Maths Chapter 15 – Properties of Triangles Exercise 15.2

### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 15 – Properties of Triangles Exercise 15.2

Exercise 15.2 Page No: 15.12

**1. Two angles of a triangle are of measures 150 ^{o} and 30^{o}. Find the measure of the third angle.**

**Solution:**

Given two angles of a triangle are of measures 150^{o} and 30^{o}

Let the required third angle be x

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

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

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

x = 180^{o} â€“ 135^{o}

x = 45^{o}

Therefore the third angle is 45^{o}

**2. One of the angles of a triangle is 130 ^{o}, and the other two angles are equal. What is the measure of each of these equal angles?**

**Solution:**

Given one of the angles of a triangle is 130^{o}

Also given that remaining two angles are equal

So let the second and third angle be x

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

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

130^{o} + 2x = 180^{o}

2x = 180^{o }â€“ 130^{o}

2x = 50^{o}

x = 50/2

x = 25^{o}

Therefore the two other angles are 25^{o} each

**3. The three angles of a triangle are equal to one another. What is the measure of each of the angles?**

**Solution:**

Given that three angles of a triangle are equal to one another

So let the each angle be x

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

x + x + x = 180^{o}

3x = 180^{o}

x = 180/3

x = 60^{o}

Therefore angle is 60^{o} each

**4. If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.**

**Solution:**

Given angles of the triangle are in the ratio 1: 2: 3

So take first angle as x, second angle as 2x and third angle as 3x

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

x + 2x + 3x = 180^{o}

6x = 180^{o}

x = 180/6

x = 30^{o}

2x = 30^{o} Ã— 2 = 60^{o}

3x = 30^{o} Ã— 3 = 90^{o}

Therefore the first angle is 30^{o}, second angle is 60^{o} and third angle is 90^{o}.

**5. The angles of a triangle are (x âˆ’ 40) ^{o}, (x âˆ’ 20)^{o} and (1/2 âˆ’ 10)^{o}. Find the value of x.**

**Solution:**

Given the angles of a triangle are (x âˆ’ 40)^{o}, (x âˆ’ 20)^{o} and (1/2 âˆ’ 10)^{o}.

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

(x âˆ’ 40)^{o} + (x âˆ’ 20)^{o} + (1/2 âˆ’ 10)^{o} = 180^{o}

x + x + (1/2) â€“ 40^{o} â€“ 20^{o }â€“ 10^{o} = 180^{o}

x + x + (1/2) â€“ 70^{o} = 180^{o}

(5x/2) = 180^{o }+ 70^{o}

(5x/2) = 250^{o}

x = (2/5) Ã— 250^{o}

x = 100^{o}

Hence the value of x is 100^{o}

**6. The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10 ^{o}. Find the three angles.**

**Solution:**

Given that angles of a triangle are arranged in ascending order of magnitude

Also given that difference between two consecutive angles is 10^{o}

Let the first angle be x

Second angle be x + 10^{o}

Third angle be x + 10^{o} + 10^{o}

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

x + x + 10^{o} + x + 10^{o} +10^{o} = 180^{o}

3x + 30 = 180

3x = 180 – 30

3x = 150

x = 150/3

x = 50^{o}

First angle is 50^{o}

Second angle x + 10^{o} = 50 + 10 = 60^{o}

Third angle x + 10^{o} +10^{o} = 50 + 10 + 10 = 70^{o}

**7. Two angles of a triangle are equal and the third angle is greater than each of those angles by 30 ^{o}. Determine all the angles of the triangle**

**Solution:**

Given that two angles of a triangle are equal

Let the first and second angle be x

Also given that third angle is greater than each of those angles by 30^{o}

Therefore the third angle is greater than the first and second by 30^{o} = x + 30^{o}

The first and the second angles are equal

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

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

3x + 30 = 180

3x = 180 – 30

3x = 150

x = 150/3

x = 50^{o}

Third angle = x + 30^{o} = 50^{o} + 30^{o} = 80^{o}

The first and the second angle is 50^{o} and the third angle is 80^{o}.

**8. If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.**

**Solution:**

Given that one angle of a triangle is equal to the sum of the other two

Let the measure of angles be x, y, z

Therefore we can write above statement as x = y + z

x + y + z = 180^{o}

Substituting the above value we get

x + x = 180^{o}

2x = 180^{o}

x = 180/2

x = 90^{o}

If one angle is 90^{o} then the given triangle is a right angled triangle

**9. If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.**

**Solution:**

Given that each angle of a triangle is less than the sum of the other two

Let the measure of angles be x, y and z

From the above statement we can write as

x > y + z

y < x + z

z < x + y

Therefore triangle is an acute triangle

**10. In each of the following, the measures of three angles are given. State in which cases the angles can possibly be those of a triangle:**

**(i) 63 ^{o}, 37^{o}, 80^{o}**

**(ii) 45 ^{o}, 61^{o}, 73^{o}**

**(iii) 59 ^{o}, 72^{o}, 61^{o}**

**(iv) 45 ^{o}, 45^{o}, 90^{o}**

**(v) 30 ^{o}, 20^{o}, 125^{o}**

**Solution:**

(i) 63^{o} + 37^{o }+ 80^{o} = 180^{o}

Angles form a triangle

(ii) 45^{o}, 61^{o}, 73^{o} is not equal to 180^{o}

Therefore not a triangle

(iii) 59^{o}, 72^{o}, 61^{o} is not equal to 180^{0}

Therefore not a triangle

(iv) 45^{o}+ 45^{o}+ 90^{o} = 180^{o}

Angles form a triangle

(v) 30^{o}, 20^{o}, 125^{o} is not equal to 180^{o}

Therefore not a triangle

**11. The angles of a triangle are in the ratio 3: 4: 5. Find the smallest angle**

**Solution:**

GivenÂ that angles of a triangle are in the ratio: 3: 4: 5

Therefore let the measure of the angles be 3x, 4x, 5x

We know that sum ofÂ the angles of a triangle =180^{o}

3x + 4x + 5x = 180^{o}

12x = 180^{o}

x = 180/12

x = 15^{o}

Smallest angle = 3x

= 3 Ã— 15^{o}

= 45^{o}

Therefore smallest angle = 45^{o}

**12. Two acute angles of a right triangle are equal. Find the two angles.**

**Solution:**

Given that acute angles of a right angled triangle are equal

We know that Right triangle: whose one of the angle is a right angle

Let the measure of angle be x, x, 90^{o}

x + x + 90^{o}= 180^{o}

2x = 90^{o}

x = 90/2

x = 45^{o}

The two angles are 45^{o} and 45^{o}

**13. One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?**

**Solution:**

Given one angle of a triangle is greater than the sum of the other two

Let the measure of the angles be x, y, z

From the question we can write as

x > y + zÂ or

y > x + zÂ Â or

z > x + y

x or y or z > 90^{o} which is obtuse

Therefore triangle is an obtuse angle

**14. In the six cornered figure, (fig. 20), AC, AD and AE are joined. Find âˆ FAB + âˆ ABC + âˆ BCD + âˆ CDE + âˆ DEF + âˆ EFA.**

**Solution:**

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

Therefore in âˆ†ABC, we have

âˆ CAB + âˆ ABC + âˆ BCA = 180^{o }â€¦â€¦.. (i)

Inâ–³ ACD, we have

âˆ DAC + âˆ ACD + âˆ CDA = 180^{o }â€¦â€¦.. (ii)

In â–³ADE, we have

âˆ EAD + âˆ ADE + âˆ DEA =180^{o} â€¦â€¦â€¦. (iii)

In â–³AEF, we have

âˆ FAE + âˆ AEF + âˆ EFA = 180^{o} â€¦â€¦â€¦. (iv)

Adding (i), (ii), (iii), (iv) we get

âˆ CAB + âˆ ABC + âˆ BCA + âˆ DAC + âˆ ACD + âˆ CDA + âˆ EAD + âˆ ADE + âˆ DEA + âˆ FAE + âˆ AEF +âˆ EFA = 720^{o}

Therefore âˆ FAB + âˆ ABC + âˆ BCD + âˆ CDE + âˆ DEF + âˆ EFA = 720^{o}

**15. Find x, y, z (whichever is required) from the figures (Fig. 21) given below: **

**Solution:**

(i) In â–³ABC and â–³ADE we have,

âˆ ADE = âˆ ABC [corresponding angles]

x = 40^{o}

âˆ AED = âˆ ACB (corresponding angles)

y = 30^{o}

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

x + y + z = 180^{o} (Angles of â–³ADE)

Which means: 40^{o} + 30^{o} + z = 180^{o}

z = 180^{o} – 70^{o}

z = 110^{o}

Therefore, we can conclude that the three angles of the given triangle are 40^{o}, 30^{o}Â and 110^{o}

(ii) Â We can see that in â–³ADC, âˆ ADC is equal to 90^{o}.

(â–³ADCÂ is a right triangle)

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

Which means: 45^{o} + 90^{o} + y = 180^{o} (Sum of the angles of â–³ADC)

135^{o} + y = 180^{o}

y = 180^{o} â€“ 135^{o}.

y = 45^{o}.

We can also say that in â–³ ABC, âˆ ABC + âˆ ACB + âˆ BACÂ is equal to 180^{o}.

(Sum of the angles of â–³ABC)

40^{o} + y + (x + 45^{o}) = 180^{o}

40^{o} + 45^{o} + x + 45^{o} = 180^{o}Â (y = 45^{o})

x = 180^{o} â€“130^{o}

x = 50^{o}

Therefore, we can say that the required angles are 45^{o} and 50^{o}.

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

Therefore, for â–³ABD:

âˆ ABD +âˆ ADB + âˆ BAD = 180Â° (Sum of the angles of â–³ABD)

50^{o} + x + 50^{o} = 180^{o}

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

x = 180^{o} â€“ 100^{o}

x = 80^{o}

For â–³ ABC:

âˆ ABC + âˆ ACB + âˆ BAC = 180^{o} (Sum of the angles of â–³ABC)

50^{o} + z + (50^{o} + 30^{o}) = 180^{o}

50^{o} + z + 50^{o} + 30^{o} = 180^{o}

z = 180^{o} â€“ 130^{o}

z = 50^{o}

Using the same argument for â–³ADC:

âˆ ADC + âˆ ACD + âˆ DAC = 180^{o} (Sum of the angles of â–³ADC)

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

y + 50^{o} + 30^{o} = 180^{o} (z = 50^{o})

y = 180^{o} â€“ 80^{o}

y = 100^{0}

Therefore, we can conclude that the required angles are 80^{o}, 50^{o} and 100^{o}.

(iv) In â–³ABC and â–³ADE we have:

âˆ ADE = âˆ ABC (Corresponding angles)

y = 50^{o}

Also, âˆ AED = âˆ ACBÂ (Corresponding angles)

z = 40^{o}

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

We can write as x + 50^{o} + 40^{o} = 180^{o} (Angles of â–³ADE)

x = 180^{o} â€“ 90^{o}

x = 90^{o}

Therefore, we can conclude that the required angles are 50^{o}, 40^{o} and 90^{o}.

**16. If one angle of a triangle is 60 ^{o} and the other two angles are in the ratio 1: 2, find the angles.**

**Solution:**

Given that one of the angles of the given triangle is 60^{o}.

Also given that the other two angles of the triangle are in the ratio 1: 2.

Let one of the other two angles be x.

Therefore, the second one will be 2x.

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

60^{o} + x + 2x = 180^{o}

3x = 180^{o} â€“ 60^{o}

3x = 120^{o}

x = 120^{o}/3

x = 40^{o}

2x = 2 Ã— 40^{o}

2x = 80^{o}

Hence, we can conclude that the required angles are 40^{o} and 80^{o}.

**17. If one angle of a triangle is 100 ^{o} and the other two angles are in the ratio 2: 3. Find the angles.**

**Solution:**

Given that one of the angles of the given triangle is 100^{o}.

Also given that the other two angles are in the ratio 2: 3.

Let one of the other two angle be 2x.

Therefore, the second angle will be 3x.

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

100^{o} + 2x + 3x = 180^{o}

5x = 180^{o} â€“ 100^{o}

5x = 80^{o}

x = 80/5

x = 16

2x = 2 Ã—16

2x = 32^{o}

3x = 3Ã—16

3x = 48^{o}

Thus, the required angles are 32^{o} and 48^{o}.

**18. In â–³ABC, if 3âˆ A = 4âˆ B = 6âˆ C, calculate the angles.**

**Solution:**

We know that for the given triangle, 3âˆ A = 6âˆ C

âˆ A = 2âˆ C â€¦â€¦. (i)

We also know that for the same triangle, 4âˆ B = 6âˆ C

âˆ B = (6/4) âˆ C â€¦â€¦ (ii)

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

Therefore, we can say that:

âˆ A + âˆ B + âˆ C = 180^{o} (Angles of â–³ABC)â€¦â€¦ (iii)

On putting the values of âˆ A and âˆ B in equation (iii), we get:

2âˆ C + (6/4) âˆ C + âˆ C = 180^{o}

(18/4)Â âˆ C = 180^{o}

âˆ C = 40^{o}

From equation (i), we have:

âˆ A = 2âˆ C = 2 Ã— 40

âˆ A = 80^{o}

From equation (ii), we have:

âˆ B = (6/4) âˆ C = (6/4) Ã— 40^{o}

âˆ B = 60^{o}

âˆ A = 80^{o}, âˆ B = 60^{o}, âˆ C = 40^{o}

Therefore, the three angles of the given triangle are 80^{o}, 60^{o}, and 40^{o}.

**19. Is it possible to have a triangle, in which**

**(i) Two of the angles are right?**

**(ii) Two of the angles are obtuse?**

**(iii) Two of the angles are acute?**

**(iv) Each angle is less than 60 ^{o}?**

**(v) Each angle is greater than 60 ^{o}?**

**(vi) Each angle is equal to 60 ^{o}?**

**Solution:**

(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.

(ii) No, because as we know that the sum of all three angles of a triangle is always 180^{o}. If there are two obtuse angles, then their sum will be more than 180^{o}, which is not possible in case of a triangle.

(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.

(iv) No, because if each angle is less than 60^{o}, then the sum of all three angles will be less than 180^{o}, which is not possible in case of a triangle.

(v) No, because if each angle is greater than 60^{0}, then the sum of all three angles will be greater than 180^{0}, which is not possible.

(vi) Yes, if each angle of the triangle is equal to 60^{0}, then the sum of all three angles will be 180^{0}, which is possible in case of a triangle.

**20. Inâ–³ ABC, âˆ A = 100 ^{o}, AD bisects âˆ A and AD âŠ¥ BC. Find âˆ B**

**Solution:**

Given that in â–³ABC, âˆ A = 100^{o}

Also given that AD âŠ¥ BC

Consider â–³ABD

âˆ BAD = 100/2Â Â (AD bisects âˆ A)

âˆ BAD = 50^{o}

âˆ ADB = 90^{o}Â Â (AD perpendicular to BC)

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

Thus,

âˆ ABD + âˆ BAD + âˆ ADB = 180^{o}Â (Sum of angles of â–³ABD)

Or,

âˆ ABD + 50^{o} + 90^{o} = 180^{o}

âˆ ABD =180^{o} â€“Â 140^{o}

âˆ ABD = 40^{o}

**21. In â–³ABC,Â âˆ A = 50 ^{o}, âˆ B = 70^{o}Â and bisector of âˆ CÂ meets AB in D. Find the angles of the triangles ADC and BDC**

**Solution:**

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

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

âˆ A + âˆ B + âˆ C = 180^{o} (Sum of angles of â–³ABC)

50^{o} + 70^{o} + âˆ CÂ = 180^{o}

âˆ C= 180^{o} â€“120^{o}

âˆ C = 60^{o}

âˆ ACD = âˆ BCD =âˆ C2 (CD bisects âˆ C and meets AB in D.)

âˆ ACDÂ = âˆ BCDÂ = 60/2= 30^{o}

Using the same logic for the given â–³ACD, we can say that:

âˆ DACÂ + âˆ ACDÂ + âˆ ADCÂ = 180^{o}

50^{o} + 30^{o} + âˆ ADC = 180^{o}

âˆ ADC = 180^{o}â€“ 80^{o}

âˆ ADC = 100^{o}

If we use the same logic for the given â–³BCD, we can say that

âˆ DBCÂ + âˆ BCDÂ + âˆ BDC = 180^{o}

70^{o} + 30^{o} + âˆ BDCÂ = 180^{o}

âˆ BDCÂ = 180^{o} â€“ 100^{o}

âˆ BDCÂ = 80^{o}

Thus,

For â–³ADC: âˆ A = 50^{o}, âˆ D = 100^{o}Â âˆ C = 30^{o}

â–³BDC: âˆ B = 70^{o}, âˆ D = 80^{o}Â âˆ C = 30^{o}

**22. In âˆ†ABC, âˆ A = 60 ^{o}, âˆ B = 80^{o}, and the bisectors of âˆ BÂ and âˆ C, meet at O. Find**

**(i) âˆ C**

**(ii) âˆ BOC**

**Solution:**

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

Hence, for â–³ABC, we can say that:

âˆ A + âˆ B + âˆ C = 180^{o} (Sum of angles of âˆ†ABC)

60^{o} + 80^{o} + âˆ C= 180^{o}.

âˆ C = 180^{o} â€“ 140^{o}

âˆ C = 40^{o}.

(ii)For â–³OBC,

âˆ OBC = âˆ B/2 = 80/2 (OB bisects âˆ B)

âˆ OBC = 40^{o}

âˆ OCB =âˆ C/2 = 40/2 (OC bisects âˆ C)

âˆ OCB = 20^{o}

If we apply the above logic to this triangle, we can say that:

âˆ OCB + âˆ OBC + âˆ BOC = 180^{o} (Sum of angles of â–³OBC)

20^{o} + 40^{o} + âˆ BOC = 180^{o}

âˆ BOC = 180^{o} â€“ 60^{o}

âˆ BOC = 120^{o}

**23. The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.**

**Solution:**

Given bisectors of the acute angles of a right triangle meet at O

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

Hence, for â–³ABC, we can say that:

âˆ A + âˆ B + âˆ C = 180^{o}

âˆ AÂ + 90^{o}Â + âˆ CÂ = 180^{o}

âˆ AÂ + âˆ CÂ = 180^{o} â€“ 90^{o}

âˆ AÂ + âˆ CÂ = 90^{o}

For â–³OAC:

âˆ OACÂ = âˆ A/2Â (OA bisects âˆ A)

âˆ OCAÂ Â = âˆ C/2 (OC bisects âˆ C)

On applying the above logic to â–³OAC, we get

âˆ AOC + âˆ OAC + âˆ OCAÂ = 180^{o} (Sum of angles of â–³AOC)

âˆ AOC + âˆ A/2 + âˆ C/2 = 180^{o}

âˆ AOC + (âˆ A + âˆ C)/2 = 180^{o}

âˆ AOC + 90/2 = 180^{o}

âˆ AOC = 180^{o} â€“ 45^{o}

âˆ AOC = 135^{o}

**24. Â In â–³ABC, âˆ A = 50 ^{o} and BC is produced to a point D. The bisectors of âˆ ABCÂ and âˆ ACD meet at E. Find âˆ E.**

**Solution:**

In the given triangle,

âˆ ACD = âˆ A + âˆ B. (Exterior angle is equal to the sum of two opposite interior angles.)

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

Therefore, for the given triangle, we know that the sum of the angles = 180^{o}

âˆ ABC + âˆ BCA + âˆ CAB = 180^{o}

âˆ A + âˆ B + âˆ BCA = 180^{o}

âˆ BCA = 180^{o} â€“ (âˆ A + âˆ B)

But we know that EC bisects âˆ ACD

Therefore âˆ ECA = âˆ ACD/2

âˆ ECA = (âˆ A + âˆ B)/2 [âˆ ACD = (âˆ A + âˆ B)]

But EB bisects âˆ ABC

âˆ EBC = âˆ ABC/2 = âˆ B/2

âˆ EBC = âˆ ECA + âˆ BCA

âˆ EBC = (âˆ A + âˆ B)/2 + 180^{o} – (âˆ A + âˆ B)

If we use same steps for â–³EBC, then we get,

âˆ B/2 + (âˆ A + âˆ B)/2 + 180^{o} – (âˆ A + âˆ B) + âˆ BEC = 180^{o}

âˆ BEC = âˆ A + âˆ B – (âˆ A + âˆ B)/2 – âˆ B/2

âˆ BEC = âˆ A/2

âˆ BEC = 50^{o}/2

= 25^{o}

**25. In âˆ†ABC, âˆ B = 60Â°, âˆ C = 40Â°, ALÂ âŠ¥ BC and AD bisects âˆ AÂ such that L and D lie on side BC. Find âˆ LAD**

**Solution:**

We know that the sum of all angles of a triangle is 180Â°

Consider â–³ABC, we can write as

âˆ A + âˆ B + âˆ C = 180^{o}

âˆ A + 60^{o} + 40^{o} = 180^{o}

âˆ A = 80^{0}

But we know that âˆ DAC bisects âˆ A

âˆ DAC = âˆ A/2

âˆ DAC = 80^{o}/2

If we apply same steps for the â–³ADC, we get

We know that the sum of all angles of a triangle is 180Â°

âˆ ADC + âˆ DCA + âˆ DAC = 180^{o}

âˆ ADC + 40^{o} + 40^{o} = 180^{o}

âˆ ADC = 180^{o }– 80^{o} = 100^{0}

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

Therefore we have

âˆ ADC = âˆ ALD+ âˆ LAD

But here AL perpendicular to BC

100^{o} = 90^{o} + âˆ LAD

âˆ LAD = 10^{o}

**26. Line segments AB and CD intersect at O such that AC âˆ¥ DB. It âˆ CAB = 35 ^{o} and âˆ CDB = 55^{o}. Find âˆ BOD.**

**Solution:**

We know that AC parallel to BD and AB cuts AC and BD at A and B, respectively.

âˆ CAB = âˆ DBA (Alternate interior angles)

âˆ DBA = 35^{o}

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

Hence, for â–³OBD, we can say that:

âˆ DBO + âˆ ODB + âˆ BOD = 180Â°

35^{o} + 55^{o} + âˆ BOD = 180^{o} (âˆ DBO = âˆ DBA and âˆ ODB = âˆ CDB)

âˆ BOD = 180^{o} – 90^{o}

âˆ BOD = 90^{o}

**27. In Fig. 22, âˆ†ABCÂ is right angled at A, Q and R are points on line BC and P is a point such that QP âˆ¥ AC and RP âˆ¥ AB. Find âˆ P**

**Solution:**

In the given triangle, AC parallel to QP and BR cuts AC and QP at C and Q, respectively.

âˆ QCA = âˆ CQP (Alternate interior angles)

Because RP parallel to AB and BR cuts AB and RP at B and R, respectively,

âˆ ABC = âˆ PRQ (alternate interior angles).

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

Hence, for âˆ†ABC, we can say that:

âˆ ABC + âˆ ACB + âˆ BAC = 180^{o}

âˆ ABC + âˆ ACB + 90^{o} = 180^{o} (Right angled at A)

âˆ ABC + âˆ ACB = 90^{o}

Using the same logic for â–³PQR, we can say that:

âˆ PQR + âˆ PRQ + âˆ QPR = 180^{o}

âˆ ABC + âˆ ACB + âˆ QPR = 180^{o} (âˆ ACB = âˆ PQR and âˆ ABC = âˆ PRQ)

Or,

90^{o}+ âˆ QPR =180^{o} (âˆ ABC+ âˆ ACB = 90^{o})

âˆ QPR = 90^{o}