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**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}

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

x + x + (x/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. It 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 angles 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}.

Â

**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.

**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 = 100^{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 = 140^{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 LA)

âˆ OCAÂ Â = âˆ C/2 (OC bisects LC)

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 A2 + âˆ C2 = 180^{o}

âˆ AOC + âˆ A + âˆ C2 = 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}

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 = 90^{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 ABC = âˆ PRQ and âˆ QCA = âˆ CQP)

Or,

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

âˆ QPR = 90^{o}