The document RD Sharma Solutions - Ex-15.2, (Part - 1), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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**Q1. Two angles of a triangle are of measures 150 ^{∘} and 30^{∘}. Find the measure of the third angle.**

**Solution.**

Let the third angle be x

Sum of all the angles of a triangle = 180^{∘}

105^{∘}+ 30^{∘}+x = 180^{∘}

135^{∘}+ x = 180^{∘}

x = 180^{∘ }– 135^{∘}

x = 45^{∘}

Therefore the third angle is 45^{∘}

**Q2. One of the angles of a triangle is 130 ^{∘}, and the other two angles are equal What is the measure of each of these equal angles?**

**Solution.**

Let the second and third angle be x

Sum of all the angles of a triangle = 180^{∘}

130∘+ x+x = 180^{∘}

130^{∘}+2x = 180^{∘}

2x = 180^{∘}– 130^{∘}

2x = 50^{∘}

x = 50/2

x = 25^{∘}

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

**Q3. The three angles of a triangle are equal to one another. What is the measure of each of the angles?**

**Solution.**

Let the each angle be x

Sum of all the angles of a triangle = 180^{∘}

x+x+x = 180^{∘}

3x = 180^{∘}

x = 1803

x = 60^{∘}

Therefore angle is 60^{∘} each

**Q4. If the angles of a triangle are in the ratio 1 : 2 : 3, determine three angles.**

**Solution.**

If angles of the triangle are in the ratio 1:2:3 then take first angle as ‘x’, second angle as ‘2x’ and third angle as ‘3x’

Sum of all the angles of a triangle = 180^{∘}

x+2x+3x = 180^{∘}

6x = 180^{∘}

x = 180/6

x=30^{∘}

2x = 30^{∘}×2 = 60^{∘}

3x = 30^{∘}×3 = 90^{∘}

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

**Q5. The angles of a triangle are ** **Find the value of x.**

**Solution.**

Sum of all the angles of a triangle = 180^{∘}

Hence we can conclude that x is equal to 100^{∘}

**Q6. The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10 ^{∘}. Find the three angles.**

**Solution.**

Let the first angle be x

Second angle be x+10^{∘}

Third angle be x+10^{∘}+ 10^{∘}

Sum of all the angles of a triangle = 180^{∘}

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

3x+30 = 180

3x = 180-30

3x = 150

x = 150/3

x = 50^{∘}

First angle is 50

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

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

**Q7. Two angles of a triangle are equal and the third angle is greater than each of those angles by 30**^{∘}**. Determine all the angles of the triangle**

**Solution.**

Let the first and second angle be x

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

The first and the second angles are equal

Sum of all the angles of a triangle=180∘

x+x+x+30^{∘} = 180^{∘}

3x+30 = 180

3x=180-30

3x = 150

x = 150/3

x = 50^{∘}

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

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

**Q8. If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.**

**Solution.**

One angle of a triangle is equal to the sum of the other two

x = y+z

Let the measure of angles be x,y,z

x+y+z = 180^{∘}

x+x = 180^{∘}

2x = 180^{∘}

x = 180^{∘}/2

x = 90^{∘}

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

**Q9. If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.**

**Solution.**

Each angle of a triangle is less than the sum of the other two

Measure of angles be x,y and z

x>y+z

y<x+z

z<x+y

Therefore triangle is an acute triangle

**Q10. 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 ^{∘}, 37^{∘}, 80^{∘}**

**(ii) 45 ^{∘}, 61^{∘}, 73^{∘}**

**(iii) 59 ^{∘}, 72^{∘}, 61^{∘}**

**(iv) 45 ^{∘}, 45^{∘}, 90^{∘}**

**(v) 30 ^{∘}, 20^{∘}, 125^{∘}**

**Solution.**

(i) 63^{∘}, 37^{∘}, 80^{∘} = 180^{∘}

Angles form a triangle

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

Therefore not a triangle

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

Therefore not a triangle

(iv) 45^{∘}, 45^{∘}, 90^{∘} = 180

Angles form a triangle

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

Therefore not a triangle

**Q11. The angles of a triangle are in the ratio 3: 4 : 5. Find the smallest angle**

**Solution.**

Given that

Angles of a triangle are in the ratio: 3: 4: 5

Measure of the angles be 3x, 4x, 5x

Sum of the angles of a triangle = 180^{∘}

3x+4x+5x = 180^{∘}

12x = 180^{∘}

x = 180^{∘}/12

x = 15^{∘}

Smallest angle = 3x

= 3 x 15^{∘}

= 45^{∘}

**Q12. Two acute angles of a right triangle are equal. Find the two angles.**

**Solution.**

Given acute angles of a right angled triangle are equal

Right triangle: whose one of the angle is a right angle

Measured angle be x,x,90^{∘}

x+x+180^{∘} = 180^{∘}

2x = 90^{∘}

x = 90^{∘}/2

x = 45^{∘}

The two angles are 45^{∘} and 45^{∘}

**Q13. 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.**

Angle of a triangle is greater than the sum of the other two

Measure of the angles be x,y,z

x>y+z or

y>x+z or

z>x+y

x or y or z>90^{∘} which is obtuse

Therefore triangle is an obtuse angle

**Q14. AC, AD and AE are joined. Find ∠FAB+∠ABC+∠BCD+∠CDE+∠DEF+∠EFA**

**Solution.**

∠FAB+∠ABC+∠BCD+∠CDE+∠DEF+∠EFA

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

Therefore in ΔABC,we have

∠CAB+∠ABC+∠BCA = 180^{∘}—(i)

In ΔACD,we have

∠DAC+∠ACD+∠CDA = 180^{∘}—(ii)

In ΔADE,we have

∠EAD+∠ADE+∠DEA = 180^{∘}—(iii)

In ΔAEF,we have

∠FAE+∠AEF+∠EFA = 180^{∘}—(iv)

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

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

Therefore

∠FAB+∠ABC+∠BCD+∠CDE+∠DEF+∠EFA = 720^{∘}