The document RD Sharma Solutions - Ex-15.2, (Part - 2), 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.

All you need of Class 7 at this link: Class 7

**Q15.Find x,y,z(whichever is required) from the figures given below**

**Solution.**

(i) In Î”ABC and Î”ADE we have :

âˆ ADE = âˆ ABC (corresponding angles)

x = 40^{âˆ˜}

âˆ AED = âˆ ACB (corresponding angles)

y = 30^{âˆ˜}

We know that the sum of all the three angles of a triangle is equal to 180âˆ˜

x +y +z = 180^{âˆ˜} (Angles of A ADE)

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

z = 180^{âˆ˜}âˆ’70^{âˆ˜}

z = 110^{âˆ˜}

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

(ii) We can see that in Î”ADC, âˆ ADC is equal to 90^{âˆ˜}.

(Î”ADC is a right triangle)

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

Which means : 45^{âˆ˜}+ 90^{âˆ˜} +y = 180^{âˆ˜} (Sum of the angles of Î”ADC)

135^{âˆ˜} + y = 180^{âˆ˜}

y = 180^{âˆ˜} â€“ 135^{âˆ˜}.

y = 45^{âˆ˜}.

We can also say that in Î”ABC, âˆ ABC+âˆ ACB+âˆ BAC is equal to 180^{âˆ˜}.

(Sum of the angles of A ABC)

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

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

x = 180^{âˆ˜} â€“130^{âˆ˜}

x = 50^{âˆ˜}

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

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

Therefore, for Î”ABD:

âˆ ABD+âˆ ADB+âˆ BAD = 180^{âˆ˜} (Sum of the angles of Î”ABD)

50^{âˆ˜} + x + 50^{âˆ˜} = 180^{âˆ˜}

100^{âˆ˜} +x = 180^{âˆ˜}

x = 180^{âˆ˜} â€“100^{âˆ˜}

x = 80^{âˆ˜}

For Î”ABC:

âˆ ABC+âˆ ACB+âˆ BAC = 180^{âˆ˜} (Sum of the angles of Î”ABC)

50^{âˆ˜} + z + (50^{âˆ˜} + 30^{âˆ˜}) = 180^{âˆ˜}

50^{âˆ˜} + z + 50^{âˆ˜} + 30^{âˆ˜} = 180^{âˆ˜}

z = 180^{âˆ˜} â€“ 130^{âˆ˜}

z = 50^{âˆ˜}

Using the same argument for Î”ADC:

âˆ ADC+âˆ ACD+âˆ DAC = 180^{âˆ˜} (Sum of the angles of Î”ADC)

y +z +30^{âˆ˜} =180^{âˆ˜}

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

y = 180^{âˆ˜ }â€“ 80^{âˆ˜}

y = 100^{âˆ˜}

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

(iv) In Î”ABC and Î”ADE we have :

âˆ ADE = âˆ ABC (Corresponding angles)

y = 50^{âˆ˜}

Also, âˆ AED =âˆ ACB (Corresponding angles)

z = 40^{âˆ˜}

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

Which means : x+50^{âˆ˜} +40^{âˆ˜} = 180^{âˆ˜} (Angles of Î”ADE)

x = 180^{âˆ˜}â€“ 90^{âˆ˜}

x = 90^{âˆ˜}

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

**Q16. If one angle of a triangle is 60âˆ˜ and the other two angles are in the ratio 1 :2, find the angles**

**Solution.**

We know that one of the angles of the given triangle is 60^{âˆ˜}. (Given)

We also know 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^{âˆ˜}.

60^{âˆ˜} +x + 2x = 180^{âˆ˜}

3x =180^{âˆ˜}â€“ 60^{âˆ˜}

3x = 120^{âˆ˜}

x = 120^{âˆ˜}/3

x = 40^{âˆ˜}

2x = 2 x 40

2x = 80^{âˆ˜}

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

**Q17. It one angle of a triangle is 100âˆ˜ and the other two angles are in the ratio 2 : 3. find the angles.**

**Solution.**

We know that one of the angles of the given triangle is 100^{âˆ˜}.

We also know 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^{âˆ˜}.

100^{âˆ˜} + 2x + 3x = 180^{âˆ˜}

5x = 180^{âˆ˜}â€“ 100^{âˆ˜}

5x = 80^{âˆ˜}

x = 80^{âˆ˜}/5

2x = 2 x 16

2x = 32^{âˆ˜}

3x = 3 x 16

3x = 48^{âˆ˜}

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

**Q18. In Î”ABC, 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 Câ€”(i)

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

We know that the sum of all three angles of a triangle is 180âˆ˜.

Therefore, we can say that:

âˆ A+âˆ B+âˆ C=180^{âˆ˜} (Angles of Î”ABC)â€”(iii)

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

From equation (i), we have:

angleA = 2âˆ C = 2Ã—40

angleA = 80^{âˆ˜}

From equation (ii), we have:

angleB = 60^{âˆ˜}

angleA = 80^{âˆ˜}, angleB = 60^{âˆ˜}, angleC = 40^{âˆ˜}

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

**Q19. 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 ^{âˆ˜}?**

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

**(vi) Each angle is equal to 60 ^{âˆ˜}**

**Give reasons in support of your answer in each case.**

**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^{âˆ˜}. If there are two obtuse angles, then their sum will be more than 180^{âˆ˜}, 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^{âˆ˜}, then the sum of all three angles will be less than 180^{âˆ˜}, which is not possible in case of a triangle.

Proof:

Let the three angles of the triangle be âˆ A, âˆ B and âˆ C.

As per the given information,

âˆ A < 60^{âˆ˜} â€¦ (i)

âˆ B< 60^{âˆ˜} â€¦(ii)

âˆ C< 60^{âˆ˜} â€¦ (iii)

On adding (i), (ii) and (iii), we get :

âˆ A + âˆ B + âˆ C < 60^{âˆ˜}+ 60^{âˆ˜}+ 60^{âˆ˜}

âˆ A + âˆ B + âˆ C < 180^{âˆ˜}

We can see that the sum of all three angles is less than 180^{âˆ˜}, which is not possible for a triangle.

Hence, we can say that it is not possible for each angle of a triangle to be less than 60^{âˆ˜}.

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

Proof:

Let the three angles of the triangle be âˆ A, âˆ B and âˆ C. As per the given information,

âˆ A > 60^{âˆ˜} â€¦ (i)

âˆ B>60^{âˆ˜} â€¦(ii)

âˆ C> 60^{âˆ˜} â€¦ (iii)

On adding (i), (ii) and (iii), we get:

âˆ A + âˆ B + âˆ C > 60^{âˆ˜}+ 60^{âˆ˜}+ 60^{âˆ˜}

âˆ A + âˆ B + âˆ C > 180^{âˆ˜}

We can see that the sum of all three angles of the given triangle are greater than 180^{âˆ˜}, which is not possible for a triangle.

Hence, we can say that it is not possible for each angle of a triangle to be greater than 60^{âˆ˜}.

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

Proof:

Let the three angles of the triangle be âˆ A, âˆ B and âˆ C. As per the given information,

âˆ A = 60^{âˆ˜} â€¦ (i)

âˆ B = 60^{âˆ˜} â€¦(ii)

âˆ C = 60^{âˆ˜} â€¦ (iii)

On adding (i), (ii) and (iii), we get:

âˆ A + âˆ B + âˆ C = 60^{âˆ˜}+ 60^{âˆ˜}+ 60^{âˆ˜}

âˆ A + âˆ B + âˆ C = 180^{âˆ˜}

We can see that the sum of all three angles of the given triangle is equal to 180^{âˆ˜}, which is possible in case of a triangle. Hence, we can say that it is possible for each angle of a triangle to be equal to 60^{âˆ˜}.

**Q20. In Î”ABC, âˆ A = 100 ^{âˆ˜}, AD bisects âˆ A and AD perpendicular BC. Find âˆ B**

**Solution.**

Consider Î”ABD

(AD bisects âˆ A)

âˆ BAD = 50^{âˆ˜}

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

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

Thus,

âˆ ABD+âˆ BAD+âˆ ADB = 180^{âˆ˜} (Sum of angles of Î”ABD)

Or,

âˆ ABD + 50^{âˆ˜} + 90^{âˆ˜} = 180^{âˆ˜}

âˆ ABD = 180^{âˆ˜}â€“ 140^{âˆ˜}

âˆ ABD = 40^{âˆ˜}

**Q21. In Î”ABC, âˆ A = 50 ^{âˆ˜}, âˆ B=100^{âˆ˜} and bisector of âˆ C 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^{âˆ˜}.

Therefore, for the given Î”ABC, we can say that :

âˆ A + âˆ B + âˆ C = 180^{âˆ˜} (Sum of angles of Î”ABC)

50âˆ˜ + 70âˆ˜ + âˆ C = 180^{âˆ˜}

âˆ C = 180^{âˆ˜} â€“120^{âˆ˜}

âˆ C = 60^{âˆ˜}

(CD bisects âˆ C and meets AB in D.)

Using the same logic for the given Î”ACD, we can say that :

âˆ DAC + âˆ ACD + âˆ ADC = 180^{âˆ˜}

50^{âˆ˜} +30^{âˆ˜} +âˆ ADC = 180^{âˆ˜}

âˆ ADC = 180^{âˆ˜} â€“ 80^{âˆ˜}

âˆ ADC = 100^{âˆ˜}

If we use the same logic for the given Î”BCD, we can say that

âˆ DBC +âˆ BCD +âˆ BDC = 180^{âˆ˜}

70^{âˆ˜} + 30^{âˆ˜} + âˆ BDC = 180^{âˆ˜}

âˆ BDC = 180^{âˆ˜} â€“ 100^{âˆ˜}

âˆ BDC = 80^{âˆ˜}

Thus,

For Î”ADC: âˆ A = 50^{âˆ˜}, âˆ D = 100^{âˆ˜} âˆ C = 30^{âˆ˜}

Î”BDC: âˆ B = 70^{âˆ˜}, âˆ D = 80^{âˆ˜} âˆ C = 30^{âˆ˜}

**Q22. In Î”ABC, âˆ A=60**^{âˆ˜}**, âˆ B=80**^{âˆ˜}**, and the bisectors of âˆ B and âˆ C, meet at O. Find**

**(i) âˆ C**

**(ii) âˆ BOC**

**Solution.**

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

Hence, for Î”ABC, we can say that :

âˆ A + âˆ B + âˆ C = 180^{âˆ˜} (Sum of angles of Î”ABC)

60^{âˆ˜}+ 80âˆ˜ + âˆ C = 180^{âˆ˜}.

âˆ C = 180^{âˆ˜} â€“ 140^{âˆ˜}

âˆ C = 140^{âˆ˜}.

For Î”OBC,

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

âˆ OCB + âˆ OBC + âˆ BOC = 180^{âˆ˜} (Sum of angles of Î”OBC)

20^{âˆ˜}+ 40^{âˆ˜} +âˆ BOC = 180^{âˆ˜}

âˆ BOC = 180^{âˆ˜} â€“ 60^{âˆ˜}

âˆ BOC = 120^{âˆ˜}

**Q23. The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.**

**Solution.**

We know that the sum of all three angles of a triangle is 180âˆ˜.

Hence, for Î”ABC , we can say that :

âˆ A + âˆ B + âˆ C = 180^{âˆ˜}

âˆ A + 90^{âˆ˜} + âˆ C = 180^{âˆ˜}

âˆ A + âˆ C = 180^{âˆ˜} â€“ 90^{âˆ˜}

âˆ A + âˆ C = 90^{âˆ˜}

For Î”OAC :

(OA bisects LA)

(OC bisects LC)

On applying the above logic to Î”OAC, we get :

âˆ AOC+ âˆ OAC+âˆ OCA = 180^{âˆ˜} (Sum of angles of Î”AOC)

**Q24. In Î”ABC, âˆ A = 50 ^{âˆ˜} and BC is produced to a point D. The bisectors of âˆ ABC 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^{âˆ˜}.

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

âˆ ABC+ âˆ BCA + âˆ CAB = 180^{âˆ˜} (Sum of all angles of Î”ABC)

âˆ A + âˆ B + âˆ BCA = 180^{âˆ˜}

âˆ BCA = 180Â°- (âˆ A + âˆ B)

If we use the same logic for Î”EBC , we can say that :

âˆ EBC+ âˆ ECB + âˆ BEC = 180^{âˆ˜} (Sum of all angles of Î”EBC)

**Q25. In Î”ABC, âˆ B=60 ^{âˆ˜}, âˆ C=40^{âˆ˜}, AL perpendicular BC and AD bisects âˆ A 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 180^{âˆ˜}

Therefore, for Î”ABC, we can say that :

âˆ A + âˆ B + âˆ C = 180^{âˆ˜}

Or,

âˆ A + 60^{âˆ˜} + 40âˆ˜ = 180^{âˆ˜}

âˆ A = 80^{âˆ˜}

If we use the above logic on Î”ADC, we can say that :

âˆ ADC + âˆ DCA + âˆ DAC = 180^{âˆ˜} (Sum of all the angles of Î”ADC)

âˆ ADC + 40^{âˆ˜} + 40^{âˆ˜} = 180^{âˆ˜}

âˆ ADC = 180^{âˆ˜} + 80^{âˆ˜}

âˆ ADC = âˆ ALD + âˆ LAD(Exterior angle is equal to the sum of two Interior opposite angles.)

100^{âˆ˜} = 90^{âˆ˜}+ âˆ LAD (AL perpendicular to BC)

âˆ LAD = 90^{âˆ˜}

**Q26. Line segments AB and CD intersect at O such that AC perpendicular DB. It âˆ CAB = 35 ^{âˆ˜} and âˆ CDB = 55^{âˆ˜} **

**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^{âˆ˜}

We also know that the sum of all three angles of a triangle is 180âˆ˜.

Hence, for Î”OBD, we can say that :

âˆ DBO + âˆ ODB+ âˆ BOD = 180^{âˆ˜}

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

âˆ BOD =180^{âˆ˜} âˆ’ 90^{âˆ˜}

âˆ BOD = 90^{âˆ˜}

**Q27. In Fig. 22, Î”ABC is right angled at A, Q and R are points on line BC and P is a point such that QP perpendicular to AC and RP perpendicular to 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^{âˆ˜}.

Hence, for Î”ABC, we can say that :

âˆ ABC + âˆ ACB + âˆ BAC = 180^{âˆ˜}

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

âˆ ABC + âˆ ACB = 90^{âˆ˜}

Using the same logic for Î”PQR, we can say that :

âˆ PQR + âˆ PRQ + âˆ QPR = 180^{âˆ˜}

âˆ ABC + âˆ ACB + âˆ QPR =180âˆ˜ (âˆ ABC = âˆ PRQ and âˆ QCA =âˆ CQP)

Or,

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

âˆ QPR = 90^{âˆ˜}