Ex-15.2, (Part - 2), Properties Of Triangles, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics PDF Download

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

Solution.

(i) In ΔABC and ΔADE we have :

∠ADE = ∠ABC (corresponding angles)

x = 40^∘

∠AED = ∠ACB (corresponding angles)

y = 30^∘

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

x +y +z = 180^∘ (Angles of A ADE)

Which means : 40^∘ +30^∘  + z = 180^∘

z = 180^∘−70^∘

z = 110^∘

Therefore, we can conclude that the three angles of the given triangle are 40^∘, 30^∘  and 110^∘.

(ii) We can see that in ΔADC, ∠ADC  is equal to 90^∘.

(ΔADC  is a right triangle)

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

Which means : 45^∘+ 90^∘ +y = 180^∘ (Sum of the angles of ΔADC)

135^∘  + y = 180^∘

y = 180^∘ – 135^∘.

y = 45^∘.

We can also say that in ΔABC,  ∠ABC+∠ACB+∠BAC  is equal to 180^∘.

(Sum of the angles of A ABC)

40^∘ + y + (x + 45^∘) = 180^∘

40^∘ + 45^∘ + x + 45^∘ = 180^∘                                                 (y = 45^∘)

x = 180^∘ –130^∘

x = 50^∘

Therefore, we can say that the required angles are 45^∘ and 50^∘.

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

Therefore, for ΔABD:

∠ABD+∠ADB+∠BAD = 180^∘ (Sum of the angles of ΔABD)

50^∘ + x + 50^∘  = 180^∘

100^∘  +x = 180^∘

x = 180^∘ –100^∘

x = 80^∘

For ΔABC:

∠ABC+∠ACB+∠BAC = 180^∘ (Sum of the angles of ΔABC)

50^∘ + z + (50^∘ + 30^∘) = 180^∘

50^∘ + z + 50^∘ + 30^∘ = 180^∘

z = 180^∘ – 130^∘

z = 50^∘

Using the same argument for ΔADC:

∠ADC+∠ACD+∠DAC = 180^∘ (Sum of the angles of ΔADC)

y +z +30^∘ =180^∘

y + 50^∘ + 30^∘ =180^∘    (z = 50^∘)

y = 180^∘ – 80^∘

y = 100^∘

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

(iv) In ΔABC and ΔADE we have :

∠ADE = ∠ABC (Corresponding angles)

y = 50^∘

Also, ∠AED =∠ACB  (Corresponding angles)

z = 40^∘

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

Which means : x+50^∘ +40^∘ = 180^∘  (Angles of ΔADE)

x = 180^∘– 90^∘

x = 90^∘

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

Q16. If one angle of a triangle is 60∘ 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^∘. (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^∘.

60^∘ +x + 2x = 180^∘

3x =180^∘– 60^∘

3x = 120^∘

x = 120^∘/3

x = 40^∘

2x = 2 x 40

2x = 80^∘

Hence, we can conclude that the required angles are 40^∘ and 80^∘.

Q17. It one angle of a triangle is 100∘ 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^∘.

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^∘.

100^∘ + 2x + 3x = 180^∘

5x = 180^∘– 100^∘

5x = 80^∘

x = 80^∘/5

2x = 2 x 16

2x = 32^∘

3x = 3 x 16

3x = 48^∘

Thus, the required angles are 32^∘  and 48^∘.

Q18. In ΔABC, if 3∠A = 4∠B = 6∠C, calculate the angles.

Solution.

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

∠A = 2∠C—(i)

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

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

Therefore, we can say that:

∠A+∠B+∠C=180^∘ (Angles of ΔABC)—(iii)

On putting the values of ∠Aand∠B in equation (iii), we get :

From equation (i), we have:

 angleA = 2∠C = 2×40

 angleA = 80^∘

From equation (ii), we have:

angleB = 60^∘

 angleA = 80^∘,  angleB = 60^∘,  angleC = 40^∘

Therefore, the three angles of the given triangle are 80^∘, 60^∘, and 40^∘.

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^∘?

(v) Each angle is greater than 60^∘?

(vi) Each angle is equal to 60^∘

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

Proof:

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

As per the given information,

∠A < 60^∘  … (i)

∠B< 60^∘   …(ii)

∠C< 60^∘   … (iii)

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

∠A + ∠B + ∠C < 60^∘+ 60^∘+ 60^∘

∠A + ∠B + ∠C < 180^∘

We can see that the sum of all three angles is less than 180^∘, 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^∘.

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

Proof:

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

∠A > 60^∘  … (i)

∠B>60^∘   …(ii)

∠C> 60^∘   … (iii)

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

∠A + ∠B + ∠C > 60^∘+ 60^∘+ 60^∘

∠A + ∠B + ∠C > 180^∘

We can see that the sum of all three angles of the given triangle are greater than 180^∘, 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^∘.

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

Proof:

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

∠A = 60^∘  … (i)

∠B = 60^∘   …(ii)

∠C = 60^∘   … (iii)

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

∠A + ∠B + ∠C = 60^∘+ 60^∘+ 60^∘

∠A + ∠B + ∠C = 180^∘

We can see that the sum of all three angles of the given triangle is equal to 180^∘, 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^∘.

Q20. In ΔABC, ∠A = 100^∘, AD bisects ∠A and AD perpendicular BC. Find ∠B

Solution.

Consider ΔABD

      (AD bisects ∠A)

∠BAD = 50^∘

∠ADB = 90^∘          (AD perpendicular to BC)

We know that the sum of all three angles of a triangle is 180^∘.

Thus,

∠ABD+∠BAD+∠ADB = 180^∘  (Sum of angles of ΔABD)

Or,

∠ABD + 50^∘ + 90^∘ = 180^∘

∠ABD = 180^∘– 140^∘

∠ABD = 40^∘

Q21. In ΔABC,  ∠A = 50^∘, ∠B=100^∘  and bisector of ∠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^∘.

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

∠A + ∠B + ∠C = 180^∘ (Sum of angles of ΔABC)

50∘ + 70∘ + ∠C  = 180^∘

∠C = 180^∘ –120^∘

∠C = 60^∘

 (CD bisects ∠C  and meets AB in D.)

Using the same logic for the given ΔACD, we can say that :

∠DAC  + ∠ACD  + ∠ADC  = 180^∘

50^∘ +30^∘ +∠ADC = 180^∘

∠ADC = 180^∘ – 80^∘

∠ADC = 100^∘

If we use the same logic for the given ΔBCD, we can say that

∠DBC   +∠BCD   +∠BDC   = 180^∘

70^∘ + 30^∘ + ∠BDC   = 180^∘

∠BDC   = 180^∘ – 100^∘

∠BDC   = 80^∘

Thus,

For ΔADC: ∠A = 50^∘, ∠D = 100^∘   ∠C = 30^∘

ΔBDC: ∠B = 70^∘, ∠D = 80^∘   ∠C = 30^∘

Q22. In ΔABC, ∠A=60^∘, ∠B=80^∘,  and the bisectors of ∠B  and ∠C,  meet at O. Find

(i) ∠C

(ii) ∠BOC

Solution.

We know that the sum of all three angles of a triangle is 180^∘.

Hence, for ΔABC, we can say that :

∠A + ∠B + ∠C = 180^∘ (Sum of angles of ΔABC)

60^∘+ 80∘ + ∠C = 180^∘.

∠C = 180^∘ – 140^∘

∠C = 140^∘.

For ΔOBC,

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

∠OCB + ∠OBC + ∠BOC = 180^∘ (Sum of angles of ΔOBC)

20^∘+ 40^∘  +∠BOC  = 180^∘

∠BOC = 180^∘ – 60^∘

∠BOC = 120^∘

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 180∘.

Hence, for ΔABC , we can say that :

∠A + ∠B + ∠C = 180^∘

∠A  + 90^∘  + ∠C  = 180^∘

∠A  + ∠C  = 180^∘ – 90^∘

∠A  + ∠C = 90^∘

For ΔOAC   :

                  (OA bisects LA)

                   (OC bisects LC)

On applying the above logic to ΔOAC, we get :

∠AOC+ ∠OAC+∠OCA   = 180^∘    (Sum of angles of ΔAOC)

Q24. In ΔABC, ∠A = 50^∘ and BC is produced to a point D. The bisectors of ∠ABC  and ∠ACD meet at E. Find ∠E  .

Solution.

In the given triangle,

∠ACD = ∠A + ∠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^∘.

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

∠ABC+ ∠BCA + ∠CAB = 180^∘ (Sum of all angles of ΔABC)

∠A + ∠B + ∠BCA = 180^∘

∠BCA = 180°- (∠A + ∠B)

If we use the same logic for ΔEBC , we can say that :

∠EBC+ ∠ECB + ∠BEC = 180^∘ (Sum of all angles of ΔEBC)

Q25. In ΔABC, ∠B=60^∘, ∠C=40^∘, AL  perpendicular BC and AD bisects ∠A  such that L and D lie on side BC. Find ∠LAD 

Solution.

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

Therefore, for ΔABC, we can say that :

∠A + ∠B + ∠C = 180^∘

Or,

∠A + 60^∘ + 40∘ = 180^∘

∠A = 80^∘

If we use the above logic on ΔADC, we can say that :

∠ADC + ∠DCA + ∠DAC = 180^∘ (Sum of all the angles of ΔADC)

∠ADC + 40^∘ + 40^∘ = 180^∘

∠ADC = 180^∘ + 80^∘

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

100^∘ = 90^∘+ ∠LAD       (AL perpendicular to BC)

∠LAD = 90^∘

Q26. Line segments AB and CD intersect at O such that AC perpendicular DB. It ∠CAB = 35^∘  and ∠CDB = 55^∘ . Find ∠BOD.  

Solution.

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

∠CAB = ∠DBA (Alternate interior angles)

∠DBA = 35^∘

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

Hence, for ΔOBD, we can say that :

∠DBO + ∠ODB+ ∠BOD = 180^∘

35^∘ + 55^∘ + ∠BOD = 180^∘ (∠DBO =∠DBA and ∠ODB =∠CDB)

∠BOD =180^∘ − 90^∘

∠BOD = 90^∘

Q27. In Fig. 22, ΔABC   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 ∠P 

Solution.

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

∠QCA = ∠CQP (Alternate interior angles)

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

∠ABC = ∠PRQ (alternate interior angles).

We know that the sum of all three angles of a triangle is 180^∘.

Hence, for ΔABC, we can say that :

∠ABC + ∠ACB + ∠BAC = 180^∘

∠ABC + ∠ACB + 90^∘ = 180^∘ (Right angled at A)

∠ABC + ∠ACB = 90^∘

Using the same logic for ΔPQR, we can say that :

∠PQR + ∠PRQ + ∠QPR = 180^∘

∠ABC + ∠ACB + ∠QPR =180∘ (∠ABC = ∠PRQ and ∠QCA =∠CQP)

Or,

90^∘ + ∠QPR = 180^∘ (∠ABC+ ∠ACB = 90^∘)

∠QPR = 90^∘