The document Hots Questions- Lines and Angles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.

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**Question 1. If a ray CD stands on a line AB, then prove that ∠ACD + ∠BCD = 180º. Solution: **Let us draw CE ⊥ AB.

∴ ∠ACE = 90º and

∠BCE = 90º

Now, ∠ACD = ∠ACE + ∠ECD = 90º + ∠ECD …(1)

∠BCD = ∠BCE - ∠ECD = 90º - ∠ECD …(2)

Adding (1) and (2), we have: ∠ACD + ∠BCD

= [90º + ∠ECD] + [90º - (∠ECD)]

= 90º + 90º + ∠ECD - ∠ECD

= 180º**Note: **Above example may be stated as: “The sum of the angles of linear pair is 180°.”

OR

“Sum of all the angles formed on the same side of a line at a given point on the line is 180°.”

**Question 2. Two lines AB and CD intersect at a point O. Prove that: ∠AOD = ∠ BOC.**

**Solution:** Since, OA stands on the given line CD.

∴ ∠AOC + ∠AOD = 180º ...(1)

Again, OD stands on the given line AB.

∴ ∠AOD + ∠BOD = 180º ...(2)

From (1) and (2), we have: ∠AOC + ∠AOD = ∠AOD + ∠BOD

or ∠AOC = ∠BOD

**Question 3. In the following figure, AOB is a straight line. Find ∠AOC and ∠BOD.**

**Solution: **Since AOB is a straight line.

∴ The sum of all the angles on the same side of AOB at a point on it is 180º.

∠AOC + ∠COD + ∠DOB = 180º

∴ x + 60º + (2x - 15)º = 180º

or 3x + 60º - 15º = 180º

or 3x = 180º - 60º + 15º = 135º

or x = (135^{0}/3) = 45º

Now 2x - 15 = 2(45) - 15 = 75º

or ∠AOC = 45º and ∠BOD = 75º

**Question 4. In the following figure, p : q : r = 2 : 3 : 4. If AOB is a straight line, then find the values of p, q and r.**

**Solution: **Let p = 2x, q = 3x and r = 4x [∵ p : q : r = 2 : 3 : 4]

∵ AOB is a straight line.

∴ ∠AOC + ∠COD + ∠DOB = 180º

or 2x + 3x + 4x = 180º

or 9x = 180º

or x = (180^{0}/9) = 20º

So, 2x = 2 x 20º = 40º

∴ 3x = 3 × 20° = 60°

4x = 4 × 20° = 80°

Thus, p = 40°, q = 60°, r = 80°

**Question 5. In the figure, AB || CD. GE and HF are the bisectors of ∠AEF and ∠EFD respectively. Show that GE || FH**

OR**If two parallel lines are intersected by a transversal, then show that the bisectors of a pair of alternate interior angles are parallel. Solution: **∵ AB || CD and EF is a transversal

∴ ∠AEF = ∠EFD [Interior alternate angles]

⇒ (1/2) ∠AEF =(1/2)∠EFD

⇒ ∠GEF = ∠HFE [∵ GF and HF are angle bisectors (given)]

But they are the angles of a pair of interior alternate angles.

∴ GE || FH.

**Question 6. In the figure, AB || CD. EG and FH are bisectors of ∠PEB and ∠EFD respectively. Show that EG || FH.**

**OR**

**If two parallel lines are intersected by a transversal then prove that the bisectors of any pair of corresponding angles are parallel. Solution**: ∵ AB || CD and EF is a transversal.

∴ ∠BEP = ∠EFD [corresponding angles]

⇒ (1/2) ∠BEP = (1/2)∠EFD

⇒ ∠PEG = ∠EFH [∵ EG and FH are the angle bisectors of ∠BEP and ∠EFD respectively]

But they form pair of corresponding angles.

∴ EG || FH.

**Question 7. If the arms of an angle are respectively parallel to the arms of another angle, then show that the two angles are either equal or supplementary. Solution:** Let we have two angles ∠ABC and ∠DEF such that BA || ED and BC || EF in the same sense or in the opposite sense.

∴ We can have the following three cases:

∵ BA || ED and BC is a transversal,

∴ ∠1= ∠2 [Corresponding angles] …(1)

Again, BC || EF and DE is a transversal,

∴ ∠3= ∠2 [Corresponding angles] …(2)

From (1) and (2), we have ∠1= ∠3

i.e. ∠ABC = ∠DEF.

**Case II:** [Both pairs of arms are parallel in opposite sense.]

∵ BA || ED and BC is a transversal,

∴ ∠1= ∠2 [Corresponding angles] …(1)

Again BC || EF and ED is a transversal, [Alternate interior angles]

∴ ∠3= ∠2 …(2)

From (1) and (2), we have ∠1= ∠3

i.e. ∠ABC = ∠DEF**Case III:** [One pair of arms are parallel in the same sense and other pair parallel in opposite sense.]

∵ BA || ED and BC is a transversal.

∴ ∠1= ∠2 [Interior alternate angles] …(1)

Again BC || EF and DE is a transversal.

∴ ∠3 + ∠2 = 180º [Sum of interior opposite angles]

⇒ ∠3 + ∠1 = 180º [From (1)]

i.e. ∠DEF + ∠ABC = 180º

Hence, ∠ABC and ∠DEF are supplementary.

**Question 8. Prove that the sum of the angles of a triangle is 180º. Solution: **Let us consider ΔABC and through A draw DE || BC.

∵ BC || DE and AB is a transversal.

∴ ∠4= ∠1 [Interior alternate angles] …(1)

Again, BC || DF and AC is a transversal.

∴ ∠5= ∠2 [Interior alternate angles] …(2)

Adding (1) and (2), we have ∠4 + ∠5= ∠1 + ∠2

Adding ∠3 on both sides, we have ∠4 + ∠5 + ∠3 = ∠1 + ∠2 + ∠3

Since, ΔAF is a straight line,

∴ ∠4 + ∠3 + ∠5 = 180º ∠1 + ∠2 + ∠3 = 180°

i.e. ∠ABC + ∠BCA + ∠BAC = 180º

∴ The sum of angles of a triangle is 180º.

**Question 9. Prove the following statement: “If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.” Solution:** Let us consider a triangle ABC such that its side BC is produced to D, forming exterior ∠ACD.

∵ Sum of the angles of a triangle = 180º

∴ ∠1 + ∠2 + ∠3 = 180º …(1)

Since, BCD is a straight line.

∴ ∠2 + ∠4 = 180º …(2)

From (1) and (2), we have ∠2 + ∠4= ∠1 + ∠2 + ∠3

⇒ ∠4= ∠1 + ∠3 i.e.

Exterior ∠ACD = [Sum of the interior opposite angles]

**REMEMBER**

An exterior angle of a triangle is greater than either of the interior opposite angles.

**Question 10. The angles of a triangle are in the ratio 2 : 3 : 5. Find the measure of each angle of the triangle. Solution: **Let the angles of the given triangle measure (2x)º, (3x)º and (5x)º.

∵ Sum of the angles of a triangle is 180º.

∴ (2x)º + (3x)º + (5x)º = 180º ⇒ 10x = 180º

or x = (180

∴ 2x = 2 x 18 = 36º

3x = 3 x 18 = 54º

5x = 5 x 18 = 90º

∴ The measures of the angles of the given triangle are 36º, 54º and 90º.

**Question 11. In a triangle, the bisectors of ∠B and ∠C intersect each other at a point O. Prove that ∠BOC = 90º + 1 2 ∠A. Solution: **In a ΔABC, we have: ∠A + ∠B + ∠C = 180º [By angle sum property]

Again, in ΔOBC, we have ∠1 + ∠2 + ∠BOC = 180º [By angle sum property]

⇒ (∠1 + ∠2) + ∠BOC = 180º

**Question 12. In a ΔABC, if ∠A + ∠B = 150º and ∠B + ∠C = 100. Find the measure of each angle of the triangle. Solution: **We have ∠ A + ∠ B = 150º …(1)

∠B + ∠C = 100º …(2)

Adding (1) and (2), we get

∴ ∠A + 2∠B + ∠C = 150 + 100º = 250º

⇒ (∠A + ∠B + ∠C) + ∠B = 250º

⇒ 180º + ∠B = 250º [∵ ∠ A + ∠B + ∠ C = 180º]

∴ ∠B = 250º - 180º = 70º.

Now, ∠A + ∠B = 150º

⇒ ∠A = 150º - ∠B = 150º - 70º = 80º

Also ∠B + ∠C = 100º

⇒ ∠C = 100 - ∠B = 100 - 70º = 30º

Thus, ∠A = 80º, ∠B = 70º and ∠C = 30º.

**Question 13. In a triangle, if ∠A = 2∠B = 6∠C, find the measures of ∠A, ∠B and ∠C. Solution:** Let ∠A = 2∠B = 6∠C = x

∴ ∠A= x 2∠B= x

⇒ ∠B = x 2

6∠C = x

⇒ ∠C = x 6

We know that ∠A + ∠B + ∠C = 180º (using angle sum property)

or 6x^{0} + 3x^{o }+ x^{o} = 6 x 180^{o}

^{⇒ }10x^{o} = 6 x 180^{o}

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