Class 9 Maths Chapter 6 HOTS Questions - Lines and Angles

Q1. 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°.”

 Q2. Two lines AB and CD intersect each other at a point O such that ∠AOC : ∠AOD = 5 : 7. Find all the angles.

Solution: Let ∠AOC = 5k and ∠AOD = 7k, where k is some constant.
Here, ∠AOC and ∠AOD form a linear pair.
∴ ∠AOC + ∠AOD = 180º
⇒ 5k + 7k = 180º
⇒ 12k = 180º
⇒ k = 15º
∴ ∠AOC = 5k = 5 × 15º = 75º
∠AOD = 7k = 7 × 15º = 105º
Now, ∠BOD = ∠AOC = 75º (Vertically opposite angles)
∠BOC = ∠AOD = 105º (Vertically opposite angles)

Q3. 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°/3) = 45º
Now 2x - 15 = 2(45) - 15 = 75º
or ∠AOC = 45º and ∠BOD = 75º 

Q4. 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°/9) = 20º
So, 2x = 2 x 20º = 40º
∴ 3x = 3 × 20° = 60°
4x = 4 × 20° = 80°
Thus, p = 40°, q = 60°, r = 80°

Q5. In the figure, AB || CD. EG and FH are bisectors of ∠PEB and ∠EFD respectively.
 Show that EG || FH.

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.

Q6. 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.

Q7. 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.

Q8. 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

Q9. In a ΔABC, if ∠A + ∠B = 150º and ∠B + ∠C = 100º. 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.

Q10. 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  6xº + 3xº + xº = 6 x 180º 

^⇒ 10xº = 6 x 180º