**Question 1.In the adjoining figure, ∠ACD is an exterior angle of ∆ABC. ∠B = 40°, ∠A = 70°. Find the measure of ∠ACD.**

Solution:

∠A = 70° , ∠B = 40° [Given]

∠ACD is an exterior angle of ∆ABC. [Given]

∴ ∠ACD = ∠A + ∠B

= 70° + 40°

∴ ∠ACD = 110°

**Question 2.In ∆PQR, ∠P = 70°, ∠Q = 65°, then find ∠R.Solution:**∠P = 70°, ∠Q = 65° [Given]

In ∆PQR,

∠P + ∠Q + ∠R = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ 70° + 65° + ∠R = 180°

∴ ∠R = 180° – 70° – 65°

∴ ∠R = 45°

**Question 3.The measures of angles of a triangle are x°, (x – 20)°, (x – 40)°. Find the measure of each angle.Solution:**The measures of the angles of a triangle are x°, (x – 20)°, (x – 40)°. [Given]

∴ x°+ (x – 20)° + (x – 40)° = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ 3x – 60 = 180

∴ 3x = 180 + 60

∴ 3x = 240

∴ x = 240

∴ x = 80°

∴ The measures of the remaining angles are

x – 20° = 80° – 20° = 60°,

x – 40° = 80° – 40° = 40°

∴ The measures of the angles of the triangle are 80°, 60° and 40°.

**Question 4.The measure of one of the angles of a triangle is twice the measure of its smallest angle and the measure of the other is thrice the measure of the smallest angle. Find the measures of the three angles.Solution:**

Let the measure of the smallest angle be x°.

One of the angles is twice the measure of the smallest angle.

∴ Measure of that angle = 2x°

Another angle is thrice the measure of the smallest angle.

∴ Measure of that angle = 3x°

∴ The measures of the remaining two angles are 2x° and 3x°.

Now, x° + 2x° + 3x° = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ 6x = 180

∴ x = 180

∴ x° = 30°

The measures of the remaining angles are 2x° = 2 x 30° = 60°

3x° = 3 x 30° = 90°

The measures of the three angles of the triangle are 30°, 60° and 90°.

**Question 5.In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.Solution:**

i. ∠NET = 100° and ∠EMR = 140°

∠EMN + ∠EMR = 180°

∴ z +140° =180°

∴ z = 180° -140°

∴ z = 40°

ii. Also, ∠NET + ∠NEM = 180° [Angles in a linear pair]

∴ 100° + y = 180°

∴ y = 180° – 100°

∴ y = 80°

iii. In ∆ENM,

∴ ∠ENM + ∠NEM + ∠EMN = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ x +80°+ 40°= 180°

∴ x = 180° – 80° – 40°

∴ x = 60°

∴ x = 60°, = 80°, z = 40°

**Question 6.In the adjoining figure, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.Solution:**

i. ∠B AD = 70°, ∠DER = 40° [Given]

line AB || line DE and seg AD is their transversal.

∴ ∠EDA = ∠BAD [Alternate Angles]

∴ ∠EDA = 70° ….(i)

In ∆DRE,

∠EDR + ∠DER + ∠DRE = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ 70°+ 40° +∠DRE = 180° [From (i) and D – R – A]

∴ ∠DRE = 180° -70° -40°

∴ ∠DRE = 70°

ii. ∠DRE + ∠ARE = 180° [Angles in a linear pair]

∴ 70° + ∠ARE = 180°

∴ ∠ARE = 180°-70°

∴ ∠ARE =110°

∴ ∠DRE = 70°, ∠ARE = 110°

**Question 7.In ∆ABC, bisectors of ∠A and ∠B intersect at point O. If ∠C = 70°, find the measure of ∠AOB.Solution:**

∠OAB = ∠OAC = – ∠BAC ….(i) [Seg AO bisects ∠BAC]

∠OBA = ∠OBC = – ∠ABC …..(ii) [Seg RO bisects ∠ABC]

In AABC,

∠BAC + ∠ABC + ∠ACB = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ ∠BAC + ∠ABC + 70° = 180°

∴ ∠BAC + ∠ABC = 180°- 70°

∴ ∠BAC + ∠ABC = 110°

∴ (∠BAC) + (∠ABC) = x 110° [MuItiplying both sides by ]

∴ ∠OAB + ∠OBA = 55° ….(iii) [From (i) and (ii)]

In AOAB,

∠OAB + ∠OBA + ∠AOB = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ 55° + ∠AOB = 180° [From (iii)]

∴ ∠AOB = 180°- 55°

∴ ∠AOB = 125°

**Question 8.In the adjoining figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. Prove that m ∠PTQ = 90°.Given: line AB || line CD and line PQ is the transversal.ray PT and ray QT are the bisectors of ∠BPQ and ∠PQD respectively.To prove: m∠PTQ = 90°Solution:**

**Question 9.Using the information in the adjoining figure, find the measures of ∠a, ∠b and ∠c.Solution:**

i. ∠c + 100° = 180° [Angles in a linear pair]

∴ ∠c = 180° – 100°

∴ ∠c = 80°

ii. ∠b = 70° [Vertically opposite angles]

iii. ∠a + ∠b +∠c = 180° [Sum of the measures of the angles of a triangle is 180°]

∠a + 70° + 80° = 1800

∴ ∠a = 180° – 70° – 80°

∴ ∠a = 30°

∴ ∠a = 30°, ∠b = 70°,∠ c = 80°

**Question 10.In the adjoining figure, line DE || line GF, ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively. Prove that,ii. EF = FGSolution:**

i. ∠DEG = ∠FEG = x° ….(i) [Ray EG bisects ∠DEF]

∠GFD = ∠GFM = y° …..(ii) [Ray FG bisects ∠DFM]

line DE || line GF and DF is their transversal. [Given]

∴ ∠EDF = ∠GFD [Alternate angles]

∴ ∠EDF = y° ….(iii) [From (ii)]

line DE || line GF and EM is their transversal. [Given]

∴ ∠DEF = ∠GFM [Corresponding angles]

∴ ∠DEG + ∠FEG = ∠GFM [Angle addition property]

∴ x°+ x° = y° [From (i) and (ii)]

∴ 2x° = y°

ii. line DE || line GF and GE is their transversal. [Given]

∴ ∠DEG = ∠FGE …(iv) [Alternate angles]

∴ ∠FEG = ∠FGE ….(v) [From (i) and (iv)]

∴ In ∆FEG,

∠FEG = ∠FGE [From (v)]

∴ EF = FG [Converse of isosceles triangle theorem]

**Intext Questions and Activities**

**Question 1. Can you give an alternative proof of the above theorem by drawing a line through point R and parallel to seg PQ in the above figure? (Textbook pg. no. 25)Solution:**

Yes.

Construction: Draw line RM parallel to seg PQ through a point R.

Proof:

seg PQ || line RM and seg PR is their transversal. [Construction]

∴ ∠PRM = ∠QPR ……..(i) [Alternate angles]

seg PQ || line RM and seg QR is their transversal. [Construction]

∴ ∠SRM = ∠PQR ……..(ii) [Corresponding angles]

∴ ∠PRM + ∠SRM = ∠QPR + ∠PQR [Adding (i) and (ii)]

∴ ∠PRS = ∠PQR + ∠QPR [Angle addition property]

**Question 2. Observe the given figure and find the measures of ∠PRS and ∠RTS. (Textbook pg. no.25)Solution:**

∠PRS is an exterior angle of ∆PQR.

So from the theorem of remote interior angles,

∠PRS = ∠PQR + ∠QPR

= 40° + 30°

∴ ∠PRS = 70°

∴ ∠TRS=70° …[P – T – R]

In ∆RTS,

∠TRS + ∠RTS + ∠TSR = 180° …[Sum of the measures of the angles of a triangle is 180°]

∴ 70° + ∠RTS + 20° = 180°

∴ ∠RTS + 90° = 180°

∴ ∠RTS = 180°

∴ ∠RTS = 90°