In this chapter we will solve questions related to angle sum property of triangle.

All the questions are to the standard of grade 09.

**Question 01**

Given below is the image of triangle PQR in which line QT perpendicular to side PR. Find the measure of angle x and y.

**Solution**

Consider triangle QTR.

We know that sum of internal angle of triangle measures 180 degree.

∠TQR + ∠QTR +∠x = 180

40 + 90 + ∠x = 180

130 + ∠x = 180

∠x = 180 – 130

∠x = 50 degree.

Using the exterior angle theorem in triangle PSR;

∠y = ∠SPR + ∠x

∠y = 30 + 50

∠y = 80 degree

Hence, we calculated the value of ∠x and ∠y.

**Question 02**

Given below is the triangle ABC with interior angles ∠x, ∠y and ∠z. Here lines OB & OC bisect angles ∠EBC & ∠DCB respectively. Prove that ∠BOC = y/2 + z/2

**Solution**

We know that in straight line, sum of alternate angle measures 180 degree.

Since AE is a straight line, we can write;

∠ y + ∠EBC = 180

∠EBC = 180 – y

Similarly AD is a straight line, we can write;

∠z + ∠BCD = 180

∠BCD = 180 – ∠z

OB is angle bisector of ∠EBC.

∠CBO = 1/2 ( ∠CBE )

∠CBO = 1/2 ( 180 – y )

OC is angle bisector of ∠BCD

∠OCB = 1/2 ( ∠ BCD )

∠ OCB = 1/2 ( 180 – ∠z )

Now consider triangle OBC.

We know that sum of all angles of triangle measure 180 degree.

∠BOC + ∠OBC + ∠BCO = 180

∠BOC + 1/2 ( 180 – y ) + 1/2 ( 180 – ∠z ) = 180

∠BOC + 90 – y/2 + 90 – z/2 = 180

∠BOC + 180 – y/2 – z/2 = 180

∠BOC = y/2 + z/2

Hence Proved.

**Question 03**

Given below is triangle ACD in which all the sides are extended outside. Find all the internal angles of triangle.

**Solution**

BD is a straight line and sum of all alternate angles in straight line measures 180 degree.

∠BCA + ∠ DCA = 180

∠BCA + 105 = 180

∠BCA = 180 – 105

∠BCA = 75 degree

∠EAF & ∠BAC are vertically opposite angles.

We know that vertically opposite angles are equal in measure. So we can write;

∠EAF = ∠BAC = 45

We know that sum of all angles of triangle measure 180 degree. So we can write;

∠BCA + ∠BAC + ∠ABC = 180

75 + 45 + ∠ABC = 180

∠ ABC = 180 – 120

∠ ABC = 60 degree

Hence, we calculated all the internal angles of triangle.

**Question 04**

Study the below image and find the value of x.

**Solution**

CE is a straight line and we know that in straight line sum of alternate angle measure 180 degree.

∠BAC + ∠BAE = 180

∠BAC + 120 = 180

∠BAC = 180 – 120

∠BAC = 60 degree

Now using exterior angle theorem we can write;

∠ ACD = ∠ ABC + ∠ BAC

112 = x + 60

x = 112 – 60

x = 52 degree

Hence, angle x measures 52 degree.

**Question 05**

Study the below image and find the measure of angle x.

**Solution**

BD is a straight line.

We know that adjacent angles in straight line measures 180 degree.

∠ACB + ∠ACD = 180

∠ACB + 110 = 180

∠ACB = 180 – 110

∠ACB = 70 degree

Now using external angle theorem we can write;

∠ABE = ∠x + ∠ACB

120 = ∠x + 70

∠x = 120 – 70

∠x = 50 degree

Hence, angle x measures 50 degrees.

**Question 06**

In the below image, line AB & CD are parallel to each other. Find the measure of ∠x

**Solution**

Line AB & CD are parallel to each other intersected by transversal AD.

In this case we know that alternate interior angles are equal.

∠ADC = ∠BAD

∠ADC = 52

Now consider triangle CED.

We know that in triangle sum of all internal angle measure 180 degree.

40 + x + ∠ADC = 180

40 + x + 52 = 180

x = 180 – 92

x = 88 degree

Hence, x measures 88 degrees.

**Question 07**

In the below triangle ADC, line AB divides ∠DAC in ratio 1 : 3 and length of side AB = DB. Find the measure of ∠x.

**Solution**

Since AB divide ∠DAC in ratio 1: 3, we can do following assumption;

∠DAB = y and ∠BAC = 3y

We know that angle opposite to equal sides are equal.

Since AB = DB;

then, ∠BDA= ∠DAB

∠BDA = y

DA is a straight line, so we can write;

∠DAC + 108 = 180

(y + 3y) + 108 = 180

4y = 180 – 108

4y = 72

y = 18

Now apply angle sum property in triangle ADC.

x + ∠DAC + ∠ADC = 180

x + 4y + y = 180

x + 5y = 180

x + 5(18) = 180

x = 180 – 90

x = 90

Hence value of angle x is 90 degrees.