# Questions on angle sum property of triangle – grade 09

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.

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.