Given below are collection of questions related to lines and angles.
All questions are to the standard of grade 09 math.
Question 01
Consider the below figure in which ∠BOD = 40 degree and ∠ AOC + ∠ BOE = 70 degree. Find measure of ∠BOE & ∠COE.
Solution
It’s given that ∠BOD = 40 degree.
Note that ∠BOD and ∠COA are vertically opposite angle.
∠ BOD = ∠COA = 40 degree
It’s given in question that ∠ AOC + ∠ BOE = 70 degree.
Putting the value of ∠AOC we get;
40 + ∠ BOE = 70 degree
∠ BOE = 70 – 40
∠ BOE = 30 degree
We know that line AB is a straight line and adjacent angle in a straight line measures 180 degree.
∠ AOC + ∠ COE + ∠ BOE = 180
40 + ∠COE + 30 = 180
∠COE = 180 – 70
∠COE = 110 degree
Hence, measure of ∠COE is 110 degree.
Question 02
In the below figure ∠PQR = ∠PRQ. Prove that ∠PQS = ∠PRT.
Solution
We know that in straight line adjacent angle measures 180 degree.
∠ SQP + ∠PQR = 180 degree
Similarly QRT is also a straight line. So we can write;
∠QRP + ∠TRP = 180 degree
Combining the above two expression.
∠ SQP + ∠PQR = ∠QRP + ∠TRP
We know that;
∠PQR = ∠PRQ ( given )
Rewriting the above expression;
∠ SQP + ∠PQR = ∠PQR + ∠TRP
Cancelling ∠PQR on both sides.
∠ SQP = ∠TRP
Hence Proved.
Question 03
Consider the below figure in which line PQ is a straight line. The line OR intersect PQ at right angle. Prove that \mathtt{\angle ROS=\frac{1}{2}( \angle QOS-\angle POS)}
Solution
Taking the right side of the expression;
= 1/2 (∠QOS – ∠POS)
∠QOS can be written as follows;
∠QOS = ∠QOR + ∠ROS
Similarly ∠POS is written as follows;
∠POS = ∠POR – ∠SOR
Putting both the values in main expression.
= 1/2 ( ∠QOR + ∠ROS – ( ∠POR – ∠SOR) )
= 1/2 ( ∠QOR + ∠ROS – ∠POR + ∠SOR)
= 1/2 ( 90 + ∠ROS – 90 + ∠SOR)
= 1/2 ( 2∠ROS )
= ∠ROS
Hence proved the given expression.
Question 04
Consider the below figure in which ∠XYZ = 64 degree. Note that ray YQ bisect ∠ZYP. Find the measure of ∠XYQ and ∠QYP.
Solution
Note that XYP is a straight line.
We know that in straight line adjacent angle measures 180 degree.
∠XYZ + ∠ZYP = 180 degree
64 + ∠ZYP = 180
∠ZYP = 180 – 64
∠ZYP = 116 degree
We know that YQ bisect ∠PYZ.
∠PYQ = ∠QYZ
∠ZYP can be written as;
∠PYQ + ∠QYZ = 116
2 ∠PYQ = 116
∠PYQ = 58 degree
Now we have got all the required data, let’s calculate the value of ∠XYQ.
∠XYQ + ∠QYP = 180
∠XYQ + 58 = 180
∠XYQ = 180 – 58
∠XYQ = 122 degree
Hence, we got all the required values.
Question 05
In the below figure, line PQ & RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7, find the measure of all the angles.
Solution
Note that PQ is a straight line and we know that adjacent angle in straight line measures 180 degree.
∠POR + ∠ROQ = 180 degree
It’s given that, ∠POR : ∠ROQ = 5 : 7
Let ∠POR = 5x and ∠ROQ = 7x
Putting this value in above equation;
5x + 7x = 180 degree
12x = 180
x = 180 / 12
x = 15
Putting the value of x to find angle measurement.
∠POR = 5x
∠POR = 12 (5) = 60 degree
Similarly ∠ ROQ = 7x
∠ ROQ = 7(12) = 72 degree.
∠POS = ∠ ROQ = 72 (vertically opposite angle)
∠ POR = ∠ SOQ = 60 degree
Hence, we got the value of all angles.