Class 9 Maths: Lines and Angles – Extra Questions with Answers
Q. 1 If an angle measures 35 degrees, what is the measure of its complementary angle?
Check Solution
Ans: C
Solution: Complementary angles sum to 90 degrees.
Answer: 55 degrees
Q. 2 Two lines AB and CD intersect at point O. If $\angle AOC$ is 60 degrees, find the measures of $\angle AOD$, $\angle COB$, and $\angle BOD$.
Check Solution
Ans: A
Solution:
$\angle AOD = 180^\circ – \angle AOC = 180^\circ – 60^\circ = 120^\circ$
$\angle COB = \angle AOC = 60^\circ$ (Vertical angles)
$\angle BOD = \angle AOD = 120^\circ$ (Vertical angles)
Answer: $\angle AOD = 120^\circ$, $\angle COB = 60^\circ$, $\angle BOD = 120^\circ$
Q. 3 The angles of a triangle are in the ratio $1:2:3$. The smallest angle of the triangle is
Check Solution
Ans: B
Solution: Let the angles be $x, 2x, 3x$. The sum of angles in a triangle is $180^\circ$.
Therefore, $x+2x+3x=180^\circ$, which means $6x=180^\circ$.
So, $x=30^\circ$. The smallest angle is $x$.
Answer: 30
Q. 4 Find the measure of an angle which is $30^\circ$ more than its complement.
Check Solution
Ans: D
Solution:
Let the angle be $x$.
Its complement is $90^\circ – x$.
Given that $x = (90^\circ – x) + 30^\circ$.
$2x = 120^\circ$
$x = 60^\circ$
Answer: 60
Q. 5 The angles of a triangle are in the ratio 3 : 5 : 4. The largest angle of the triangle is ______ degrees.
Check Solution
Ans: C
Solution: The sum of the angles in a triangle is 180 degrees. Let the angles be 3x, 5x, and 4x.
3x + 5x + 4x = 180
12x = 180
x = 15
The angles are 3(15) = 45, 5(15) = 75, and 4(15) = 60. The largest angle is 75 degrees.
Answer: 75
Q. 6 Fill in the blank: The supplement of an angle measuring 115 degrees is ____.
Check Solution
Ans: B
Solution: 180 – 115 = 65
Answer: 65
Q. 7 If two lines intersect and one angle formed is 60 degrees, what are the measures of the other three angles?
Check Solution
Ans: C
Solution:
When two lines intersect, they form two pairs of vertical angles. Vertical angles are equal. Also, adjacent angles formed by intersecting lines are supplementary, meaning they add up to 180 degrees.
Answer: 60 degrees, 120 degrees, 120 degrees
Q. 8 Find the angle which is five times its complement.
Check Solution
Ans: B
Solution: Let the angle be x. Its complement is 90-x. We are given that x = 5(90-x). Solving for x, x = 450 – 5x, 6x = 450, x = 75.
Answer: 75
Q. 9 Two angles measure $(30^\circ – b)$ and $(140^\circ + b)$. If each one is the supplement of the other, then the value of $b$ is?
Check Solution
Ans: D
Solution:
Supplementary angles add up to 180 degrees. So,
$(30^\circ – b) + (140^\circ + b) = 180^\circ$
$170^\circ = 180^\circ$
This is incorrect. However, we know that
$(30^\circ – b) + (140^\circ + b) = 180$
$170 – b + b = 180$.
$(30-b) + (140+b) = 180$. Then $170 = 180$. There appears to be an error.
Consider each being supplementary to the other.
$30-b + 140+b = 180$
$170 = 180$, which is not possible.
However, we are given that these are supplements. Thus
$(30-b) + (140+b) = 180$.
$170=180$
This makes no sense.
$30-b + 140+b = 180$, giving 170=180, which isn’t true, and implies b has no solution.
Let’s consider the angles individually:
If $30-b$ is supplementary to $140+b$,
$30-b + 140+b = 180$
$170=180$, no solution.
If $30-b = 180 – (140+b)$
$30-b = 40-b$
$30=40$ impossible.
If $140+b = 180-(30-b)$
$140+b=150+b$, which is not possible.
Perhaps the question means that the angles are equal to each other.
$30-b + 140+b = 180$
$170 = 180$, which is wrong
The angles are supplementary.
$30-b+140+b = 180$
$170=180$. Not a consistent equation.
If we let the angles be x and y:
$x=30-b$
$y=140+b$
$x+y=180$
$(30-b) + (140+b) = 180$
$170 = 180$
The question is wrong.
$30-b + 140+b = 180$
$170=180$
This can’t happen.
Answer: No solution
Q. 10 In Figure, if lines $PQ$ and $RS$ intersect at a point $T$ such that $\angle PRT = 45^\circ$, and $\angle RPT = 90^\circ$ and $\angle TSQ = 70^\circ$, find $\angle SQT$.
Check Solution
Ans: D
Solution:
In $\triangle PRT$, $\angle PTR = 180^\circ – \angle PRT – \angle RPT = 180^\circ – 45^\circ – 90^\circ = 45^\circ$.
Since $\angle PTR$ and $\angle STQ$ are vertical angles, $\angle STQ = 45^\circ$.
In $\triangle STQ$, $\angle SQT = 180^\circ – \angle TSQ – \angle STQ = 180^\circ – 70^\circ – 45^\circ = 65^\circ$.
Answer: $65^\circ$
Next Topic: Triangles
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