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Question 1: In a $\triangle ABC$, if $2\angle A = 3\angle B = 6\angle C$, calculate $\angle A, \angle B, \angle C$.
Solution: Let $2\angle A = 3\angle B = 6\angle C = k$. Then $\angle A = k/2, \angle B = k/3, \angle C = k/6$.
Sum $= 180^\circ \Rightarrow k/2 + k/3 + k/6 = 180^\circ \Rightarrow (3k+2k+k)/6 = 180^\circ \Rightarrow k = 180^\circ$.
$\therefore \angle A = 90^\circ, \angle B = 60^\circ, \angle C = 30^\circ$.
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Question 2: If two parallel lines are intersected by a transversal, prove that the bisectors of any pair of alternate interior angles are parallel.
Solution: Let lines be $AB \parallel CD$ and transversal $t$. Alternate angles $\angle AQR = \angle QRD$. Their halves are equal: $\frac{1}{2}\angle AQR = \frac{1}{2}\angle QRD$. These halves form alternate angles for the bisectors. Hence, bisectors are parallel.
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Question 3: In a figure, if $PQ \parallel RS$, $\angle MXQ = 135^\circ$ and $\angle MYR = 40^\circ$, find $\angle XMY$.
Solution: Draw a line through $M$ parallel to $PQ$ (and $RS$).
Angle on left $= 180^\circ - 135^\circ = 45^\circ$ (Co-interior).
Angle on right $= 40^\circ$ (Alternate).
$\angle XMY = 45^\circ + 40^\circ = 85^\circ$.
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Question 4: If the angles of a triangle are in the ratio $2:3:4$, find the angles.
Solution: Let angles be $2x, 3x, 4x$.
$2x + 3x + 4x = 180^\circ \Rightarrow 9x = 180^\circ \Rightarrow x = 20^\circ$.
Angles are $40^\circ, 60^\circ, 80^\circ$.
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Question 5: In the figure, lines $AB$ and $CD$ intersect at $O$. If $\angle AOC + \angle BOE = 70^\circ$ and $\angle BOD = 40^\circ$, find $\angle BOE$ and reflex $\angle COE$.
Solution: $\angle AOC = \angle BOD = 40^\circ$ (Vertically opposite).
$\angle BOE = 70^\circ - 40^\circ = 30^\circ$.
$\angle COE = 180^\circ - (40^\circ + 30^\circ) = 110^\circ$.
Reflex $\angle COE = 360^\circ - 110^\circ = 250^\circ$.
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Question 6: If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.
Solution: Since bisectors are parallel, corresponding angles formed by them are equal. Double of these are the original corresponding angles. Since original corresponding angles are equal, the lines are parallel.
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Question 7: The exterior angles obtained on producing the base of a triangle both ways are $104^\circ$ and $136^\circ$. Find all the angles of the triangle.
Solution: Interior base angles are $180^\circ - 104^\circ = 76^\circ$ and $180^\circ - 136^\circ = 44^\circ$.
Third angle $= 180^\circ - (76^\circ + 44^\circ) = 180^\circ - 120^\circ = 60^\circ$.
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Question 8: In $\triangle PQR$, sides $QP$ and $RQ$ are produced to points $S$ and $T$ respectively. If $\angle SPR = 135^\circ$ and $\angle PQT = 110^\circ$, find $\angle PRQ$.
Solution: $\angle QPR = 180^\circ - 135^\circ = 45^\circ$. $\angle PQR = 180^\circ - 110^\circ = 70^\circ$.
$\angle PRQ = 180^\circ - (45^\circ + 70^\circ) = 180^\circ - 115^\circ = 65^\circ$.
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Question 9: It is given that $\angle XYZ = 64^\circ$ and $XY$ is produced to point $P$. Ray $YQ$ bisects $\angle ZYP$. Find $\angle XYQ$.
Solution: $\angle ZYP = 180^\circ - 64^\circ = 116^\circ$.
Since $YQ$ bisects $\angle ZYP$, $\angle ZYQ = \angle QYP = 58^\circ$.
$\angle XYQ = \angle XYZ + \angle ZYQ = 64^\circ + 58^\circ = 122^\circ$.
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Question 10: Prove that if two lines intersect each other, then the vertically opposite angles are equal.
Solution: Let lines $AB$ and $CD$ intersect at $O$. Ray $OA$ stands on line $CD$, so $\angle AOC + \angle AOD = 180^\circ$. Ray $OD$ stands on line $AB$, so $\angle AOD + \angle BOD = 180^\circ$.
From both, $\angle AOC = \angle BOD$. Similarly, $\angle AOD = \angle BOC$.