In the given figure, STM and QM are Tangents to the circle. QRS is a straight line and SR = TR. If Angle TSR = 25°, find the value of angle QMT.
Answers
The value of angle QMT is 80°.
Step by step explanation:
Given:
STM and QM are the two tangents to the circle.
QRS is a straight line.
SR = TR.
Angle TSR = 25°.
To find:
The value of angle QMT.
Solution:
Step 1: Find the value of angle RTS.
As per the given information, SR = TR.
According to the property of the isosceles triangle, when two opposite sides of a triangle are equal, their angles are equal too.
therefore, ∠TSR = ∠RTS
i.e,
∠RTS = 25°
Step 2: Find the value of angle OTS.
Here, Join the radius OT. O is the center of the circle above the line QRS.
Join radius OR and OQ.
Like the radius, OT is perpendicular to the tangent STM of the circle, ∠OTS = 90°.
Step 3: Find the value of angle OTR.
As ∠OTS = 90° and ∠RTS = 25°, the value of ∠OTR is:
∠OTS = ∠OTR + ∠RTS
90° = ∠OTR + 25°
∠OTR = 90° - 25°
∠OTR = 65°
Step 4: Find the value of angle ROT.
Here,
In triangle OTR,
OT = OR
so,
∠OTR + ∠TRO + ∠ROT = 180°
65° + 65° + ∠ROT = 180°
∠ROT = 180° - 130°
∠ROT = 50°
Step 5: Find the value of angle ORQ.
here,
∠ORQ = 65° - 50°
∠ORQ = 15°
Step 6: Find the value OQR.
According to the property of the isosceles triangle, when two opposite sides of a triangle are equal, their angles are equal too.
So in triangle OQR,
OQ = OR
therefore,
∠ORQ = ∠OQR
∠OQR = 15°
Step 7: Find the value of angle SQM.
Here,
As the radius OQ is perpendicular to the tangent QM, angle OQM = 90°.
also,
∠OQR = 15°
so,
∠SQM = 90° - 15°
∠SQM = 75°
Step 8: Find the value of angle QMT.
In triangle SQM, the sum of all the angles of a triangle is equal to 180°.
so,
∠SQM + ∠QMS + ∠MSQ = 180°
75° + ∠QMS + 25° = 180°
∠QMS = 180° - 100°
∠QMS or ∠QMT = 80°.
Hence,
The value of angle QMT is 80°.
