In the figure, AB is diameter of circle with center O and QC is a tangent to the circle at C.
If ∠CAB = 30°, Find (i) ∠CQA (ii) ∠CBA
The answers are
(i) 30°
(ii) 60°
Please answer accordingly!
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Answers
The measure of angles CQA and CBA is 30° and 60°, respectively.
Given:
Angle CAB=30°
To find:
Angles CQA and CBA
Solution:
The radii OA, OC, and OB are of equal length.
So, angle OAC=angle OCA=30°.
Similarly, angle OBC=Angle OCB.
(i) The line CQ is the tangent, so the angle OCQ=90° as the tangent and radius are perpendicular.
Angle ACQ=Angle OCA+angle OCQ
Angle ACQ=30°+90°
Angle ACQ=120°
Now, in ΔACQ,
Angle ACQ+angle CAQ+angle CQA=180°
Using values,
120°+30°+Angle CQA=180°
150°+angle CQA=180°
Angle CQA=180°-150°
Angle CQA=30°
(ii) Since AB is the diameter, angle ACB=90°. (Angle in the semi-circle)
In ΔACB,
Angle ACB+angle CAB+angle CBA=180°
Using values,
90°+30°+angle CBA=180°
120°+Angle CBA=180°
Angle CBA=180-120°
Angle CBA=60°
Therefore, the measure of angles CQA and CBA is 30° and 60°, respectively.
Answer:
The angle of ∠CQA is 30 ° and ∠CBA is 60 °.
Step-by-step explanation:
Explanation :
Given, AB is a diameter of the circle with centre O
QC is a tangent
∠CAB = 30 °
Step 1:
Here we have ,OC=OA (radius of circle)
Therefore , ∠OAC=∠OCA = 30° (∠OAC = 30 given )
∴In triangle ΔACO
∠CAO+∠ACO+∠AOC= 180 (sum of angle of a triangle is 180°)
⇒30 +30 + ∠AOC = 180 (∠CAB=30 and ∠CAB = 90)
⇒60+∠AOC= 180
⇒∠AOC = 180 -60 = 120°
Step2:
From figure ,
∠AOC +∠BOC= 180
120 + ∠BOC = 180
⇒∠BOC = 180 -120 = 60°
Step2:
OB=OC (radius of circle )
So , the opposite angle be equal
Therefore ,∠CBO = ∠OCB (let the angle be x )
Now in triangle ΔOBC
∠COB +∠CBO+∠OCB = 180 (Sum of angle of a triangle )
120 + x+x = 180
⇒2x = 180-120 = 60°
⇒x = 30 ° = ∠CBA
Step3:
In right angle triangle Δ OCQ
∠OCQ+∠COQ+∠CQB = 180
60+90 + ∠CQB = 180 (∠BOC=∠COQ)
⇒∠CQB = 180 - 150 = 30 (∠CQB = ∠CQA)
Final answer :
Hence , the angle of ∠CQA is 30 ° and ∠CBA is 60 °.
