The tangent at a point A of a circle with centre 'O' intersects the diameter PQ of the circle
(when extended) at the point B. If |BAQ = 105º, find APQ.
Answers
Answer: ∠APQ = 105º
Step-by-step explanation:
Given data:
A tangent AB is drawn to a circle with centre O that intersects the diameter PQ of the circle which is extended to point B.
Angle BAQ = 105º ….. (i)
To find: ∠APQ
Solution:
Step 1:
Join points O and A just as shown in the figure attached below.
Since the inscribed angle is formed by drawing a line from each end of the diameter to any point on the semicircle is 90°, therefore,
Angle PAQ = 90° ….. (ii)
Step 2:
From the figure attached below, we have
∠PAB = ∠BAQ + ∠PAQ
⇒ ∠ PAB = 105º + 90° = 195° …. [Substituting values from (i) & (ii)] … (iii)
Step 3:
Since the radius from the centre of the circle to the point of tangency is perpendicular to the tangent line.
∴ ∠BAO = 90° …… (iv)
We have from the figure,
∠PAO = ∠PAB – ∠BAO
⇒ ∠PAO = 195° - 90° = 105º …. [Substituting values from (iii) & (iv)] ….. (v)
Step 4:
OA = OP ….. [radius of the circle]
Since angles opposite to equal sides are also equal
∴ ∠APO = ∠PAO = 105º …. [from (v)]
Thus, ∠APO = ∠APQ = 105º
Step-by-step explanation:
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