In Fig. 9.9, BOA is a diameter of a circle and the tangent at a point P meets BA extended at T. If ∠PBO = 30°, then ∠PTA is equal to 30°. Write ‘True’ or ‘False’ and give reasons for your answer
Solution:
Given, BOA is a diameter of a circle
The tangent at a point P meets BA extended at T.
We have to determine if ∠PBO = 30°, then ∠PTA is equal to 30°
We know that the radius of a circle is perpendicular to the tangent at the point of contact.
So, ∠OPT = 90°
We know that the angle in a semicircle is always equal to 90°
So, ∠BPA = 90°
In triangle PAB,
We know that the sum of all three interior angles of a triangle is always equal to 180°
∠PBA + ∠PAB + ∠BPA = 180°
From the figure,
∠PBA = 30°
30° + ∠PAB + 90° = 180°
120° + ∠PAB = 180°
∠PAB = 180° - 120°
∠PAB = 60°
We know that ∠PAB = ∠OAP = 60°
From the figure,
OP = OA = OB = radius
In triangle OPA,
∠OPA = ∠OAP
Also, ∠OPT = ∠OPA + ∠APT
90° = 60° + ∠APT
∠APT = 90° - 60°
∠APT = 30°
Therefore, the measure of angle APT is equal to 30°
✦ Try This: AB is a diameter of a circle and AC is its chord such that ∠BAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.2 Sample Problem 1
In Fig. 9.9, BOA is a diameter of a circle and the tangent at a point P meets BA extended at T. If ∠PBO = 30°, then ∠PTA is equal to 30°. Write ‘True’ or ‘False’ and give reasons for your answer
Summary:
The statement “In Fig. 9.9, BOA is a diameter of a circle and the tangent at a point P meets BA extended at T. If ∠PBO = 30°, then ∠PTA is equal to 30°” is true
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