Q1 of 52 Page 10

In figure 1, O is the center of a circle, PQ is a chord and PT is the tangent at P. If ∠ POQ = 70°, then calculate ∠TPQ. (CBSE 2011,2017 C)

Given: PT is the tangent at P.


∠POQ = 70°


Since OP and OQ are the radii of the circle, therefore POQ forms an isosceles triangle.


Therefore, ∠ OPQ and ∠ PQO are of the same measure.


Let ∠ OPQ = ∠1 and ∠ PQO = ∠2. Thus ∠1 = ∠2 ( = x, say)


As sum of all angles of a triangle is 180°, therefore,


∠ OPQ + ∠ PQO + ∠ POQ = 180°


x + x + 70° = 180°


∴ 2x = 110°


⇒ x = 55°


Now, PT is tangent at P, therefore ∠ TPO = 90°


But, ∠ TPO = ∠ TPQ + ∠ QPO


∴ ∠ TPQ = ∠ TPO - ∠ QPO


= 90° - 55°


= 35°

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