Multiple choice

Draw a circle with center $O$ and radius $6$ cm. Take a point $P$ outside the circle at a distance of $10$ cm from $O$. Draw tangents to the circle from point $P$. Let the tangents intersect the circle in points $A$ and $B$. Find the approximate value of $\angle OPB$.

  1. $37^o$
  2. $53^o$
  3. $45^o$

  4. None of these
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A Correct answer
Explanation

In the right triangle formed by the center O, the point P, and the tangent point A, OA = 6, OP = 10. Thus, AP = sqrt(10^2 - 6^2) = 8. In triangle OAP, sin(angle OPA) = OA/OP = 6/10 = 0.6. Angle OPA = arcsin(0.6) approx 36.87 degrees. Since the two tangents are symmetric, angle OPB is the same.

AI explanation

The tangent at a point on a circle is perpendicular to the radius at that point, making triangle OPB a right triangle with the right angle at B. In this triangle, the hypotenuse OP is 10 cm and the side OB (the radius) is 6 cm. Using the trigonometric ratio for the right triangle, sine of angle OPB equals opposite side OB divided by hypotenuse OP, giving sin(OPB) equals 6 divided by 10. The inverse sine of 0.6 gives the value of angle OPB as approximately 37 degrees.

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