Multiple choice

Draw a circle with centre $C$ and radius $3\ cm$. Take a point $P$ outside the circle such that $CP = 6\ cm$. Construct tangents $PA$ and $PB$ from this point to the circle, where $A$ and $B$ are the intersection points of the tangents. Then $m \angle ACB = $ ?

  1. $60^\circ$
  2. $120^\circ$
  3. $30^\circ$
  4. $90^\circ$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In right triangle OAP (where O is the center), OA = 3 and OP = 6. Thus, sin(angle OPA) = 3/6 = 1/2, so angle OPA = 30 degrees. In triangle OAP, angle AOP = 60 degrees. Since the two tangents are symmetric, angle AOB = 2 * 60 = 120 degrees.

AI explanation

The tangent at a point on a circle is perpendicular to the radius at that point, so triangles PAC and PBC are right-angled at A and B. Using the trigonometric ratio for the right triangle PAC, cosine of angle APC equals adjacent side AC divided by hypotenuse CP, giving cos(APC) equals 3 divided by 6. This simplifies to cos(APC) equals 0.5, which means angle APC is 60 degrees. The quadrilateral PACB has angles of 90 degrees at A and B, so angles A and B plus angle P plus angle C must sum to 360 degrees, yielding 90 plus 90 plus 60 plus angle C equals 360. Solving for angle ACB gives 180 plus 60 plus angle C equals 360, meaning the measure of angle ACB is 120 degrees.

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