In figure, AB is a diameter of a circle with center O and AC is a tangent. If $∠ A O D = 60 °,$ find $∠ A C D .$

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30 o

90 o

50 o

60 o

answer is C.

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Detailed Solution

Given that AB is a diameter of a circle with center O and AC is a tangent.
We have to find

∠ A C D .

The triangle is composed of three interior angles and the sum of the three angles of a triangle is equal to 180 degrees.
Since AB is the straight line. So,

∠ A O D + ∠ B O D = 180 ° ⇒ 60 ° + ∠ B O D = 180 ° ⇒ ∠ B O D = 180 ° − 60 ° ∴ ∠ B O D = 120 °

In the triangle BOD, OB = OD, because they are the radius of the circle. Thus,

∠ O B D = ∠ O D B

Because angles opposite to the equal sides are equal. Also,

∠ B A C = 90 °

. Since the tangent to a circle is perpendicular to the tangent. Then by using the angle sum property of the triangle,

∠ O B D + ∠ O D B + ∠ B O D = 180 ° ⇒ 120 ° + 2 ∠ O B D = 180 ° ⇒ 2 ∠ O B D = 60 ° ⇒ ∠ O B D = 30 °

Now, in the triangle ABC,

∠ A B C + ∠ B A C + ∠ B C A = 180 ° ⇒ 30 ° + 90 ° + ∠ B C A = 180 ° ⇒ ∠ B C A = 60 °

The required value is

∠ A C D = 60 °

Therefore, the correct option is 3.

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