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Question
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 = ______.
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Solution
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.
Explanation:
AB is the diameter of the circle
AC is a chord and ∠BAC = 30°.
The tangent at C intersects the extended diameter AB at D.
Step 1: Apply the alternate segment theorem
The angle between a tangent (CD) and a chord through the point of contact (BC) is equal to the angle in the alternate segment (∠BAC).
∠BCD = ∠BAC = 30°
Step 2: Find the angles in ΔABC
Since AB is the diameter of the circle, the angle it subtends at any point on the semicircle is a right angle.
∠ACB = 90°
Using the angle sum property in the right-angled ΔABC:
∠ABC = 180° – (∠ACB + ∠BAC)
∠ABC = 180° – (90° + 30°) = 60°
Step 3: Use the exterior angle property for ΔBCD
For ΔBCD, the side DB is extended to A, making ∠ABC the exterior angle at vertex B. The exterior angle of a triangle is equal to the sum of the two opposite interior angles. §
∠ABC = ∠BDC + ∠BCD
Substitute the known values into the equation:
60° = ∠BDC + 30°
∠BDC = 30°
Step 4: Conclude with the isosceles triangle property
In ΔBCD, we have established that two angles are equal:
∠BCD = 30°
∠BDC = 30°
Because the base angles are equal, ΔBCD is an isosceles triangle. Therefore, the sides opposite to these equal angles must also be equal:
BC = BD
