Question

In the figure below, all the vertices of the small triangle are on the circle and all the sides of the larger triangle touch the circle at these points.

Find all angles of the small triangle.

Answer 1

Using angle sum property in ΔABC:

∠ABC + ∠BCA + ∠CAB = 180°

⇒ 30° + 70° + ∠CAB = 180°

⇒ 100° + ∠CAB = 180°

⇒ ∠CAB = 180° −100° = 80°

We know that lengths of tangents drawn from a point outside the circle are equal.

∴ BQ = BR

AQ = AP

CP = CR

In ΔBQR, BQ = BR

We know that angles opposite to equal sides are equal in measure.

∠BRQ = ∠BQR

By angle sum property:

∠BQR + ∠QRB + ∠RBQ = 180°

⇒ 2∠BQR + 30° = 180°

⇒ 2∠BQR = 150°

⇒ ∠BQR = ∠BRQ = 75°

Similarly, in ΔRPC:

∠CRP = ∠CPR (As CP = CR)

By angle sum property:

∠CRP + ∠RPC + ∠PCR = 180°

⇒ 2∠CRP + 70° = 180°

⇒ 2∠CRP = 110°

⇒ ∠CRP = ∠CPR = 55°

In ΔAQP:

∠APQ = ∠AQP (As AQ = AP)

By angle sum property:

∠APQ + ∠PQA + ∠QAP = 180°

⇒ 2∠APQ + 80° = 180°

⇒ 2∠APQ = 100°

⇒ ∠AQP =∠APQ = 50°

We know that each angle between a chord and the tangent at one of its ends in a circle is equal to the angle in the segment on the other side of the chord.

∴ ∠BQR = ∠RPQ = 75°

∠RPC = ∠RQP = 55°

∠AQP = ∠QRP = 50°

Hence, the measures of all the angles of smaller triangle are 75°, 55° and 50°.