Mathematics
In the given figure, PA and PB are tangents to a circle with centre O and ΔABC has been inscribed in the circle such that AB = AC. If ∠BAC = 72°, calculate
(i) ∠AOB
(ii) ∠APB.
Answer
(i) AB = AC [ABC is a isosceles triangle]
∠BAC = 72°
∠ABC = ∠ACB [Angles opposite to equal sides in a triangle are equal]
Sum of angles of a triangle :
⇒ ∠ABC + ∠ACB + ∠BAC = 180°
⇒ 2∠ABC + 72° = 180°
⇒ 2∠ABC = 108°
⇒ ∠ABC = 54°
⇒ ∠ABC = ∠ACB = 54°.
Arc AB subtends ∠AOB at center and ∠ACB on the remaining part of the circle.
⇒ ∠AOB = 2∠ACB = 2(54°) = 108°.
Hence, ∠AOB = 108°.
(ii) ∠A = ∠B = 90° [∵The tangent at any point of a circle is perpendicular to the radius through the point of contact.]
⇒ ∠AOB + ∠OAP + ∠OBP + ∠APB = 360° [Angle Sum Property of a Quadrilateral]
⇒ 108° + 90° + 90° + ∠APB = 360°
⇒ 288° + ∠APB = 360°
⇒ ∠APB = 360° - 288°
⇒ ∠APB = 72°.
Hence, ∠APB = 72°.
Related Questions
In the given figure, PQ is a diameter of a circle with centre O and PT is a tangent at P. QT meets the circle at R. If ∠POR = 72°, find ∠PTR.
In the given figure, O is the centre of the circumcircle of ΔABC. Tangents at A and B intersect at T. If ∠ATB = 80° and ∠AOC = 130°, calculate ∠CAB.
Show that the tangent lines at the end points of a diameter of a circle are parallel.
Prove that the tangents at the extremities of any chord make equal angles with the chord.
