Example 1: Find ∠OAB ,∠ACB and ∠AOB if ∠ABO = 50° as shown in the figure.
Solution 1: ΔAOB is an isosceles triangle
∠OAB = ∠ABO = 50°
∠AOB = 180°− 2∠ABO = 180° − 100° = 80°
Angle at Centre is twice the angle on circle subtended by same arc
∠ACB = \(\frac{1}{2}\)×∠AOB =\(\frac{1}{2}\)×80 = 40°
Example 2: AB is a tangent to the circle O. Find ∠ABC and ∠BDC if the angle subtended by the chord at the centre is 100°.
Solution 2: Join B and C with the centre O
∠BOC = 100° Given
Angle at Centre is twice the angle on circle subtended by same arc
∠BDC = \(\frac{1}{2}\)×∠BOC = \(\frac{1}{2}\)×100°=50°
∠ACB = ∠BDC = 50° Tangent Theorem
Example 3:
TA and TB are the tangents to the circle O. Find ∠OAB, ∠AOB and ∠ACB
Solution 3: ΔTAB is an isosceles triangle
∠TAB = \(\frac{ 180°−50°}{2}\) = \(\frac{130°}{2}\) = 65°
∠OAB=90°−∠=90°−65°=25°
Alternate Segment Theorem
∠ACB=∠=65°
∠AOB=2∠ACB=2×65°=130°
Verify Results by checking sum of angles in ΔAOB
25°+25°+130°=180°
Sum of angles in quadrilateral TAOB is 360°
∠OAT + ∠OBT = 90°+90° = 180°
∠AOB = 180°−50°=130°
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