Q16 of 141 Page 464

In the given figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

Given: and


In ΔOBD


OB = OD = DB


∴ Δ OBD is equilateral


∴ ∠ODB = ∠DBO = ∠BOD = 60°


Consider ΔDEB and ∠BEC


Here,


BE = BE (common)


∠CEB = ∠DEB (right angle)


CE = DE ( OE is perpendicular bisector)


∴ By SAS congruency



ΔDEB ∠BEC


∴∠DEB = ∠EBC (C.P.C.T)


∴∠EBC = 60°


Now, in ΔABC


∠EBC = 60°


∠ACB = 90° (angle in semicircle)


By angle sum property


∠EBC + ∠ACB + ∠CAB = 180°


60° + 90° + ∠CAB = 180°


∠CAB = 180° – 60° – 90° = 30°


∴∠CAB = 30°


More from this chapter

All 141 →
14

In the given figure, O is the centre of a circle, ∠ACB = 40°. Find ∠OAB.

 15

In the given figure, O is the centre of a circle, ∠OAB = 30° and ∠OCB = 55°. Find ∠BOC and ∠AOC.

 17

In the given figure, ABCD is a cyclic quadrilateral. A circle passing through A and B meets AD and BC in the points E and F respectively. Prove that EF || DC.

 18

In the given figure, AOB is a diameter of the circle and C, D, E are any three points on the semicircle. Find the value of ∠ACD + ∠BED.