Question

In the given figure, O is the centre of the circle, BD=OD and CD is perpendicular to AB. Find angle CAB​

Answer 1

Answer = Given : O is the center of the circle BD=OD and CD | AD

To determine : ∆CAB

Construction = join OC

Determination : BD=OD

:. ∆DOE = ∆DBE

| Angle opposite to equale side of triangle are equal

In angle ∆OED and angle ∆BED,

∆DOE =∆DEB | from above

∆DEO = ∆DBE

| Each =90° (given ) as CD | AB

OD=BD | given

:. ∆OED = ∆BED

| AAS congruence rule

:. OE = BE | C.P.C.T

Now, in ∆CEO and ∆CEB ,

CE = CE | common

∆CEO and ∆CEB | each 90°

OE = BE | proved above

:. ∆CEO =~ ∆CEB

|SAS congruence rule

:. CO = CB .... (1)

∆ACB = 90°

| Angle in a semi circule is a right angle

| OA = OB = OC ....(2)

| The mid point of the hypotenure of a right angled triangle is equidistant from its vertices

Frome (1) and (2)

OB= OC = BC

:. ∆OBC is equilateral triangle is 60°

= 2 ∆ BAC = 60°

The angle subtanded by an arc of circle at the center is twice the angle substended by it at any point on the remaining part of circle

= ∆BAC = 30°

= ∆CAB = 30°

Step-by-step explanation:

Answer 2

Answer:

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