Question

in the given figure AB is a diameter and DC is tangent which meets AB produced at point C .if angle DAC=x°,find in terms of x°. 1)angle DCB 2)angle DBC

Answer 1

Answer:

∠DCB=90-2x and ∠DBC=90°+x

Step-by-step explanation:

Given AB is a diameter and DC is tangent which meets AB produced at point C. If angle DAC=x°

We have to find ∠DCB and ∠DBC.

∠ADB is the angle subtended in the semicircle ∴∠ADB=90°

By theorem, An angle between a tangent and a chord through the point of contact is equal to the angle made in triangle

∠BDC=∠DAB=x

By exterior angle property of triangle

∠DBC=∠ADB+∠BAD=90°+x

Now, ∠DCB+∠BDC+∠DBC=180°

⇒ ∠DCB+x+90+x=180°        

⇒ ∠DCB+90+2x=180

⇒ ∠DCB=90-2x