Question

B. In the lig C & D are on the semicircle described on BA as diameter, BAD = 70' and DBC = 30 Calculate (ABD, (BCD and Bac =​

Answer 1

Answer:

Step-by-step explanation:

here ∠ADB =90°  (The angle subtended by an arc of a circle at its centre is twice of the angle it subtends anywhere on the circle’s circumference.)

Then in triangle ADB

∠ADB+∠BAD+∠ABD =180°

90°+70°+∠ABD =180°

∠ABD =180°-160°=20°

And here

∠BCD+∠BAD =180° (the opposite angles in a cyclic quadrilateral are supplementary i.e 180° )

∠BCD=180°-∠BAD

∠BCD=180°-70°=110°

triangle BDC

∠BCD+∠BDC+∠DBC=180°

110°+∠BDC+30°=180°

∠BDC=180°-140°=40°