Question

ABC is a right angled triangle , right angled at B if AD=CD and angle DBC =20 , find angle BAC​

Answer 1

Answer:

Step-by-step explanation:

SOLUTION 1)

Given: ∠DBC=20°

In ΔBDC,

∠DBC+∠DCB+∠BCD=180°

∠DCB=180°-(90°+20°)= 70°

Now,

In ΔABC,

∠ABC+∠BCA+∠CAB=180°

90°+70°+∠CAB=180°

160°+∠CAB=180°

∠CAB=180°-160°=30°

 

SOLUTION 3)  

Given: AB and AC are two equal words of a circle such that the centre of the circle lies on the bisector of ∠BAC.

⇒ OA is the bisector of ∠BAC.

 

Again, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing  

the angle.

∴ P divides BC in the ratio = 6 : 6 = 1 : 1.

⇒ P is mid-point of BC.

⇒ OP ⊥ BC.

In ΔABP,  

By pythagoras theorem,

 

Now,  

In right triangle OBP,

Equating (i) and (ii), we get

Putting the value of AP in (i), we get,