Question

for ∆ABC, C= π/3 ( a+b-c) (a+b+c)​

Answer 1

Answer:

In ABD, applying sine rule, we get

=

...(i)

and in ACD, applying sine rule, we get

...(i)

From Eqs. (i) and (ii).

Answer 2

Correct option is

A

6

1

In triangles ABD and ACD, we have

sinB

AB

=

sin∠BAD

BD

and

sinC

AD

=

sin∠CAD

CD

⇒

sinB

sinC

=

sin∠BAD

sin∠CAD

.

CD

BD

⇒

sin

3

π

sin

4

π

=

sin∠BAD

sin∠CAD

.

3

1

⇒

sin∠CAD

sin∠BAD

=

3

1

×

1/

2

3

/2

=

6

1