Question

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Answer 1

Answer:

Given,

Δ=a

2

−(b−c)

2

Now as we know that s=

2

a+b+c

or, a=2s−(b+c)

putting this value of a in Δ=a

2

−(b+c)

2

, we get

Δ=[2s−(b+c)]

2

−(b−c)

2

Δ=[4s

2

+(b+c)

2

−4s(b+c)]−(b−c)

2

Δ=4s

2

−4s(b+c)+[(b+c)

2

−(b−c)

2

]

Δ=4s

2

−4s(b+c)+4bc

Δ=4s

2

−4sb−4sc+4bc

Δ=4s(s−b)−4c(s−b)

Δ=(4s−4c)(s−b)

Δ=4(s−c)(s−b)

4

1

=

Δ

(s−c)(s−b)

...............1

Now as we know that tan

2

A

=

s(s−a)

(s−b)(s−c)

(s−b)(s−c)

=tan

2

A

s(s−a)

multiply both side by

(s−b)(s−c)

(s−b)(s−c)= tan

2

A

s(s−a)(s−b)(s−c)

(s−b)(s−c)= tan

2

A

Δ

Δ

(s−b)(s−c)

= tan

2

A

................2

by using 1 and 2

4

1

= tan

2

A

tanA=

1−tan

2

2

A

2tan

2

A

=

1−(

4

1

)

2

2(

4

1

)

=

15

8