in triangle ABC , A = a²-(b²-c²) then tanA =?
Answers
Step-by-step explanation:
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
ANSWER
Given,
Δ=a² −(b−c)²
Now as we know that s= a+b+c/2
or, a=2s−(b+c)
putting this value of a in Δ=a²−(b+c) ², we get
Δ=[2s−(b+c)] ²−(b−c)²
Δ=[4s² +(b+c) ²−4s(b+c)]−(b−c)²
Δ=4s²−4s(b+c)+[(b+c)²−(b−c)²]
Δ=4s²−4sb−4sc+4bc
Δ=4s(s−b)−4c(s−b)
Δ=(4s−4c)(s−b)
Δ=4(s−c)(s−b)
Now as we know that tan A/2
=✓(s−b)(s−c)/s(s-a)
✓(s-b)(s-c)=tan A/2✓s(s−a)
multiply both side by ✓(s−b)(s−c)
(s−b)(s−c)= tan A/2 Δ
(s−b)(s−c)/∆= tan A/2................2
by using 1 and 2 =1/4 = tan A/2
8/15