Question

In an acute angled triangle abc if tan(a+b-c)=root 3 and sec (b+c-a)=1 find angle a,b,c

Answer 1

Answer:

a = 90, b = 30, c = 60

Step-by-step explanation:

Given In an acute angled triangle abc if tan(a+b-c)=root 3 and sec (b+c-a)=1 find angle a,b,c

Now tan(a + b - c) = √3

        tan(a + b - c) = tan 60

          a + b - c = 60------------------(1)

          a + b + c = 180 (angle sum property)--------------(2)

      Sec (b + c - a) = 1

      sec(b + c - a) = sec 0

          b + c - a = 0---------------------(3)

   We get b + c = a

Substituting in eqn (1)

  a + b - c = 60

b + c + b - c = 60

   2 b = 60

     b = 30

a + b + c = 180

a + a = 180 ( b + c = a)

2 a = 180

a = 90

 a + b - c = 60

90 + 30 - c = 60

 120 - c = 60

   c = 60

a = 90, b = 30, c = 60

Answer 2

Answer: I hope this attachment helps

Step-by-step explanation: