Question

in an acute angled triangle ABC if tan (A+B-C)=1 and sec(B+C-A)=2 find the value of A,B,C.

Answer 1

Answer:

Step-by-step explanation:

Ans is here

By applying trigno

Answer 2

Answer:

The value of A is 60, B is 52.5  and C is 67.5.

Step-by-step explanation:

Given,

$tan(A+B-C) = 1$

$\implies tan(A+B-C) = tan 45^{\circ}$

$\implies A+B-C= 45$ --------(1),

Also, given,

$sec(B+C-A) = 2$

$\implies sec(B+C-A) = sec 60^{\circ}$

$\implies B+C-A= 60$ --------(2),

Equation (1) + Equation (2),

2 B = 105 ⇒ B = 52.5

Now, triangle ABC is a triangle,

Where, A, B and C are the interior angles of the triangle ABC,

⇒ A + B + C = 180

⇒ A + B = 180 - C -------(3),

From equation (1) and (3),

180 - C - C = 45

⇒ 180 - 45 = 2∠C ⇒ C = 67.5

Now, from equation (3),

We will get,

A + 52.5 = 180 - 67.5

⇒ A = 60