Question

5. In an acute angled AABC, sec (B + C - A) =
1
2 and tan (C + A - B) = Find the three
3
2.
angles of AABC.
C -​

Answer 1

Step-by-step explanation:

We have ,

tan( A + B - C) = 1 and sec (B + C - A) = 2

⇒ tan( A + B - C) = tan45

∘

and sec (B + C - A) = sec60

∘

⇒ A + B - C =45

∘

and B + C - A = 60

∘

⇒ ( A + B - C) + (B + C - A) = 45

∘

+60

∘

⇒ 2B = 105

∘

⇒B=52

2

1

∘

Putting B=52

2

1

∘

in B + C - A = 60

∘

, we get

52

2

1

∘

+C−A=60

∘

⇒C−A=7

2

1

∘

...(i)

Also , in ΔABC , we have

A + B + C = 180

∘

⇒A+52

2

1

∘

+C=180

∘

[∵B=52 21 ∘ ]

⇒C+A=127 21 ∘

...(ii)

Adding and subtracting (i) and (ii) , we get ,

2C = 135 ∘

and 2A = 120 ∘

⇒C=67 21 ∘

and A = 60 ∘

Hence , A = 60 ∘ , B = 52 21

and C=67 21 ∘