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

1 thought on “5. In an acute angled AABC, sec (B + C – A) =<br />1<br />2 and tan (C + A – B) = Find the three<br />3<br />2.<br />angles of AAB”

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 ∘

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