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 and C.​

Answer 1

Aɴsᴡᴇʀ

We have,

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

$\bold{ \implies \tan (A + B - C) = \tan45 \degree } \\ \bold{and } \\ \bold{ \: \: \: \: \: \: \: \sec(B + C - A) = \sec 60 \degree}$

$\bold{ \implies (A + B - C) = 45 \degree} \\ \bold{and} \\ \: \: \: \: \: \: \: \: \: \bold{(B + C - A) = 60 \degree}$

$\bold{ \implies(A + B - C) + (B + C - A) =45 \degree + 60 \degree } \\ \\ \bold{\implies \cancel{ A} + B - \cancel{C} + B + \cancel{ C} - \cancel{ A} = 105 \degree} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\implies2 B = 105 \degree} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\implies} \boxed{ \red{\bold{ B = \frac{105 }{2} \degree= 52 \frac{1}{2} \degree }}} ..........(i)\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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now, putting B = 105/2° in (B + C - A) = 60°, we get

$\bold{\implies \frac{105}{2} \degree +C - A = 60 \degree } \: \: \\ \\ \bold{ \implies C - A = 60 \degree - \frac{105}{2} \degree } \: \: \\ \\ \bold{\implies C - A = \frac{120 \degree - 105 }{2} \degree} \\ \\ \bold{ \implies} \boxed{ \bold{ C - A = \frac{15 }{2} \degree } }..........(ii)$

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Also , In ∆ABC , we have

A+B+C = 180°

(Angle sum property :- sum of all interior angles of triangle is 180°)

from eq.(i) ,we get

• B = 105/2°

$\bold{ \implies A + \frac{105}{2} \degree+ C = 180 \degree} \: \\ \\ \bold{\implies A + C = 180 \degree - \frac{105}{2} \degree } \: \: \\ \\ \bold{ \implies A + C = \frac{360 \degree - 105 \degree}{2} } \: \: \: \\ \\ \bold{ \implies} \boxed{\bold{C + A = \frac{255 }{2} \degree }}.......(iii)$

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now, we have 2 equations

• C - A = 15/2° ..........(ii)
• C + A = 255/2° ..........(iii)

adding (ii) and (iii)

$\bold{ \implies C - \cancel{A} + C + \cancel{A} = \frac{15}{2} \degree + \frac{255}{2} \degree } \\ \\ \bold{ \implies 2C = \frac{15 \degree+ 255 \degree}{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies C = \cancel{\frac{270}{4} \degree}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies}\boxed{ \red{\bold{C = \frac{135}{2} \degree =67 \frac{1}{2} \degree} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

now, putting C = 135/2° in (iii)

$\bold { \implies \frac{135}{2} \degree+ A = \frac{255 }{2} \degree } \\ \\ \bold{ \implies A = \frac{255}{2} \degree - \frac{135}{2} \degree } \\ \\ \bold{\implies A = \frac{255 \degree - 135 \degree}{2} } \\ \\ \bold { \implies A = \frac{120}{2} \degree} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\implies }\boxed{ \red{ \bold{ A = 60 \degree}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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hence,

$\bold{• \: \: \: \: A = 60 \degree} \: \: \\ \\ \bold{• \: \: \: \:B = 52\frac{1}{2} \degree } \\ \\ \bold{• \: \: \: \: C = 67 \frac{1}{2} \degree }$