Question

IfA, B, C are acute angles such that
sin (B+C-A)= cos(C+A-B)=tan(A+B-C)=1 then (A, B, C) is​

Answer 1

$\huge\boxed{\fcolorbox{black}{pink}{answer}}$

Sin(b+c-a) =1

Sin90=1

Or b+c-a=90…………… (i)

Similarly,

Cos(c+a-b) =1

Cos 0 = 1

Or c+a-b = 0……………… (ii)

Tan(a+b-c)=1

Tan 45 =1

Or a+b - c =45……………(iii)

On adding (i) and (ii),

B+c-a+c+a-b=90+0

Or 2c=90

Or [c =45]

Now,

b+c-a=90

Or b+45-a=90

Or b-a=45………….(iv)

Also,

a+b-45=45

Or a+b=90…………(v)

On solving (iv) and (v),

A+b+b-a=90+45

Or 2b=135

Or b=67.5

Putting b=67.5 in equation (v)

A+67.5 =90

Or a=22.5

So,

$\blue { \tt{ a=22.5}}$

$\blue { \tt{b= 67.5 }}$

$\blue { \tt{c=45 degrees }}$

Answer 2

Answer:

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