Question

tan(A+B-C)=1 , sin(B+C-A) = 1, cot(C+A-B) = 1 ; find a, b and c​

Answer 1

Answer:a=22.5,b=67.5,c=45

Step-by-step explanation: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,[[ a=22.5, b= 67.5 and c=45 degrees]]

Hope it helps

Answer 2

Heya!!

Tan ( A + B - C ) = Tan ( 45° )

Becoz Tan ( 45° ) = 1

A + B - C = 45 ..... equation 1

Sin ( B + C - A ) = Sin ( 90° )

Becoz Sin ( 90° ) = 1

B + C - A = 90 .... equation 2

Cot ( C + A - B ) = Cot ( 45° )

Becoz Cot ( 45° )= 1

C + A - B = 45 .... equation 3

Now, Add Equation 1 and 3 We get

2A = 90

A = 45°

Add equation 2 and 3 we get

2C = 135

C = 67.5°

Now, put value of C and A in equation 1 we get

B = 67.5°

Have a nice day ahead.