Question

if A B C are positive acute angle and sin (B+C-A)=cos (C+A-B)=tan (A+B-C)=1 then A B C are

Answer 1

As sin (B+C-A)=1; cos (C+A-B)=1 and tan (A+B-C)=1; that implies B+C-A=90 degrees ..........eqn 1 C+A-B=0 degrees .............eqn 2 A+B-C=45 degrees ...........eqn 3 Solving the three equations we get A=22.5 degrees; B=67.5 degrees and C=45 degrees.

Answer 2

Answer:

The answer is $A=22.50^{\circ}$, $\quad B=67.50^{\circ}$ and $\quad C=45^{\circ}$.

Step-by-step explanation:

Given:

If A B C is a positive acute angle and

$sin (B+C-A)=cos (C+A-B)=tan (A+B-C)=1.$

To find:

The values of A, B, and C.

Step 1

Let, $\sin (B+C-A)=1$

$\Longrightarrow B+C-A$

$=\sin ^{1}(1)=90^{\circ} .......................(1)$

$\cos (C+A-B)=1$

$\Longrightarrow C+A-B$

$=\cos ^{-1}(1)=0^{\circ} .....................(2)$

$\tan (A+B-C)=1$

$\Longrightarrow A+B-C$

$=\tan ^{-1}(1)=45^{\circ} ............................(3)$

Step 2

By $(1)$, we get

$A=B+C-90^{\circ} \quad$

Putting this result in $(2)$, then

$C+(B+C-90)-B = 0^{0}$

$\Longrightarrow 2 C-90^{\circ}$

$\Longrightarrow 2 C=90^{\circ}$

$\Longrightarrow C=45^{\circ}$

Hence, $C=45^{\circ}$.

Step 3

Putting the value of $\mathrm{C}$ in $(1)$

$B+45^{\circ}-A=90^{\circ}$

$\Longrightarrow B-A=45^{\circ}....................(4)$

Putting the value of  $\mathrm{C}$ in $(3)$

$A+B-45^{\circ}=45^{\circ}$

$\Longrightarrow A+B=90^{\circ}.....................(5)$

Step 4

equating the values $(4)+(5)$ then

$&2 B=135^{\circ}$

$\Longrightarrow B=67.50^{\circ}$

$&A+B=90^{\circ}$

$\Longrightarrow A+67.50^{\circ}=90^{\circ}$

$&\Longrightarrow A=22.50^{\circ}$

Therefore, $A=22.50^{\circ}$

$\quad B=67.50^{\circ}$

and $\quad C=45^{\circ}$.

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