Question

If sin (A+B+C)=1, tan(A-B)=1upon root 3 and sec(A+C)=2 find A,B,C

Answer 1

∠ A =  60°, ∠ B = 30° and ∠ C = 0°

Step-by-step explanation:

We have,

$\sin (A+B+C)=1$, $\tan (A-B)=\dfrac{1}{\sqrt{3}}$ and

$\sec (A+C)=2$

To find, ∠ A, ∠ B and ∠ C = ?

∴$\sin (A+B+C)=1=\sin 90$

⇒ A + B + C = 90°        .......(1)

$\tan (A-B)=\dfrac{1}{\sqrt{3}}=\tan 30$

⇒ A - B = 30°              .......(2)

and

$\sec (A+C)=2=\sec 60$

⇒ A + C = 60°            .......(3)

From (1) - (3),

A + B + C - A - C = 90° - 60°  

⇒ B = 30°  

Put B = 30°  in (2), we get

A - 30°  = 30°

⇒ A = 60°

And C = 0°

Hence, ∠ A =  60°, ∠ B = 30° and ∠ C = 0°

Answer 2

Answer:

go through the attachment