Question

if A,B,Care acute angles tan A =1/2 tanB =1/5 tanC=1/8 then A+B+C=​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

if A,B,Care acute angles tan A =1/2 tanB =1/5 tanC=1/8 then A+B+C=​π/4.

Step-by-step explanation:

We are given that A, B, and C are acute angles and tanA=1/2, tanB=1/5 and tanC=1/8.

We are asked to find A+B+C

We can find the value of A+B+C by using the trigonometric formula.

$tan(A+B+C)=\frac{tanA+tanB+tanC-tanAtanBtanC}{1-tanAtanB-tanBtanC-tanCtanA}$

Substitute the values of tanA=1/2, tanB=1/ and tanC=1/8. We get,

$tan(A+B+C)=\frac{\frac{1}{2}+\frac{1}{5}+\frac{1}{8}-\frac{1}{2}\frac{1}{5}\frac{1}{8}}{1-\frac{1}{2}\frac{1}{5}-\frac{1}{5}\frac{1}{8}-\frac{1}{8}\frac{1}{5}}$

Perform addition on numerator.

$tan(A+B+C)=\frac{\frac{40+16+10}{80}-\frac{1}{80}}{1-\frac{1}{10}-\frac{1}{40}-\frac{1}{16}}$

Perform subtraction on numerator and denominator.

$tan(A+B+C)=\frac{\frac{66}{80}-\frac{1}{80}}{1-\frac{8+2+5}{80}}$

On simplification. We get,

$tan(A+B+C)=\frac{\frac{65}{80}-\frac{1}{80}}{1-\frac{15}{80}}$

$tan(A+B+C)=\frac{\frac{65}{80}}{\frac{65}{80}}$

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

Take tan inverse on both sides

$A+B+C=tan^{-1} 1(1)$

$A+B+C=45^0$$=$π/4

Therefore, the value of A+B+C=π/4.