Question

In triangle ABC,sin(A-B)=3/5,sinA=2sinB then a+b/c=?​

Answer 1

Step-by-step explanation:

Given: $Sin(A-B)=\frac{3}{5}$ and $SinA=2SinB$

To find:$\frac{a+b}{c}$

For calculation of the value,

⇒ $Tan (\frac{A-B}{2} )=\frac{1}{3}$

⇒ $\frac{a-b}{a+b}$ $cot \frac{C}{2}$ $=$∠$90$°

⇒$\frac{a+b}{c} = \frac{cos(\frac{A-B}{2})}{sin\frac{C}{2} }$

⇒$\frac{\frac{3}{\sqrt{10}} }{\frac{1}{\sqrt{2} } }$

⇒ $\frac{3}{\sqrt{5} }$

The value of $\frac{a+b}{c}$ is $\frac{3}{\sqrt{5} }$.