sin t 5 Solutionsin t xx5 Here opposite x and hypotenuse x5

Solution

sin t= x/(x+5)

Here opposite= x   and hypotenuse= x+5

Therefore adjacent = sqrt((x+5)²-x²)= sqrt(x²+10x+25-x²) = sqrt(10x+25)

tan t= oppsite/adjacent

tant = x/sqrt(10x+25)

t=tan^-1(x/sqrt(10x+25))

Therefore tan^-1(x/sqrt(10x+25))=t