Answer:
tanθ = [tex]\frac{5}{12}[/tex]
Step-by-step explanation:
• tanθ = [tex]\frac{sin0}{cos0}[/tex]
given sinθ = [tex]\frac{5}{13}[/tex] , we have to find cosθ
Now sinθ = [tex]\frac{5}{13}[/tex] = [tex]\frac{opposite}{hypotenuse}[/tex]
This is a right triangle with opposite side 5 and hypotenuse 13
To find the adjacent side , a , use Pythagoras' identity
• a² + b² = c² ( c is the hypotenuse and a, b the legs )
let a = adjacent , b = 5 and c = 13 , then
a² + 5² = 13²
a² + 25 = 169 ( subtract 25 from both sides )
a² = 144 ( take square root of both sides )
[tex]\sqrt{a^2}[/tex] = [tex]\sqrt{144}[/tex]
a = 12
Thus the adjacent side is 12
Then cosθ = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{12}{13}[/tex]
Then
tanθ = [tex]\frac{\frac{5}{13} }{\frac{12}{13} }[/tex] = [tex]\frac{5}{13}[/tex] × [tex]\frac{13}{12}[/tex] = [tex]\frac{5}{12}[/tex]
