Option a:
sin A for the triangle is [tex]\frac{3}{5}[/tex].
Solution:
The given triangle is a right triangle.
The adjacent side to angle A is AC.
The opposite side to angle A is BC.
Hypotenuse is AB.
AC = 8, BC = 6 and AB = 10.
Using trigonometric ratio formulas,
[tex]$\sin \theta=\frac{\text { Opposite side }}{\text { Hypotenuse }}[/tex]
[tex]$\sin A=\frac{BC}{AB}[/tex]
[tex]$\sin A=\frac{6}{10}[/tex]
Divide both numerator and denominator by 2, we get
[tex]$\sin A=\frac{6\div2}{10\div 2}[/tex]
[tex]$\sin A=\frac{3}{5}[/tex]
Hence sin A for the triangle is [tex]\frac{3}{5}[/tex].
Option a is the correct answer.
