here's the solution,
we know :
[tex] \sin( \theta) = \dfrac{perpendicular}{hypotenuse} [/tex]
So,
[tex] \dfrac{p}{h} = \dfrac{12}{13} [/tex]
so.. let the perpendicular be 12x and hypotenuse be 13x
now,
by applying pythagoras theorem,
- [tex]b {}^{2} = h {}^{2} - p {}^{2} [/tex]
where,
- b = base
- h = hypotenuse
- p = perpendicular
So,
- [tex]b {}^{2} = ({13x})^{2} - ({12x})^{2} [/tex]
- [tex] b {}^{2} = 169x {}^{2} - 144x {}^{2} [/tex]
- [tex] {b}^{2} = 25x {}^{2} [/tex]
- [tex]b = 5x[/tex]
so,
- [tex] \cos( \theta) = \dfrac{b}{h} [/tex]
- [tex] \cos( \theta) = \dfrac{5x}{13x} [/tex]
- [tex] \cos( \theta) = \dfrac{5}{13} [/tex]
hope it helps !!
