We know that ,
[tex] \sin( \alpha ) = \frac{opposite}{hypotenuse} [/tex]
and
[tex] \cos( \alpha ) = \frac{adjacent}{hypotenuse} [/tex]
where 'alpha' is an angle of triangle ; 'opposite' denotes the side opposite to alpha & 'adjacent' refers to the side next to the angle (but not hypotenuse)
Similarly ,
[tex] \sin(q) = \frac{opposite}{hypotenuse} = \frac{4}{5} [/tex]
Let the length of the opposite side be 4x and the length of hypotenuse be 5x. By using Pythagorean Theorem , we can find the length of base.
[tex] {base}^{2} + {(4x)}^{2} = {(5x)}^{2} [/tex]
[tex] = > {base}^{2} = 25 {x}^{2} - 16 {x}^{2} = 9 {x}^{2} [/tex]
[tex] = > base = \sqrt{9 {x}^{2} } = 3x[/tex]
Now , we have got the length of all the sides of the triangle. So,
[tex] \cos(q) = \frac{adjacent}{hypotenuse} = \frac{3x}{5x} = \frac{3}{5} [/tex]
and
[tex] \cos(p) = \frac{adjacent}{hypotenuse} = \frac{4x}{5x} = \frac{4}{5} [/tex]
So,
[tex] \cos(p) + \cos(q) = \frac{4}{5} + \frac{3}{5} = \frac{7}{5} [/tex]
