The answer is [tex] \frac{7}{5} [/tex]
Complimentary angles: the angles added up equal 90 degrees
When you have complimentary angles, the sine of one equals the cosine of the other and vice verse
Q and P are complimentary
sin(Q)=cos(P)
cos(Q)=sin(P)
Since you know that sin(Q)=[tex] \frac{4}{5} [/tex] , that also means that cos(P)=[tex] \frac{4}{5} [/tex]
Since sin(Q) is [tex] \frac{leg}{hypotenuse} [/tex] , you can use the Pythagorean Theorem to find the length of the last leg.
[tex] a^{2} + b^{2} = c^{2} [/tex]
[tex] a^{2} + (4)^{2} = (5)^{2} [/tex]
[tex] a^{2} + 16 = 25 [/tex]
[tex] a^{2} = 9 [/tex]
[tex] a = 3 [/tex]
(It's a 3,4,5 triangle)
Now you have the length of the other leg so cos(Q)=[tex] \frac{3}{5} [/tex]
cos(P)+cos(Q)
=[tex] \frac{4}{5} + \frac{3}{5} [/tex]
=[tex] \frac{7}{5} [/tex]
The answer is [tex] \frac{7}{5} [/tex]
