Question

express cos q as a fraction in simplest terms.
(there is a right triangle pqo with right angle at p, po = √11, oq = 4)

Explanation:

Step1: Find side PQ via Pythagoras

$$PQ = \sqrt{QO^2 - PO^2} = \sqrt{4^2 - (\sqrt{11})^2} = \sqrt{16-11} = \sqrt{5}$$

Step2: Define cos Q for right triangle

$\cos Q = \frac{\text{Adjacent to } Q}{\text{Hypotenuse}}$

Step3: Substitute values for cos Q

$\cos Q = \frac{PQ}{QO} = \frac{\sqrt{5}}{4}$

Answer:

$\frac{\sqrt{5}}{4}$