Since we have a right triangle we can use the pythagorean theorem to find the length of side QR. Doing so, we have:
[tex]\begin{gathered} a^2+b^2=c^2\text{ } \\ (17)^2+b^2=(28)^2\text{ (Replacing)} \\ 289+b^2=784\text{ (Raising each number to the power of 2)} \\ b^2=784-289\text{ (Subtracting 289 from both sides of the equation)} \\ b^2=495\text{ (Subtracting)} \\ b=\sqrt[]{495}\text{ (Taking the square root of both sides)} \\ b=22.248 \\ \text{The length of side QR is 22.2 (Rounded to the nearest tenth) } \end{gathered}[/tex]
Since we have the hypotenuse and the opposite length of angle R,then we can formulate the following equation with the ratio sine:
[tex]\begin{gathered} \sin (R)=\text{ }\frac{\text{ opposite length}}{\text{hypotenuse}} \\ \sin (R)=\text{ }\frac{17}{28}\text{ (Replacing)} \\ \sin ^{-1}\sin (R)=\sin ^{-1}(\frac{17}{28})\text{ (Using the inverse function of sine to find the angle)} \\ R=\text{ 37.38 \degree} \\ \text{The measure of angle R is 37.4\degree{}rounded to the nearest tenth of a degree.} \end{gathered}[/tex]
Since we have the hypotenuse and the adjacent length of angle P,then we can formulate the following equation with the ratio cosine:
[tex]\begin{gathered} \cos (P)=\frac{\text{adjacent length}}{\text{hypotenuse}} \\ cos(P)=\text{ }\frac{17}{28}\text{ (Replacing)} \\ \cos ^{-1}cos(P)=\cos ^{-1}(\frac{17}{28})\text{ (Using the inverse function of cosine to find the angle)} \\ P=52.62 \\ \text{The measure of angle P is 52.6\degree{}rounded to the nearest tenth of a degree.} \end{gathered}[/tex]
