Solution:
Given the triangle PNQ as shown below:
To evaluate the missing side PN, we use the Pythagorean theorem.
According to the Pythagorean theorem, 'the square of the length of the longest side (hypotenuse) of the triangle is the sum of the squares of its two opposite sides (opposite and adjacent).'
Thus, to evaluate PN, we have
[tex]\begin{gathered} \lvert PN\rvert^2=\lvert PQ\rvert^2+\lvert QN\rvert^2 \\ \text{where} \\ PQ=7,\text{ QN=12} \\ \text{thus,} \\ \lvert PN\rvert^2=7^2+12^2 \\ =49+144 \\ \lvert PN\rvert^2=193 \\ \lvert PN\rvert=\sqrt[]{193} \\ \Rightarrow\lvert PN\rvert=13.89244399 \\ \end{gathered}[/tex]
To evaluate the missing angles,
Angle N:
To evaluate the angle N as shown below:
Step 1: Label the triangle, using the angle N as the angle of focus.
Thus,
[tex]\begin{gathered} PQ\Rightarrow opposite(\text{side facing the angle N)} \\ QN\Rightarrow adjacent \\ PN\Rightarrow hypotenuse\text{ (longest side of the triangle)} \end{gathered}[/tex]
Step 2: Evaluate the angle N, using trigonometric ratio.
Thus,
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent}=\frac{PQ}{QN} \\ \text{where} \\ \theta=N,\text{ }PQ=7,\text{ QN=12} \\ \text{thus,} \\ \tan N=\frac{7}{12} \\ =0.5833 \\ N=\tan ^{-1}(0.5833) \\ \Rightarrow N=30.255 \end{gathered}[/tex]
Angle P:
The sum of the interior angles in a right-triangle equals 180°.
Thus,
[tex]\begin{gathered} \angle P+\angle Q+\angle N=180\text{ (sum of angles in a triangle)} \\ \text{where} \\ \angle Q=90,\text{ }\angle N=30.255 \\ \text{thus, } \\ \angle P+90+30.255=180 \\ \angle P+120.255=180 \\ \angle P=180-120.255 \\ \Rightarrow\angle P=59.745 \end{gathered}[/tex]
Hence,
[tex]\begin{gathered} PN=13.89244399 \\ \angle N=30.255 \\ \angle P=59.745 \end{gathered}[/tex]
