Answer:
(a) PT = 13 cm
(b) <PTR = [tex]39.7^{o}[/tex]
Step-by-step explanation:
(a) Applying the Pythagoras theorem to QST, we have;
[tex]QS^{2}[/tex] = [tex]ST^{2}[/tex] + [tex]QT^{2}[/tex]
= [tex]6^{2}[/tex] + [tex]8^{2}[/tex]
[tex]QS^{2}[/tex] = 36 + 64
100
QS = [tex]\sqrt{100}[/tex]
= 10
QS = 10 cm
Since QS = TR, then;
WT = [tex]\frac{1}{2}[/tex] TR
= [tex]\frac{1}{2}[/tex] x 10
WT = 5 cm
Thus, applying Pathagoras theorem to PWT;
[tex]PT^{2}[/tex] = [tex]PW^{2}[/tex] + [tex]WT^{2}[/tex]
= [tex]12^{2}[/tex] + [tex]5^{2}[/tex]
= 144 + 25
[tex]PT^{2}[/tex] = 169
PT = [tex]\sqrt{169}[/tex]
= 13
PT = 13 cm
(b) To determine <PTR, TR = 10 cm. Apply the required trigonometric function;
Cos θ = [tex]\frac{adjacent}{hypotenuse}[/tex]
Cos θ = [tex]\frac{10}{13}[/tex]
= 0.7692
θ = [tex]Cos^{-1}[/tex] 0.7692
= 39.7179
θ = [tex]39.7^{o}[/tex]
Therefore, <PTR = [tex]39.7^{o}[/tex]
