Question

PLEASE HELP ME TO ANSWER THIS QUESTION.

Answer 1

Answer:

(a) PT = 13 cm

(b) <PTR = $39.7^{o}$

Step-by-step explanation:

(a) Applying the Pythagoras theorem to QST, we have;

$QS^{2}$ = $ST^{2}$ + $QT^{2}$

      = $6^{2}$ + $8^{2}$

$QS^{2}$ = 36 + 64

       100

QS = $\sqrt{100}$

     = 10

QS = 10 cm

Since QS = TR, then;

WT = $\frac{1}{2}$ TR

      = $\frac{1}{2}$ x 10

WT = 5 cm

Thus, applying Pathagoras theorem to PWT;

$PT^{2}$ = $PW^{2}$ + $WT^{2}$

      = $12^{2}$ + $5^{2}$

      = 144 + 25

$PT^{2}$ = 169

PT = $\sqrt{169}$

    = 13

PT = 13 cm

(b) To determine <PTR, TR = 10 cm. Apply the required trigonometric function;

Cos θ = $\frac{adjacent}{hypotenuse}$

Cos θ = $\frac{10}{13}$

          = 0.7692

θ = $Cos^{-1}$ 0.7692

  = 39.7179

θ = $39.7^{o}$

Therefore, <PTR = $39.7^{o}$