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Sergio [31]
3 years ago
5

PLEASE HELP ME TO ANSWER THIS QUESTION.

Mathematics
1 answer:
Tomtit [17]3 years ago
4 0

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}

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