Question

In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
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Answer 1

Answer:

Let PR be 'x' and QR be = 25-x

Using Pythagorus Theorem,

==>PR^2 = PQ^2 + QR^2

    x^2   = (5)^2 + (25-x)^2

    x^2   = 25 + 625 + x^2 - 50x

    50x   = 650

         x   = 13

therefore, PR = 13 cm

and,          QR = (25 - 13) = 12 cm

Now,  Sin P = opposite/hypotenuse = QR/PR = 12/13

         Tan P= opposite/adjacent      = QR/PQ = 12/5

         Cos P= adjacent/hypotenuse = PQ/PR = 5/13

Step-by-step explanation:

Answer 2

Solution:

Given,

In triangle PQR,

PQ = 5cm

PR + QR = 25cm

Let us say, QR = x

Then, PR = 25 – QR = 25 – x

Using Pythagoras theorem:

PR² = PQ² + QR²

Now, substituting the value of PR, PQ and QR, we get;

(25-x)² = (5)² + (x)²

(25)²+x²-50x = 25 + x²

625 – 50 x = 25

50 x = 600

x = 12

So, QR = 12

PR = 25 – QR = 25 – 12 = 13

Therefore,

sin P = QR/PR = 12/13

cos P = PQ/pR = 5/13

tan P = QR/PQ = 12/5

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