Question

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

Answer 1

Answer:

Answer:

Sin P=12/13 , tan P=12/5

Step-by-step explanation:

Step-by-step explanation:

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

Using Pythagoras 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,

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

QR = (25 - 13) = 12 cm

Tan P= opposite/adjacent

= = QR/PQ = 12/5

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

Answer 2

in ∆PQR

PR^2 = PQ^2 + QR^2

(25 - QR)^2 = 5^2 + QR^2(PR + QR = 25)

625 + QR^2 - 50QR = 25 + QR^2

625 - 25 = 50QR

QR = 12

Now PR = 13

Now ,

sin P = 12/13

cos P= 5/13

tan P= 12/5

Hope you will understand