Question

0. 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

Given : PQ = 5 cm ,PR + QR = 25 cm

• ∴ PR = 25 - QR

⇒ PR² = PQ² + QR²  

⇒ QR² = PR2 - PQ²  = (25 - QR)² - (5)²

⇒ QR² = 625 - 50QR + QR² - 25  

⇒ 50QR = 600  

• ∴ QR = 12 cm

• ∴  PR = 25 - QR = 25 - 12 = 13 cm

Now :

• sin P = QR/PR = 12/13

• cos P = PQ/PR = 5/13

• tan P = QR/PQ = 12/5

Answer 2

Question:-

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.

Solutions:-

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.

AnsWeR:-

$PQ \div PR = \frac{5}{13}$