Question

In triangle POR, right angle at Q, PR + QR = 25cm and PQ = 5cm. Determine the value of sin P, cos P and tanP. @shreeya21

Answer 1

Given :

• PQR is a right angled triangle .
• angle Q = 90°
• PQ = 5 cm
• PR + QR = 25 cm

To find :

• Sin P
• Cos P
• Tan P

Solution:

Using PGT in triangle PQR

PR² = PQ² + QR² _____(1)

PR² = 5² + QR²

PR² - QR² = 25

(PR + QR) (PR - QR) = 25

25 (PR - QR) = 25

PR - QR = 25/25

PR - QR = 1

PR = 1 + QR

Using this in equation (1)

PR² = PQ² + QR²

(1+QR)² = 5² + QR²

1 + QR² + 2QR = 25 + QR²

1 + 2QR = 25

2QR = 25 -1

2QR = 24

QR = 24/2

QR = 12 cm

PR = 1 + QR

PR = 1+12 = 13 cm

From the figure ,

$\implies$ Sin P = $\tt \dfrac{Perpendicular}{Hypotenuse}$

= QR/PR

= 12/13

$\implies$ Cos P = $\tt \dfrac{Base}{Hypotenuse}$

= PQ/PR

= 5/13

$\implies$ Tan P =$\tt \dfrac{Perpendicular}{Base}$

= QR/PQ

= 12/5

Answer 2

Answer:

Your answer is here

Explanation:

Given:-

∠Q = 90°

PR + QR = 25cm

PQ = 5cm

To Find:-

Values of sin P, cos P and tan P.

Solution:-

Let assume the length of QR = x cm

And PR = (25 - x) cm

Now,

By using Pythagoras Theorem,

RP² = RQ² + QP²

Then,

QR = x = 12cm

PR = 25 - x = 25 - 12 = 13cm.

∴ sin P = QR/PR = 12/13

  cos P = QP/PR = 5/13 and

  tan P = QR/PQ = 12/5.

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