Question

pls answer correctly ​

Answer 1

Answer:

        sin P = 12/13

        cos P = 5/13

        tan P = 12/5

Step-by-step explanation:

Let x = QR.  Then PR+QR=25  ⇒  PR=25-x.

By Pythagoras' Theorem:

    PQ² + QR² = PR²

⇒  5² + x²  =  (25-x)²

⇒  25 + x²  =  625 - 50x + x²

⇒  50x  =  600

⇒  x = 12.

So QR = 12 cm  and  PR = 13 cm.

Then:

• sin P = QR / PR = 12/13
• cos P = PQ / PR = 5/13
• tan P = QR / PQ = 12/5

Answer 2

Answer:

According to your question

ΔPQR is a right angled at Q, and PR + QR = 25cm and PQ = 5cm.

and we have to find SinP , cosP and tanP

PR + QR = 25

PR = 25 - QR ----- (1)

and QR = x ------- (2)

So, PR = 25 - x (from 2)

Using Pythagoras theorem

(H)² = P² + B²

PR² = PQ² + QR²

(25 - x)² = 5² + x²

(25)² + (x)² - 2 × (25) × (x) = 5² + x²

625 + x² - 50x = 25 + x²

625 - 25 - 50x = x² - x²

600 - 50x = 0

600 = 50x

x = 600/50

x = 12

we get x as 12 . So, PQ = 5cm ,

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

QR = x = 12

and here , According to Sin(p)

PR = Hypotenuse = 13cm

QR = Perpendicular = 12cm

PQ = base = 5cm

$\sin(p) = \frac{prependicular}{hypotenuse} = \frac{12}{13} \\ \\ = \geqslant \cos(p) = \frac{base}{hypotenuse} = \frac{5}{13} \\ \\ = \geqslant \tan(p) = \frac{prependicular}{base} = \frac{12}{5}$