Question

In tri PQR, right angled at Q ,PR+QR=25cm &PQ=5cm determine value of sinP,cosP
& tanP

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

In ∆PQR, angle d at Q

PR + PQ = 25cm

PQ = 5 cm

★Assumption

QR = p cm

PR = (25 - p)

$\fbox{Using\;Pythagoras\;theorem}$

RP² = RQ² + QP²

(25 - p)² = p² + 5²

★Identity - (a - b)² = a² - 2ab + b²

625 - 50p + p² = p² + 25

-50p = -600

${\boxed{\sf\:{p=\dfrac{-600}{-50}}}}$

p = 12

RQ = 12 cm

RP = (25 - 12)

RP = 13 cm

★Hence

${\boxed{\sf\:{sinP=\dfrac{RQ}{RP}=\dfrac{12}{13}}}}$

${\boxed{\sf\:{cosP=\dfrac{PQ}{RP}=\dfrac{5}{13}}}}$

${\boxed{\sf\:{tanP=\dfrac{RQ}{PQ}=\dfrac{12}{5}}}}$

Answer 2

Answer:

Heya mate :)

First you have to solve for QR and PR. 

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

52 + QR2= PR2 (pythagorean theorem) 

Then substitute PR to get this equation =  52 + QR2 = (25-QR)2

Solve for QR

52 + QR2 = 625 - 50QR +QR2                  .........(QR2 cancels out)

52 = 625 - 50 QR 

-600 = -50 QR 

QR= 12 

Solve for PR using original equation.

PR= 13

Now draw the triangle on your paper for help solving the next step... 

Sin = opposite/ hypotenuse  Sin(P)= 12/13

Cos= Adjacent/ hypotenuse Cos(P) = 5/13

Tan= opposite/ adjacent  Tan(P) = 12/5

HOPE IT HELPS YOU