Question

In triangle pQR right angled at q tanp=1\√3 find the value sin p and cos p

Answer 1

Step-by-step explanation:

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