In ? PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values ofsin P, cos P and tan P

Question

In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values ofsin P, cos P and tan P

Answer 1

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 +QR^{2                  .........}(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

 

Answer 2

PR+QR=25

PR is the hypotenuse, PQ and RQ are the legs of the right angled triangle

PQ=5 (given)

Applying Pythagorean theorem we have PR² - QR² = PQ² = 25

(PR² - QR²)/(PR+QR) = PR-QR = 25/25 = 1

PR+QR = 25

PR-QR = 1

Solving for PR and QR we have PR = 13 and QR = 12

Sin(P) = QR/PR = 12/13

Cos(P) = PQ/PR = 5/13

Tan(P) = QR/PQ = 12/5

 

 

Answer 3

1. (a ± b)² = a² ± 2ab + b² 
2. Pythagorean theorem: 
  q² = p² + r²
     R                         
    | \                        
    |   \                       
p  |    \  q              
    |     \          
    |       \    
  Q    r     P

3. sin(<P) = p/q
   cos(<P) = r/q
   tan(<P) = sin(<P) / cos(<P) = p/r
~~~~~~~~~~~~~

r = 5 ,
q + p = 25 ---> p = q - 25

Let's find hypotenuse "q"
  q² = 5² + (25 - q)²
  q² = 25 + 625 - 50q + q²
- q²                            - q² 

  0 = 650 - 50q ---> q = 13

Thus, r = 5, q = 13, p = 12
sin(<P) = 12/13 ≈ 0.923
cos(<P) =  5/13 ≈ 0.385
tan(<P) = 12/5  = 2.4