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

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

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 = PQ= 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

How did you get 12 and13 sir

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

First you have to solve for QR and PR.

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

52 + QR2= PR(pythagorean theorem)

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

Solve for QR

52 + QR2 = 625 - 50QR +QR2                  .........(QRcancels 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