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

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

# 3 Answers

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 PR^{2} - QR^{2} = PQ^{2 }= 25^{
}

(PR^{2} - QR^{2})/(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

**1.** (a ± b)^{2} = a^{2} ± 2ab + b^{2}

**2.** Pythagorean theorem:

q^{2} = p^{2} + r^{2}

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"

q^{2} = 5^{2} + (25 - q)^{2}

q^{2} = 25 + 625 - 50q + q^{2}

- q^{2} - q^{2}

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

5^{2} + QR^{2}= PR^{2 }(pythagorean theorem)

**Then substitute PR** to get this equation = 5^{2} + QR^{2 }= (25-QR)^{2}

**Solve for QR**

5^{2} + QR^{2} = 625 - 50QR +QR^{2 .........}(QR^{2 }cancels out)

5^{2} =^{ }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**

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