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

Asked by Pragya Singh | 1 year ago |  169

##### Solution :-

In a given triangle PQR, right angled at Q, the following measures are

PQ = 5 cm

PR + QR = 25 cm

Now let us assume, QR = x

PR = 25-QR

PR = 25- x

According to the Pythagorean Theorem,

PR2 = PQ2 + QR2

Substitute the value of PR as x

(25- x) 2 = 5+ x2

252 + x2 – 50x = 25 + x2

625 + x2-50x -25 – x= 0

-50x = -600

x= $$\dfrac{ -600}{-50}$$

x = 12 = QR

Now, find the value of PR

PR = 25- QR

Substitute the value of QR

PR = 25-12

PR = 13

Now, substitute the value to the given problem

(1) sin p = $$\dfrac{Opposite \;side}{Hypotenuse}=\dfrac{QR}{PR}=\dfrac{12}{13}$$

(2) Cos p = $$\dfrac{Adjacent \;side}{Hypotenuse}=\dfrac{ PQ}{PR}=\dfrac{ 5}{13}$$

(3) tan p = $$\dfrac{Opposite \;side}{Adjacent \;side}=\dfrac{QR}{PQ}=\dfrac{12}{5}$$

Answered by Abhisek | 1 year ago

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