How i solve the introduction to trigonometry of ncert class 10 exercise 8.1 question no.10. Sir please help me to solve this question easily and give me the best solution of this question. 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.

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# . 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. Q.10

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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,

PR

^{2}= PQ^{2}+ QR^{2}Substitute the value of PR as x

(25- x)

^{ 2 }= 5^{2 }+ x^{2}25

^{2}+ x^{2}– 50x = 25 + x^{2}625 + x

^{2}-50x -25 – x^{2 }= 0-50x = -600

x= -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 = Opposite Side/Hypotenuse = QR/PR = 12/13

(2) Cos p = Adjacent Side/Hypotenuse = PQ/PR = 5/13

(3) tan p =Opposite Side/Adjacent side = QR/PQ = 12/5