Question

In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Detemine the values of sin P, cos P, tan P ? Class - 10 intro. to trigonometry.


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Answer 1

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Let PR be 'x' and QR be = 25-x ,

Using Pythagorus Theorem,

==>PR^2 = PQ^2 + QR^2 ,

    x^2   = (5)^2 + (25-x)^2 ,

    x^2   = 25 + 625 + x^2 - 50x ,

    50x   = 650 ,

         x   = 13 .

therefore, PR = 13 cm ,

and,          QR = (25 - 13) = 12 cm ,

Now,  Sin P = opposite/hypotenuse = QR/PR = 12/13 ,

         Tan P= opposite/adjacent      = QR/PQ = 12/5 ,

         Cos P= adjacent/hypotenuse = PQ/PR = 5/13.

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Answer 2

Answer:

⇒PQ=5cm

⇒PR+QR=25cm

⇒PR=25−QR

Now, In △PQR

⇒(PR) 2 =PQ2+QR 2

⇒(25−QR) 2 =5 2+QR 2

⇒625+QR 2−50QR=25+QR 2

⇒50QR=600

⇒QR=12cm

⇒PR=25−12=13cm

∴sinP= PR/QR= 13/12 ,cosP= PQ/PR =13/5 ,tanP=

OR/PR=12/5

Hence, the answers are sinP= 13/12 ,

cosP= 13/5 ,tanP= 5/12

Step-by-step explanation:

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