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10. In \Delta \: 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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We have,
PR + QR = 25 cm.............(i)
PQ = 5 cm
and \angle Q =90^0
According to question,
In triangle \DeltaPQR,
By using Pythagoras theorem,
\\PR^2 = PQ^2+QR^2\\ PQ^2 =PR^2-QR^2 \\ 5^2= (PR-QR)(PR+QR)\\ 25 = 25(PR-QR) \\
PR - QR = 1........(ii)

From equation(i) and equation(ii), we get;
PR = 13 cm  and QR = 12 cm.

therefore,
\\\sin P= \frac{QR}{PR}= 12/13\\ \cos P = \frac{PQ}{RP} = 5/13\\ \therefore \tan P = \frac{\sin P}{\cos P} = 12/5

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manish

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