In triangle PQR, right angled at Q, PR$+$QR=25cm and PQ=5cm.Determine the values of sinP, cosP, tanP.
Given: In triangle PQR, right angled at Q, PR$+$QR=25cm and PQ=5cm
To do: Determine the values of sinP, cosP, tanP.
Answer:
In the triangle let QR =$x$; PR $+$ QR = 25 (given); then PR = 25 $- x$
PQ = 5 cm (given)
Using Pythagoras theorem
$PQ^2 + QR^2 = PR^2$
$5^2 + x^2 = (25 - x)^2$
$25 + x^2 = 625 + x^2 - 50x$
Solving $50x = 625 - 25 = 600$
or $x = \frac{600}{50} =12$ or $QR = 12$ cm and $PR = 25 - 12 = 13$ cm
sin P = $\frac{opp}{hyp} = \frac{QR}{PR} = \frac{12}{13}$
cos P = $\frac{adj}{hyp} = \frac{PQ}{PR} = \frac{5}{13}$
tan P = $\frac{opp}{adj} = \frac{QR}{PQ} = \frac{12}{5}$
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