In A 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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Answer:
Sin P=12/13 , tan P=12/5
Step-by-step explanation:
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
IN SIMPLE FORMULA :-
Given PR + QR = 25 , PQ = 5
PR be x. and QR = 25 - x
Pythagoras theorem ,PR2 = PQ2 + QR2
x2 = (5)2 + (25 - x)2
x2 = 25 + 625 + x2 - 50x
50x = 650
x = 13
PR = 13 cm
QR = (25 - 13) cm = 12 cm
sin P = QR/PR = 12/13
cos P = PQ/PR = 5/13
tan P = QR/PQ = 12/5
Answer:
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Step-by-step explanation:
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In ΔPQR, 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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Solution:
We will use the trigonometric ratios to solve the question.
Using the Pythagoras theorem, we can find the length of all three sides. Then, we will find the required trigonometric ratios.
Given ∆PQR is right-angled at Q.
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
PQ = 5 cm
PR + QR = 25 cm
Let PR = x cm
Therefore,
QR = 25 cm - PR
= (25 - x) cm
By applying Pythagoras theorem in ∆ PQR, we obtain.
PR2 = PQ2 + QR2
x2 = (5)2 + (25 - x)2
x2 = 25 + 625 - 50x + x2
50x = 650
x = 650/50 = 13
Therefore, PR = 13 cm
QR = (25 - 13) cm = 12 cm
By substituting the values obtained above in the trigonometric ratios below we get,
sin P = side opposite to angle P / hypotenuse = QR/PR = 12/13
cos P = side adjacent to angle P / hypotenuse = PQ/PR = 5/13
tan P = side opposite to angle P / side adjacent to angle P = QR/PQ = 12/5
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