In a right triangle RQP if PQ=5cm RQ=10cm find the value of :
(i) sin2P
(ii) tan R
(iii) sin P x cos P
(iv) sin2P - cos2 P
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Step-by-step explanation:
In △PQR, right angled at Q. If PQ=5 and RQ=10cm find sin
2
P.
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Answer
PR
2
=PQ
2
+QR
2
PR
2
=25+100
PR
2
=125
PR=
125
=5
5
sin
2
P=
PR
2
QR
2
=
125
100
=
5
4
solution
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