JAMB - Mathematics (1982 - No. 42)
5 tan 25o sin 65o
5 cos 25o sin 65o
5 tan 25o cos 65o
cos 25o cos 65o
5 cosec 25o
Explanation
From \(\bigtriangleup\)QPR, < R = 180o - (25o + 90o)
180o - 115o = 65o
From \(\bigtriangleup\)PSQ, Sin 65o = \(\frac{QPR}{hyp}\) = \(\frac{PS}{5}\)
PS = 5 sin 65o
From \(\bigtriangleup\)PSR, tan = \(\frac{OPP}{adj}\) = \(\frac{PS}{QS}\)
but PS = 5 sin 65o
QS tan 25o = PS
QS tan 25o = 5 sin 65o
QS = \(\frac{5 sin 65^o}{tan 25^o}\)
= 5 tan 25o cos 65o
180o - 115o = 65o
From \(\bigtriangleup\)PSQ, Sin 65o = \(\frac{QPR}{hyp}\) = \(\frac{PS}{5}\)
PS = 5 sin 65o
From \(\bigtriangleup\)PSR, tan = \(\frac{OPP}{adj}\) = \(\frac{PS}{QS}\)
but PS = 5 sin 65o
QS tan 25o = PS
QS tan 25o = 5 sin 65o
QS = \(\frac{5 sin 65^o}{tan 25^o}\)
= 5 tan 25o cos 65o
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