In the diagram, angle QPR = 90
o, angle PSR = 90
o and PR = 5 units. Find the length of QS.
A. 5 tan 25
o sin 65
oB. 5 cos 25
o sin 65
oC. 5 tan 25
o cos 65
o 
D. cos 25
o cos 65
oE. 5 cosec 25
o Correct Answer: C
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
