Correct Answer: Option C

Explanation

From \(\bigtriangleup\)QPR, < R = 180^o - (25^o + 90^o)

180^o - 115^o = 65^o

From \(\bigtriangleup\)PSQ, Sin 65^o = \(\frac{QPR}{hyp}\) = \(\frac{PS}{5}\)

PS = 5 sin 65^o

From \(\bigtriangleup\)PSR, tan = \(\frac{OPP}{adj}\) = \(\frac{PS}{QS}\)

but PS = 5 sin 65^o

QS tan 25^o = PS

QS tan 25^o = 5 sin 65^o

QS = \(\frac{5 sin 65^o}{tan 25^o}\)

= 5 tan 25^o cos 65^o

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