Question : A tower is broken at a point P above the ground. The top of the tower makes an angle of $60^\circ$ with the ground at Q. From another point R on the opposite side Q angle of elevation of point P is $30^\circ$. If QR = 180 m, what is the total height (in meters) of the tower?
Option 1: $90$
Option 2: $45\sqrt{3}$
Option 3: $45(\sqrt{3}+1)$
Option 4: $45(\sqrt{3}+2)$
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Correct Answer: $45(\sqrt{3}+2)$
Solution : Let OP = $h$ m In $\triangle$OPQ, $\tan60°=\frac{h}{OQ}$ ⇒ $OQ=\frac{h}{\sqrt3}$ ------------------(equation 1) In $\triangle$OPR, $\tan30^\circ=\frac{h}{OR}$ ⇒ $OR=h\sqrt3$ ----------------------(equation 2) According to the question, $h\sqrt3+\frac{h}{\sqrt3}=180$ ⇒ $3h+h=180\sqrt3$ ⇒ $h=\frac{180\sqrt3}{4}$ ⇒ $h=45\sqrt3$ m Now, in $\triangle$OPQ, $\sin60^\circ=\frac{h}{PQ}$ ⇒ $PQ=45\sqrt3×\frac{2}{\sqrt3}$ ⇒ PQ = 90 m So, the total height of the tower = $(45\sqrt3+90)=45(\sqrt3+2)$ m Hence, the correct answer is $45(\sqrt3+2)$ m.
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