A horizontal park is in the shape of a triang | vedclass

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(B) Let $OP = h$. Since $OP$ is vertical,$	riangle OPA$,$	riangle OPB$,and $	riangle OPC$ are right-angled at $O$.Given $AB = 16$ and $C$ is the mi

Vedclass · Accepted Answer

$\frac{32}{\sqrt{3}}(2-\sqrt{3})$

Vedclass · Answer

(B) Let $OP = h$. Since $OP$ is vertical,$	riangle OPA$,$	riangle OPB$,and $	riangle OPC$ are right-angled at $O$.Given $AB = 16$ and $C$ is the midpoint,$AC = CB = 8$.In $	riangle OPA$,$	an 15^{\circ} = \frac{OP}{OA} \Rightarrow OA = h \cot 15^{\circ}$.In $	riangle OPC$,$	an 45^{\circ} = \frac{OP}{OC} \Rightarrow OC = h$.In $	riangle OAC$,since $\angle OCA = 90^{\circ}$ (as $OC \perp AB$ in isosceles $	riangle OAB$),we have $OA^{2} = OC^{2} + AC^{2}$.Substituting the values: $(h \cot 15^{\circ})^{2} = h^{2} + 8^{2}$.$h^{2} (\cot^{2} 15^{\circ} - 1) = 64$.Using $\cot 15^{\circ} = 2 + \sqrt{3}$,we have $\cot^{2} 15^{\circ} = (2 + \sqrt{3})^{2} = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3}$.$h^{2} (7 + 4\sqrt{3} - 1) = 64 \Rightarrow h^{2} (6 + 4\sqrt{3}) = 64$.$h^{2} = \frac{64}{2(3 + 2\sqrt{3})} = \frac{32}{3 + 2\sqrt{3}}$.Rationalizing the denominator: $h^{2} = \frac{32(2\sqrt{3} - 3)}{(2\sqrt{3})^{2} - 3^{2}} = \frac{32(2\sqrt{3} - 3)}{12 - 9} = \frac{32(2\sqrt{3} - 3)}{3} = \frac{32}{\sqrt{3}}(2 - \sqrt{3})$.

$A$ horizontal park is in the shape of a triangle $OAB$ with $AB = 16$. $A$ vertical lamp post $OP$ is erected at the point $O$ such that $\angle PAO = \angle PBO = 15^{\circ}$ and $\angle PCO = 45^{\circ}$,where $C$ is the midpoint of $AB$. Then $(OP)^{2}$ is equal to.

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