1.

If each side of a square .'a' subtends an angle 60° at the top of a tower of height 'h' metre high standing in centre of square then proof a^2=2h^2

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Answer:

Step-by-step explanation:

Consider PQRS is a square and each side of a square is = a

The height of the TOWER STANDING vertically on the CENTER of the square is (ON) = h

Proved that a^{2}=2{h}^{2}

Each side of the square subtends an angle 60° at the top of a tower N

Therefore, ∠QNR = ∠RNQ = 60°

M is the mid point of QR. now draw a line between NM and OM

In ΔNOM ∠NOM=90°

So, MN^{2}=ON^{2}+OM^{2}

we know that the perpendicular distance between the intersection point of two diagonals of a square to its side is half of the side.

Therefore  OM=\frac{a}{2}

we know that ∠QNR=60° so, ∠QNM=∠RNM=30° (as ∠QNR=60° and NM is BISECTS)

Now in ΔNMQ

tan\theta=\frac{QM}{MN}\\\\\Rightarrow{tan}30^{\circ}=\frac{QM}{MN}\\\\\Rightarrow\frac{1}{\sqrt{3}}=\frac{QM}{MN}\\\\\Rightarrow{MN}=\sqrt{3}\times{QM}\\\\\Rightarrow{MN}^{2}=3\times{QM}^{2}\\\\\Rightarrow{ON}^{2}+OM^{2}=3\times(\frac{QR}{2})^{2}\\\\\Rightarrow{h}^{2}+(\frac{a}{2})^{2}=3\times(\frac{a}{2})^{2}\\\\\Rightarrow{h}^{2}=\frac{3a^{2}}{4}-\frac{a^{2}}{4}\\\\\Rightarrow{h}^{2}=\frac{a^{2}}{2}\\\\\Rightarrow2{h}^{2}=a^{2}

Hence RHS=LHS(proved)


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