Question

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

Answer 1

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)

Answer 2

"To prove : $a^2=2h^2$

Given : a = Length of each side of the square

             h= the height of the tower standing in the centre of the square

Each side of square makes an angle$60^{o}$ at the top of the tower.

Proof:

P is the top of the tower OP . O is the centre of the square ABCD

$\angle CPB = {60}^{o}$

M is the midpoint of BC

So

$\angle MPB= \angle MPC = {30}^{o}$

In $\triangle MPB$

$cot (\angle MPB)=\frac{PM}{BM}$

$PM= BM \times cot {60}^{o}$

$\sqrt { { OP }^{ 2 }+{ OM }^{ 2 } } =\frac { BC }{ 2 } \times \sqrt { 3 } \\ \sqrt { { h }^{ 2 }+{ \left( \frac { a }{ 2}\right)}^{ 2 } } =\frac { a }{ 2 } \times \sqrt { 3 }$

Squaring on both sides we get

${ h }^{ 2 }+{ \left( \frac { a }{ 2} \right)}^{ 2 }=\frac { 3{ a }^{ 2 } }{ 4 } \\ { h }^{ 2}=\frac { 3{ a }^{ 2 } }{ 4} -\frac { { a }^{ 2 } }{ 4 } =\frac { 2{ a }^{ 2 } }{ 4 } =\frac { { a }^{ 2 }}{ 2} \\ { a }^{ 2 }=2{ h }^{ 2 }$

Hence  proved"