Each-side-of-an-equilateral-triangle-subtends-angle-of-60-at-the-top-of-a-tower-of-height-h-standing-at-the-centre-of-the-triangle-If-2a-be-the-length-of-the-side-of-the-triangle-then-a-2-h-2-

Question Number 15392 by Tinkutara last updated on 10/Jun/17

Each side of an equilateral triangle

subtends angle of 60 ° at the top of a

tower of height h standing at the centre

of the triangle . If 2 a be the length of the

side of the triangle, then \frac{a^{2}}{h^{2}} = ?

Answered by mrW1 last updated on 10/Jun/17

s = distance of center point of triangle

to the side of triangle

s = \frac{1}{3} \times 2 a \times \frac{\sqrt{3}}{2} = \frac{a}{\sqrt{3}}

h^{2} + {(\frac{a}{\sqrt{3}})}^{2} = {(2 a \times \frac{\sqrt{3}}{2})}^{2}

h^{2} + \frac{a^{2}}{3} = {3 a}^{2}

h^{2} = \frac{8}{3} a^{2}

\Rightarrow \frac{a^{2}}{h^{2}} = \frac{3}{8}

Commented by Tinkutara last updated on 10/Jun/17

How do you get line 4 ?

Commented by Tinkutara last updated on 10/Jun/17

Thanks Sir!

Commented by mrW1 last updated on 10/Jun/17

AB = BC = CA = 2 a

∵ TA = TB = TC \land ∠ ATB = ∠ BTC = ∠ CTA = 60 °

\Rightarrow Δ ABT, Δ BCT, Δ CAT = equilateral

AT = BT = AB = 2 a

TM = AT \times \sin 60 ° = 2 a \times \frac{\sqrt{3}}{2} = \sqrt{3} a

OM = \frac{1}{3} \times CM = \frac{1}{3} \times 2 a \times \frac{\sqrt{3}}{2} = \frac{a}{\sqrt{3}}

{OT}^{2} + {OM}^{2} = {TM}^{2}

h^{2} + {(\frac{a}{\sqrt{3}})}^{2} = {(\sqrt{3} a)}^{2}

h^{2} + \frac{a^{2}}{3} = {3 a}^{2}

\Rightarrow \frac{a^{2}}{h^{2}} = \frac{3}{8}

Commented by mrW1 last updated on 10/Jun/17

Facebook Tweet Pin