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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 2a be the length of the side of the triangle, then (a^2 /h^2 ) = ?

$$\mathrm{Each}\:\mathrm{side}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$$\mathrm{subtends}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{60}°\:\mathrm{at}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{tower}\:\mathrm{of}\:\mathrm{height}\:{h}\:\mathrm{standing}\:\mathrm{at}\:\mathrm{the}\:\mathrm{centre} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}.\:\mathrm{If}\:\mathrm{2}{a}\:\mathrm{be}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle},\:\mathrm{then}\:\frac{{a}^{\mathrm{2}} }{{h}^{\mathrm{2}} }\:=\:? \\ $$

Answered by mrW1 last updated on 10/Jun/17

s=distance of center point of triangle to the side of triangle s=(1/3)×2a×((√3)/2)=(a/(√3)) h^2 +((a/(√3)))^2 =(2a×((√3)/2))^2 h^2 +(a^2 /3)=3a^2 h^2 =(8/3)a^2 ⇒(a^2 /h^2 )=(3/8)

$$\mathrm{s}=\mathrm{distance}\:\mathrm{of}\:\mathrm{center}\:\mathrm{point}\:\mathrm{of}\:\mathrm{triangle} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{triangle} \\ $$$$\mathrm{s}=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{2a}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{a}}{\sqrt{\mathrm{3}}} \\ $$$$\mathrm{h}^{\mathrm{2}} +\left(\frac{\mathrm{a}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\left(\mathrm{2a}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{h}^{\mathrm{2}} +\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{3a}^{\mathrm{2}} \\ $$$$\mathrm{h}^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{3}}\mathrm{a}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{h}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{8}} \\ $$

Commented by Tinkutara last updated on 10/Jun/17

How do you get line 4?

$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:\mathrm{line}\:\mathrm{4}? \\ $$

Commented by Tinkutara last updated on 10/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Commented by mrW1 last updated on 10/Jun/17

AB=BC=CA=2a ∵TA=TB=TC ∧∠ATB=∠BTC=∠CTA=60° ⇒ΔABT,ΔBCT,ΔCAT=equilateral AT=BT=AB=2a TM=AT×sin 60°=2a×((√3)/2)=(√3)a OM=(1/3)×CM=(1/3)×2a×((√3)/2)=(a/(√3)) OT^2 +OM^2 =TM^2 h^2 +((a/(√3)))^2 =((√3)a)^2 h^2 +(a^2 /3)=3a^2 ⇒(a^2 /h^2 )=(3/8)

$$\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=\mathrm{2a} \\ $$$$\because\mathrm{TA}=\mathrm{TB}=\mathrm{TC}\:\wedge\angle\mathrm{ATB}=\angle\mathrm{BTC}=\angle\mathrm{CTA}=\mathrm{60}° \\ $$$$\Rightarrow\Delta\mathrm{ABT},\Delta\mathrm{BCT},\Delta\mathrm{CAT}=\mathrm{equilateral} \\ $$$$\mathrm{AT}=\mathrm{BT}=\mathrm{AB}=\mathrm{2a} \\ $$$$\mathrm{TM}=\mathrm{AT}×\mathrm{sin}\:\mathrm{60}°=\mathrm{2a}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{3}}\mathrm{a} \\ $$$$\mathrm{OM}=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{CM}=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{2a}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{a}}{\sqrt{\mathrm{3}}} \\ $$$$\mathrm{OT}^{\mathrm{2}} +\mathrm{OM}^{\mathrm{2}} =\mathrm{TM}^{\mathrm{2}} \\ $$$$\mathrm{h}^{\mathrm{2}} +\left(\frac{\mathrm{a}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}\mathrm{a}\right)^{\mathrm{2}} \\ $$$$\mathrm{h}^{\mathrm{2}} +\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{3a}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{h}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{8}} \\ $$

Commented by mrW1 last updated on 10/Jun/17

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