Question

can anyone say this ans please..................................





A 1.5 m tall boy is looking at the top of a temple which is 30 meter in height from a point
at certain distance. The angle of elevation from his eye to the top of the crown of the
temple increases from 30° to 60° as he walks towards the temple. Find the distance he
walked towards the temple.
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Answer 1

Answer:

$\bigstar{\bold{Distance\:walked=19\sqrt{3} \:m}}$

Step-by-step explanation:

$\Large{\underline{\rm{Given:}}}$

• Height of the boy = 1.5 m
• Height of the temple = 30 m
• Angle of elevation increases from 30° to 60°

$\Large{\underline{\rm{To\:Find:}}}$

• Distance he walked towards the temple

$\Large{\underline{\rm{Solution:}}}$

➜ Let the height of the boy be AB = 1.5 m

➜ Let the height of the tower be AC = 30 m

➜ Hence distance of AB = 30 - 1.5 = 28.5 m

➜ Consider Δ ABE

    tan 30 = AB/EB

    tan 30 = 28.5/EB

    1/√3 = 28.5/EB

    EB = 28.5√3

➜ Now consider Δ ABF

    tan 60 = AB/FB

    tan 60 = 28.5/FB

    √3 = 28.5/FB

     FB = 28.5/√3

     FB = 9.5√3

➜ Now distance the boy walked toward the temple = EF

   EF = EB - FB

➜ Substitute the data,

    EF = 28.5√3 - 9.5√3

    EF = 19√3 m

➜ Hence the boy walked 19√3 m towards the temple.

    $\boxed{\bold{Distance\:walked=19\sqrt{3} \:m}}$

$\Large{\underline{\rm{Notes:}}}$

➜ Sin A = opposite/hyptenuse

➜ Cos A = adjacent/hypotenuse

➜ Tan A = opposite/adjacent