Question

The tops of two towers x and y standing on level ground, subtend angles of 30 degree and 60 degree respectively at the center of the line joining their feet, then find x:y.

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

The figure is in attachment.

Now, In ∆ ABC,

$\mapsto\tt{\dfrac{AB}{BC}=tan 30}$

$\mapsto\tt{\dfrac{x}{BC}=tan 30}$

$\mapsto\tt{\pink{x= tan 30 \times BC}}$

Now, In ∆ DCE,

$\mapsto\tt{\dfrac{DE}{CD}=tan 30}$

$\mapsto\tt{\dfrac{y}{CD}=tan 60}$

$\mapsto\tt{\pink{y= tan 60 \times CD}}$

Now, finding the ratio of X to y.

$\leadsto\tt{\dfrac{x}{y}=\dfrac{tan 30\times BC}{tan 60 \times CD}}$

Now, from the question we can that CD=BC.

So, $\leadsto\tt{\dfrac{x}{y}=\dfrac{tan 30\times \cancel{BC}}{tan 60 \times \cancel{CD}}}$

Now, $\leadsto\tt{\dfrac{x}{y}=\dfrac{\dfrac{1}{\sqrt{3}}}{\sqrt{3}}}$

$\leadsto\tt{\dfrac{x}{y}=\dfrac{1}{\sqrt{3}\times\sqrt{3}}}$

$\leadsto\tt{\dfrac{x}{y}=\dfrac{1}{3}}$

So, $\huge\pink{\boxed{x:y = 1:3}}$

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Formulas used:-

• $\tt{tan \theta=\dfrac{Perpendicular}{Base}}$

• $\tt{tan\:30=\dfrac{1}{\sqrt{3}}}$

• $\tt{tan\:60=\sqrt{3}}$

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Answer 2

Given: The tops of two towers x and y subtend angles of 30 and 60 degree at the center.

To find: x:y.

Answer:

(Diagram for reference attached below.)

In Δ ABC,

$\sf tan\ {30}^{\circ}\ =\ \dfrac{AB}{BC}\\\\\\\dfrac{1}{\sqrt{3}}\ =\ \dfrac{x}{BC}\\\\\\\\x\ =\ \dfrac{BC}{\sqrt{3}}$

In Δ EDC,

$\sf tan\ {60}^{\circ}\ =\ \dfrac{ED}{DC}\\\\\\\sqrt{3}\ =\ \dfrac{y}{DC}\\\\\\\\y\ =\ DC\sqrt{3}$

Now, as per the question we need to find x:y. Using the values we have,

$\sf \dfrac{x}{y}\ =\ \dfrac{\dfrac{BC}{\sqrt{3}}}{DC\sqrt{3}}\\\\\\\\\dfrac{x}{y}\ =\ \dfrac{BC}{\sqrt{3}}\ \times\ \dfrac{1}{DC\sqrt{3}}$

From the question, we can tell that BC = DC. Hence,

$\sf \dfrac{x}{y}\ =\ \dfrac{BC}{\sqrt{3}}\ \times\ \dfrac{1}{BC\sqrt{3}}\\\\\\\dfrac{x}{y}\ =\ \dfrac{1}{3}$

Therefore, x:y = 1:3.