Question

From a point on the ground, 40m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The angle of elevation of the top of a water tank placed on the tower is 45°. Find the height of the tower.

Answer 1

Hey there!!

→ Let BC be the height of tower and CD be the height of the water tank.

→ Let A be the point of observation.

▶ Then, → $\angle BAC$ = 30°.

→ $\angle BAD$ = 45°

→ And AB = 40m.

▶ From right ∆ABD, we have

=> $\frac{BD}{AB}$ = tan 45°.

=> $\frac{BD}{40 m}$ = 1.

[ => tan 45° = 1 ]

=> BD = 40m.

▶ From right ∆ABC, we have

=> $\frac{BC}{AB}$ = tan 30°.

=> $\frac{BC}{40 m} = \frac{1}{ \sqrt{3} } .$

[=> tan 30° = $\frac{1}{ \sqrt{3} } .$

=> BC = $\frac{40 m }{ \sqrt{3} } .$

$= > BC = \frac{40m}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } .$

=> BC = $\huge \boxed{ \boxed{ \frac{40 \sqrt{3} }{ \sqrt{3} } m }}.$

✔✔ Hence, height of the tower = BC = $\frac{40 \sqrt{3} }{ \sqrt{3} } m .$ = 23.1 m. ✅✅

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$\huge \boxed{ \mathbb{THANKS}}$


$\huge \bf{ \#BeBrainly.}$

Answer 2

$\bold{ \mathbb{ACCORDING \: \: TO \: \: THE \: QUESTION:-}}$


From a point on the ground, 40m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The angle of elevation of the top of a water tank placed on the tower is 45°. Find the height of the tower.



$\huge{ \mathbb{SOLUTION:-}}$


Let the height of the tower above the foot of the tower ( h) m



Angle of elevation of the tower top from a point= 40 m
And



Its away from the foot of tower = 30°



According the Situation:-




h/40 = Tan 30°



Substitute the value of Tan 30°



=> h/40=1/√3



=> √3h=40



=> h=40/√3




Now,



Rationalizing The Height of the tower into Simplest Form:-

$\implies \bold{ \frac{40 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} }} \\ \\ \\ \\ \ \implies \bold{\frac{ \cancel40 \sqrt{3} }{ \cancel3}} \\ \\ \\ \\ \implies \bold{ 13.3 \times 1.732 \implies23.09}$


Therefore



Height of the tower = 23.09 metres