Draw a diagram for the following situation:
The shadow of a tower standing on a level ground is found to be 40 m longer when
Sun’s altiude is than when it is . Find the height of the tower.

Ask by Ross Bradley. in India

Mar 24,2025

Question

Draw a diagram for the following situation:
The shadow of a tower standing on a level ground is found to be 40 m longer when
Sun’s altiude is than when it is . Find the height of the tower.

Ask by Ross Bradley. in India

Mar 24,2025

UpStudy AI · Accepted Answer

$$
\begin{array}{ccccccc}
& & \text{Sun ray (at } 30^\circ \text{)} & & \text{Sun ray (at } 60^\circ \text{)} & & \
& \diagdown & & \diagdown & & \diagdown & \
\text{Sun} & \longrightarrow & \ \ \ \ \rule{0pt}{2.5ex} & \ \ \ \ \rule{0pt}{2.5ex} & & \ \ \ \ \rule{0pt}{2.5ex} & \longrightarrow \ \ \ \text{Sun} \
& & \
& & \bullet & & \bullet & & \
& & | \quad (\text{Tower of height } h) & & | \quad (\text{Tower of height } h)\
& & | & & |\
& & | & & |\[1ex]
\text{Point } A &\quad \longrightarrow \quad & \text{Shadow tip } B & &\quad \longrightarrow \quad & \text{Shadow tip } C & \
& & (s+40) \text{ m} & & & s \text{ m} & \
\end{array}
$$

Here, the vertical line represents the tower of height $$ h $$. At a Sun altitude of $$ 30^\circ $$ the shadow on the level ground is longer, and its length is $$ s + 40 $$ m, while at a Sun altitude of $$ 60^\circ $$ the shadow length is $$ s $$ m.

Since the Sun's rays create a right triangle with the tower and its shadow, we have:
- For the Sun altitude $$ 30^\circ $$:
  $$
  \tan 30^\circ = \frac{h}{s+40}
  $$
- For the Sun altitude $$ 60^\circ $$:
  $$
  \tan 60^\circ = \frac{h}{s}
  $$

We know that:
$$
\tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \tan 60^\circ = \sqrt{3}
$$

Thus, we obtain:
1. $$\displaystyle \frac{1}{\sqrt{3}} = \frac{h}{s+40} \quad \Longrightarrow \quad s+40 = h\sqrt{3}$$
2. $$\displaystyle \sqrt{3} = \frac{h}{s} \quad \Longrightarrow \quad s = \frac{h}{\sqrt{3}}$$

Substitute the value of $$ s $$ from the second equation into the first:
$$
\frac{h}{\sqrt{3}} + 40 = h\sqrt{3}
$$

Solve for $$ h $$:
$$
40 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right)
$$

Simplify the expression inside the parentheses:
$$
\sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}}
$$
Thus,
$$
40 = h \cdot \frac{2}{\sqrt{3}} \quad \Longrightarrow \quad h = 40 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3}
$$

Therefore, the height of the tower is:
$$
\boxed{20\sqrt{3} \text{ m}}
$$
The height of the tower is $$ 20\sqrt{3} $$ meters.

Draw a diagram for the following situation:
The shadow of a tower standing on a level ground is found to be 40 m longer when
Sun’s altiude is than when it is . Find the height of the tower.

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Draw a diagram for the following situation: The shadow of a tower standing on a level ground is found to be 40 m longer when Sun’s altiude is than when it is . Find the height of the tower.

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Latest Trigonometry Questions

Draw a diagram for the following situation:
The shadow of a tower standing on a level ground is found to be 40 m longer when
Sun’s altiude is than when it is . Find the height of the tower.