Question

During a certain time of the day, the sun is 60° above the horizontal. What is difference between the heights (in metres) of two building that cast shadows of 17 m and 15 m respectively during that time?

• A. $32\sqrt{3}$
• B. $2\sqrt{3}$
• C. 2
• D. $2/\sqrt{3}$

Answer 1

Answer: (B)

$2\sqrt{3}$

Answer 2

The problem involves trigonometry, specifically the concept of angles of elevation and right-angled triangles.

Let $h$ be the height of a building and $s$ be the length of its shadow. Let $\theta$ be the angle of elevation of the sun above the horizontal.

From the geometry, the building, its shadow, and the line of sight to the sun form a right-angled triangle. In this triangle:

• The height of the building ($h$) is the side opposite to the angle $\theta$.
• The length of the shadow ($s$) is the side adjacent to the angle $\theta$.

Therefore, the relationship between $h$, $s$, and $\theta$ is given by the tangent function:

$$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{s}$$

From this, the height of the building can be expressed as:

$$h = s \tan \theta$$

Given:

• Angle of elevation of the sun, $\theta = 60^\circ$.
• Length of the shadow of the first building, $s_1 = 17 \text{ m}$.
• Length of the shadow of the second building, $s_2 = 15 \text{ m}$.

First, calculate the height of the first building ($h_1$):

$$h_1 = s_1 \tan 60^\circ$$

We know that $\tan 60^\circ = \sqrt{3}$.

$$h_1 = 17 \times \sqrt{3} = 17\sqrt{3} \text{ m}$$

Next, calculate the height of the second building ($h_2$):

$$h_2 = s_2 \tan 60^\circ$$ $$h_2 = 15 \times \sqrt{3} = 15\sqrt{3} \text{ m}$$

Finally, find the difference between the heights of the two buildings:

$$\text{Difference} = h_1 - h_2$$ $$\text{Difference} = 17\sqrt{3} - 15\sqrt{3}$$ $$\text{Difference} = (17 - 15)\sqrt{3}$$ $$\text{Difference} = 2\sqrt{3} \text{ m}$$

The difference between the heights of the two buildings is $2\sqrt{3}$ metres.