Question

Represent the Situation – I with the help of a diagram.

Answer 1

Let the total height of the statue (including base) be $h + 58$ m.
Let point A be at a distance $AB = 80\sqrt{3}$ m from the base of the statue.
Let point C be the top of the statue, and B be the base of the statue.
Then in $\triangle ABC$, $\angle CAB = 60^\circ$, $AB = 80\sqrt{3}$, and $BC = h + 58$. 
We can represent this scenario with a right triangle diagram:

Answer 2

In this case: - The total height of the Statue (including base) is 240 m. - The observer is standing at a height of 40 m above the ground. - The angle of elevation to the top of the Statue from this point is $30^\circ$. Let: - $C$ be the top of the Statue - $B$ be the base of the Statue - $D$ be the observation point 40 m above the ground - $CD = 240 - 40 = 200$ m - $AD$ be the horizontal distance from point D to the Statue This forms a right triangle $\triangle DCA$ with: \[ \angle D = 30^\circ,\quad \text{opposite side } = 200\text{ m} \]

Answer 3

Using Situation I: Let $h$ be the height of the Statue (excluding the base). Given: Base height = 58 m, distance from point A = $80\sqrt{3}$ m, angle of elevation = $60^\circ$. In $\triangle ABC$, using $\tan\theta$: \[ \tan 60^\circ = \dfrac{h + 58}{80\sqrt{3}},\quad \tan 60^\circ = \sqrt{3} \Rightarrow \sqrt{3} = \dfrac{h + 58}{80\sqrt{3}} \] Multiply both sides: \[ (\sqrt{3})^2 = \dfrac{h + 58}{80} \Rightarrow 3 = \dfrac{h + 58}{80} \Rightarrow h + 58 = 240 \Rightarrow h = \boxed{182\ \text{m}} \] Height including base = $182 + 58 = \boxed{240\ \text{m}}$

Answer 4

From Situation II: Observer is at 40 m above ground, Statue height = 240 m
So, vertical height to be considered = $240 - 40 = 200$ m
Angle of elevation = $30^\circ$ Using: \[ \tan 30^\circ = \dfrac{200}{\text{horizontal distance}},\quad \tan 30^\circ = \dfrac{1}{\sqrt{3}} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{200}{x} \Rightarrow x = 200\sqrt{3} \approx \boxed{346.4\ \text{m}} \] To find $\tan \alpha$ (angle of elevation to top of base from point B): Height from base to point B = $40$ m below base (since base is at 58 m) So vertical difference = $58 - 40 = 18$ m \[ \tan \alpha = \dfrac{18}{346.4} \approx \boxed{0.052} \]