Answer:
Length of the garden = 69.28 meters.
Step-by-step explanation:
See the attached diagram.
Let CD is the pole with a height of 30 m and the elevation of the top of the pole from the endpoints of the length A and B are 60° and 30° respectively.
So, ∠ DAC = 60° and ∠ DBC = 30°
Now, Δ ACD is a right triangle and [tex]\tan 60 = \frac{CD}{AC}[/tex]
⇒ [tex]AC = \frac{CD}{\tan 60} = \frac{30}{\tan 60} = 17.32[/tex] meters.
Now, from Δ BCD which is a right triangle, [tex]\tan 30 = \frac{CD}{CB}[/tex]
⇒ [tex]CB = \frac{30}{\tan 30} = 51.96[/tex] meters.
Hence, AB = AC + CB = 17.32 + 51.96 = 69.28 meters. (Answer)
Answer:
The length of the rectangle garden is 20 [tex]\sqrt{3}[/tex] i.e 34.64 meters .
Step-by-step explanation:
Given as :
The height of the pole = H = 30 m
The two angles of elevation as 60° and 30°
Let The length of the rectangle garden = L m
Now, From figure
The measure from pole ground to first elevation 60° = x m
The measure from pole ground to second elevation 30° = (L + x ) m
Now, from triangle BOC
Tan Ф = [tex]\dfrac{\textrm perpendicular}{\textrm base}[/tex]
I.e Tan 60° = [tex]\dfrac{\textrm 30}{\textrm x}[/tex]
Or, [tex]\sqrt{3}[/tex] = [tex]\dfrac{\textrm 30}{\textrm x}[/tex]
or, x = [tex]\frac{30}{\sqrt{3} }[/tex]
I.e x = 10[tex]\sqrt{3}[/tex] meter ...1
Again From triangle BAC
Tan Ф = [tex]\dfrac{\textrm perpendicular}{\textrm base}[/tex]
I.e Tan 30° = [tex]\dfrac{\textrm 30}{\textrm L + x}[/tex]
Or, [tex]\frac{1}{\sqrt{3} }[/tex] = [tex]\dfrac{\textrm 30}{\textrm L + x}[/tex]
Or, L + x = 30 × [tex]\sqrt{3}[/tex] ....2
Put the value of x from 1 into 2
i.e L + 10[tex]\sqrt{3}[/tex] meter = 30 × [tex]\sqrt{3}[/tex]
or, L = 30 [tex]\sqrt{3}[/tex] - 10 [tex]\sqrt{3}[/tex]
Or, L = 20 [tex]\sqrt{3}[/tex] = 34.64 meters
I.e length of garden is 20 [tex]\sqrt{3}[/tex]
Hence The length of the rectangle garden is 20 [tex]\sqrt{3}[/tex] i.e 34.64 meters . answer
