Answer :
Answer:
The height of the hill is 45 feet .
Step-by-step explanation:
Refer the attached figure
Let AB be the height of hill
We are given that He determines that the angle of elevation to the top of the hill is 50°
So, [tex]\angle ACB= 50^{\circ}[/tex]
Now He then walks 40 feet farther from the base from the hill and determines that the angle of elevation to the top of the hill is now 30°
So, CD=40 feet
BD=BC+CD=BC+40
[tex]\ADB= 30^{\circ}[/tex]
In ΔACB
[tex]Tan \theta = \frac{Perpendicular}{Base}\\Tan 50^{\circ} =\frac{AB}{BC}\\1.1917 BC=AB ----1[/tex]
In ΔADB
[tex]Tan \theta = \frac{Perpendicular}{Base}\\Tan 30^{\circ} =\frac{AB}{BD}\\\frac{1}{\sqrt{3}}=\frac{AB}{BC+40}\\\frac{1}{\sqrt{3}}(BC+40)=AB----2[/tex]
So,equate 1 and 2
[tex]1.1917 BC=\frac{1}{\sqrt{3}}(BC+40)\\BC=37.59[/tex]
Substitute the value in equation 1
1.1917 (37.59)=AB
44.796=AB
Hence the height of the hill is 45 feet .
