A surveying team are trying to find the height of a hill. They take a sight on top of the hill and find that the angle of elevation is 23º27'. They move a distance of 250m on level ground directly away from the base of the hill and take a second sight. From this point, the angle of elevation is 19º46'. Draw a diagram to represent the situation and find the height of the hill to the nearest metre.
Spare the diagram, but how would I solve this?
Pi R Squared
Favorite Answer
Hi,
The height of the hill is 524 metres.
From the first point x metres away from the hill of h metres, the tangent is:
.h.
---- = tan(23+27/60)
.x
This solves to:
h = x * tan(23+27/60)
After moving 250 metres further away then:
.....h
----------- = tan(19+46/60)
x + 250
This solves to:
h = (x + 250)* tan(19+46/60)
h = xtan(19+46/60) + 250tan(19+46/60)
Since both of these expressions equal h, then they equal each other. Set them equal and solve.
x * tan(23+27/60) = xtan(19+46/60) + 250tan(19+46/60)
x * tan(23+27/60) - xtan(19+46/60) = 250tan(19+46/60)
x [tan(23+27/60) - tan(19+46/60)] = 250tan(19+46/60)
.....250tan(19+46/60)
------------------------------------------ = x
tan(23+27/60) - tan(19+46/60)
89.84
--------- = x
.0744
1207.36 = x, which is the distance from the hill to the first observation.
To find h, substitute the value of x.
.h.
------------ = tan(23+27/60)
1207.36
h = 523.72
To the nearest metre, this is 524 metres. <== ANSWER
I hope that helps!! :-)
fredoniahead
Draw your diagram and label the first distance from the hill to the surveyors x, then the distance after they move back is 250+x
The height of the hill is y.
So, equate the two and solve for x, then y will follow.
xtan(23.45)=(250+x)tan(19.77)
xtan(23.45)=250tan(19.77)+xtan(19.77)
xtan(23.45)-xtan(19.77)=250tan(19.77)
x(tan(23.45)-tan(19.77))=250tan(19.77)
x=(250tan(19.77))/(tan(23.45)-tan(19.77))
=1208.67 meters
So, the height of the hill, y, is 1208.67tan(23.45)
=524.29 m