(quite) Simple Trig Problem?

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!! :-)

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