Maths problem.........urgent plz help?

Question

In travelling acroos flat land,you notice a mountain directly in front of you.It's angle of elevation(to the peak) is 3.5 degree.After you have drive 13 miles closer to the mountain,the angle of elevation is 9 degree.Approximate the height of mountain.

Plz show me the method how to solve that

Answer 1

This means that if y is the height of the mountain and x is you distance from the mountain, tan(3.5)=y/x. Use the fact that when you are 13 miles closer, the angle is now 9 degrees: tan(9)=y/(x-13). Tan(3.5)=.0611 and tan(9)=.1583
Therefore, .0611=y/x ......y=.0611x....now substitute this into the other relationship: .1583=.0611x/(x-13).... .1583x-2.0589=.0611x
.0972x=2.0589 x=21.182 miles y=1.29 miles

Therefore, the mountain is 1.l29 miles height.

Answer 2

Draw a sketch and find the trig relation that you need.

The height stays the same. In both cases it's equal to dist times tangent of angle.

d*tan 3.5 = (d-13)*tan 9

Get d, then calc height from either eqn.

Answer 3

draw triangle x=dist from mt at 9 degree sighting
h/x=tan 9 degrees
h/(x+13)=tan 3.5 degrees

solve system for h

you do that

Answer 4

let the initialdistancebeteen you and the mountain be x
the equation is h/x=tan(3.5*)
=>h=xtan(3.5*)
after you have driven 13 miles theequation is
h/(x-13)=tan9*
=>h=(x-13)tan9*
comparing the two equations
xtan(3.5*)=(x-13)tan9*
x[tan9*-tan(3.5*)]=13tan9*
x=13tan9*/[tan9*-tan(3.5*)]
substituting in the first equation
h=[13tan9*/(tan9*-tan(3.5*))]tan(3.5*)

Answer 5

45893 feet