A hard trigonometry question...?

Question

Could anybody please help me as to what the working out would be to answer this question? Thanks in advance:

A hedgehog wishes to cross a road without being run over. He observes the angle of elevation of a lamp post on the other side of the road to be 27 from the edge of the road and 15 from a point 10m back from the road. How wide is the road? If he can run 1m/s, how long will he take to cross?


Thanks

Answer 1

Let h = height of lamp post
Let x = width of road

tan(27) = h/x ....... so h = x*tan(27)
tan(15) = h/(x+10).... so h = (x+10)tan(15)

Then x*tan(27) = (x+10)tan(15)
x*tan(27) = x*tan(15) + 10*tan(15)
x[tan(27) - tan(15)] = 10*tan(15)
x = 10*tan(15) / [tan(27) - tan(15)]
x = 11.0917 meters

So if he runs at 1m/s, it takes (barely over) 11 seconds to cross the road.

I hope this helps!

Answer 2

Hedgehogs can't measure angles!

But if it was a person measuring these angles accurately, then it can be solved.

Let w = width of the road and h = height of the lamp post. Then,

tan 27 = h/w
and tan 15 = h/(w + 10). So,

h = w*tan 27 = (w + 10)*tan 15 = w*tan 15 + 10*tan 15. So,

w = 10*tan 15/(tan 27 - tan 15) which approx = 11.09 m.

So the width of the road is 11.09 meters. Since the animal runs at 1 m/s, it will take 11.09/1 = 11.09 secs.

Answer 3

You find this information:
tan 15=h/(10+ x), h is the hight of the light, x is the whith of the road.

And you find this information: tan 27= h/x

Out of this you calculate: x= 10,8m

Then a hedgehog need 10.8 sconds to cross the road.

Answer 4

Let elevation of lampost be h and width of road to cross be x
h = x tan 27 -- (i)
h = (x+10) tan 15 --- (ii)
So x = 10. tan 15 / (tan 27 - tan 15)

Answer 5

they should be two right triangles given that the lamp post goes strait up.
Tan x = opposite/adjacent
Tan 27 = (height of lamp post)/(road distance)
Tan 15 = (height of lamp post)/(road distance+10)
you can cancel to solve for (road distance) because the (height of lamp post) can cancel out.

Answer 6

why did the hedgehog cross the road? to see his flat mate...

hogs are nocturnal, is the lam post lit? they cannot gauge distance, and use a system of mapping similar to GPS units...

and at 1m/s thats bloody fast... 60m P/m is it an olympic contender?

oh, we run a hedgehog hospital... and when our hedgehogs want to cross the road, we pick them up and carry them. they cant do math, and dont understand a calculator... although hogs do have opposing thumbs... they have no road sense.

Answer 7

It's kind of hard to explain without a diagram, the hypotenuse of the small right triangle is x. the base of the small right triangle is y.

Use law of sines on the small obtuse triangle.
(sin 12)/10 = (sin 15)/x
x = (10 sin 15)/(sin 12)
sin 117 = y/x
y = x sin 117
y = (10 sin 15 sin 117)/(sin 12)
= 11.09170229m
He takes 11.0917 seconds to cross.

Answer 8

This is the fastest method I see.
Let h = hypothenuse of triangle formed when he is at the edge of the road.

Using the sine rule
10 / sin12 = h / sin15
h = 10*sin15 / sin12
x / h = cos27
x = 10*sin15*cos27 / sin12 = 11.09 m

It will take him 11.09 sec to run across.

Answer 9

I took Algebra a million in 8th grade...and that i'm taking Geometry this year...i could think of it truly is exceptionally difficult. i'm taking Algebra 2 next year. Geometry comes after Algebra a million right here.

Answer 10

If you need to, draw a picture. It really will help you visualize the problem.

You have a total of three triangles here... two are right and one is obtuse
http://i20.photobucket.com/albums/b249/LindySoul/mathematics/problem.jpg