trigonometry problem?

Question

You are in a point on floor from where u see the top of a tree with an angle of 60º, then you walk backwards 10 meters and the angle is 30º. What is the tall of the tree and how do you get it?

Maybe this case is guessable with mental methods, but I need a procedure cuz not all the problems have easy numbers.

Where are you geting the sqr(3) from??

Answer 1

Let x meters be the distance of the first point (.i.e. the point from where the angle of elevation is 60 degrees) from the tree and let h meters be the height of the tree.

tan Θ = opposite side / adjacent side
=> tan 60 = h / x and tan 30 = h / (x + 10)
=> h = x tan 60 and h = (x + 10) tan 30
=> h = x sqrt(3) and h = (x + 10) / sqrt(3)

Equating R.H.S of both the equations
x sqrt(3) = (x + 10) / sqrt(3)

Solving
3x = x + 10
3x - x = 10
2x = 10
x = 5 meters

therefore height of tree,
h = x sqrt(3)
h = 5sqrt(3) meters

we get sqrt(3) from
tan 30 = 1 / sqrt(3) and tan 60 = sqrt(3)
These are standard values.

Hope it helps :).

Answer 2

Alright, here's how to do it....

At first you are at some position "x" from the tree looking up with an angle of 60 degrees. Now when you move back 10 meters you are actually adding 10 meters to your previous position "x". Also, keep in mind that the height of the tree stays constant so you can label that height y. Now we can create two equations. One for the initial postion and one for the new position. The two equations are

tan(30) = y / (10 + x)

tan(60) = y / x

solving for y in each we get

(10 + x ) tan(30) = y

x tan(60) = y

Equate these to each other and solve for x

x = 5

plug x back into any of the previous equations for y
and you will have your height which comes to

y = 5*sqrt(3)

Hope this helps!

Answer 3

The trigonometric function that you want to use is tangent, or tan. The opposite side is constant, and the adjacent side and angle changes.

Things you know:

tan(60)=X/Y, where X is the height of the tree, and Y is your distance.
tan(30)=X/(Y+10).

Remember SohCahToa (or Oprah Has A Huge Old A$$), and think about what details you can put into your relations.

Answer 4

Easier than it looks.
let A be the initial observation point, B be the bottom of the tree, C be the top of the tree & D be the 2nd observation point
Given:
angle BAC=60
angle BDC=30
AD=10m

angle DAC is the suplement to angle BAC so DAC=120, therefore DCA=30. since DCA=BDC, AD=AC=10
ABC is a right angle, so, since BAC=30, ACB=60 so ABC is a 30,60,90 rt triangle. AB=AC/2=5m
BC=AC*sqrt(3)/2=5sqrt(3)=8.66m=height of the tree.

Answer 5

draw a diagram, then figure out all of the angles. you can use the ratio of the sides of 30:60:90 triangles and the properties of isosceles triangles to figure this out. the answer is 5 times the square root of 3

Answer 6

tan(60) = y/x
tan(30) = y/(x + 10)

tan(60) = (y/x)
y = tan(60)x

tan(30) = y/(x + 10)
y = tan(30)(x + 10)

tan(60)x = tan(30)(x + 10)
tan(60)x = tan(30)x + 10tan(30)
tan(60)x - tan(30)x = 10tan(30)
x(tan(60) - tan(30)) = 10tan(30)

tan(60) = sqrt(3)
tan(30) = (sqrt(3))/3

x(sqrt(3) - (sqrt(3)/3)) = 10tan(30)
x((3sqrt(3) - sqrt(3))/3) = 10tan(30)
x(2sqrt(3)/3) = 10tan(30)
(2sqrt(3))/3 * x = 10tan(30)
2sqrt(3)x = 30tan(30)
sqrt(3)x = 15tan(30)
x = (15tan(30))/(sqrt(3))
x = (15 * (sqrt(3)/3))/(sqrt(3))
x = ((15sqrt(3))/3)/(sqrt(3))
x = (5sqrt(3))/(sqrt(3))
x = 5

tan(60) = y/5
y = 5tan(60)
y = 5sqrt(3)

The tree is 5sqrt(3) meters tall or about 8.66 meters tall

Answer 7

You have your ASA congruence. You can get the third angle. Do you use the Law of Sines? or the Law of Cosines?

Answer 8

5*sqrt(3) is the answer