Thread: Weird Trig Angle of Elevation problem + Azimuth Word Problem

I am having trouble with this one, it seems to work, but then it doesn't.

1.) A tree is 4.0m tall and has permanetly lean to one side. To help stop the lean in your tree you tie a 12.0m rope to the top and and anchor it to the ground at an angle of elevation of 32 degrees.

a.) How far is the rope anchored from the tree?

b.) How bad is the lean of the tree?

Also I want to verify my work for this problem. Not sure if I'm making the right approach.

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2.) In the winter you start skiing away from your cabin and chart your path as 8.0km on a 120 degree Azimuth and then 6.2km on a 40 degree Azimuth.

a.) How far are you from your cabin?

b.) What Azimuth will put you towards your cabin?

First of all, using a protactor I drew a 120 degree angle and then I used my protractor from that point to draw a 40 degree angle.

(Inbetween is what I call Angle C for this problem)

I found Angle C to be 100 degrees

120º- 40º= 80º

180º - 80º= (100º)

then I did Cosine law

c² = 8² + 6.2² - 2 (8)(6.2)cos100
c² = 64 + 38.44 +17.23
(Squared Root sign)c² = (Squared root sign)119.67
c = (10.94km)

So, if I did it CORRECTLY, I have answered a.)

Now for b.), I had to find (Angle B)

I did (*8) sinΘ/8 = sin100/10.94 (*8)

sinΘ = 0.720151921

Θ= sin-1 (0.720151921)

Θ= (46.1º) (answer to b.)

Am I on the right path for this problem?

1.) Has no Solution, height isn't tall enough in porportion to the other known side of the triangle.

Originally Posted by (?)G

I am having trouble with this one, it seems to work, but then it doesn't.

1.) A tree is 4.0m tall and has permanetly lean to one side. To help stop the lean in your tree you tie a 12.0m rope to the top and and anchor it to the ground at an angle of elevation of 32 degrees.

a.) How far is the rope anchored from the tree?

b.) How bad is the lean of the tree?

Also I want to verify my work for this problem. Not sure if I'm making the right approach.

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2.) In the winter you start skiing away from your cabin and chart your path as 8.0km on a 120 degree Azimuth and then 6.2km on a 40 degree Azimuth.

a.) How far are you from your cabin?

b.) What Azimuth will put you towards your cabin?

First of all, using a protactor I drew a 120 degree angle and then I used my protractor from that point to draw a 40 degree angle.

(Inbetween is what I call Angle C for this problem)

I found Angle C to be 100 degrees

120º- 40º= 80º

180º - 80º= (100º)

then I did Cosine law

c² = 8² + 6.2² - 2 (8)(6.2)cos100
c² = 64 + 38.44 +17.23
(Squared Root sign)c² = (Squared root sign)119.67
c = (10.94km)

So, if I did it CORRECTLY, I have answered a.)

Now for b.), I had to find (Angle B)

I did (*8) sinΘ/8 = sin100/10.94 (*8)

sinΘ = 0.720151921

Θ= sin-1 (0.720151921)

Θ= (46.1º) (answer to b.)

Am I on the right path for this problem?

The numbers in Question 1 do not make sense.

For Q2.
Your return distance is ok.
Your theta angle is ok, BUT you must add that to the reverse azimuth of the 40 degree azimuth line.

The reverse azimuth (40+180) is 220 degrees.

It says use a 12m rope to create what amounts to the hypotenuse of a right triangle. In order to come up with a solution you have to make some assumptions, that are unstated in the problem givens:

The tree is bent and we're given the 32 degree angle between the ground and the hypontenuse. This means that the length of the hypontenuse is not given at 12m, it means that it's less than or equal to 12m, so we solve for the length of the hypotenuse first, then we can solve for the anchoring distance from the tree.

sin(32) = 4/hypontenuse

rearranging: hypontenuse = 4/sin(32)

hypotenuse length then = 7.548m

there are now a couple of ways to solve for the distance between the tree and anchoring point of the rope. pythagorean theorem is the easiest: a^2 + b^2 = c^2.

let a = 4m, and c=7.548m.

solving for b then = (c^2 - a^2)^.5
putting in the numbers: (7.548^2 - 4^2)^.5 = 6.4m