Expert: Paul Klarreich - 2/9/2010

Question
for these two problems, the biggest part that i need help on is drawing the picture...

the directions are to solve the problem

Q: A guy wire to a tower makes a 67* angle with level ground. At a point 33 ft farther from the tower than the wire but on the same side of the base as the wire, the angle of the elevation to the top of the pole is 36*. Find the wire lenght (to the nearest foot)

Q2: Poiints A and B are on opposite sides of a lake. A point C is 84.5 meters from A. The measure of angle BAC is 79*20' and the measure of ACB is determined to be 33*10'. Find the distance between points A and B (to the nearest meter).

then the directions for these next problems are, Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as an angle is [0,360*]

Q1: {0,-11}
Q2: {√5,-1}

the find the angle between the given vectors to the nearest tenth of a degree
a = {2,-8} and b= {-9,-4}
and Q2: a=3i+8j, b=-6i+7j

Answer
Questioner: dillon
Country: United States
Category: Advanced Math
Private: No
Subject: trig
Question: for these two problems, the biggest part that i need help on is drawing the picture...

the directions are to solve the problem

Q: A guy wire to a tower makes a 67* angle with level ground. At a point 33 ft farther from the tower than the wire but on the same side of the base as the wire, the angle of the elevation to the top of the pole is 36*. Find the wire lenght (to the nearest foot)

>> I gather you are using  '*' for degrees.  I like it.

>> See the diagram for this.  You want R, and can find it with the law of sines.

  R         33
------- = -------
sin 36*   sin 31*



Q2: Poiints A and B are on opposite sides of a lake. A point C is 84.5 meters from A. The measure of angle BAC is 79*20' and the measure of ACB is determined to be 33*10'. Find the distance between points A and B (to the nearest meter).

>> I don't think you really need a diagram.  You have triangle ABC, in which:

angle A = 79*20'
angle C = 33*10'
and you can now find angle B.
b = AC = 84.5

Find c = AB

Use the law of sines again:
  b        c
------- = ------
sin B     sin C  

.....................................
then the directions for these next problems are, Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as an angle is [0,360*]

Q1: {0,-11}
Q2: {√5,-1}

>>>>>>>>>>>>
The magnitude is r, where  x^2 + y^2 = r^2, so that should be easy.

The direction angle  theta, satisfies  tan theta = y/x,
and you just have to work out the quadrant yourself.

For Q1:  this is just on the y-axis, so it looks like 3pi/2.
For Q2: this is in Quad 4, so expect theta between 3pi/2 and 2pi.

..........................
the find the angle between the given vectors to the nearest tenth of a degree
a = {2,-8} and b= {-9,-4}
and Q2: a=3i+8j, b=-6i+7j

>>>>>>>>>>>
Do the same stuff -- find the direction angles.  Then subtract.

(Do you know about dot-products yet?  If so, there is another way.)