Trig Problems for physics course

[pooface]

I am a very good math student and trig is something that clicks to me. However, my main problem is actually understanding the question. There are two questions which I don't understand from an English point of view.

Homework Statement

Question 1: A man surveying a mine measures a length AB=220 m, due east with a dip of 6 degrees and 45', then a length BC = 325 m due south with a dip of 10 degrees 15'. How much deeper is C than A?

I don't know what 'dip' means, but I am assuming it means 'angle of depression'.

Question 2: At a certain point the angle of elevation of a mountain peak is 48 deg 30'; at a distance of three km farther away in the same horizontal plain its angle of elevation is 39 deg 45'. Find the distance of the top of the mountain above the horizontal plain and the horizontal distance from the first point of observation to the peak.

Homework Equations

cosine law.

pythagorean theorem.

The Attempt at a Solution

Question 1: See the q1 attached picture. I deduced, using imaginary right angles for the angle of B to be 86.5 degrees. Then using cosine law I got the length of AC to be 381.176 m. So this means C is 381.176m deeper than A. I understand that the diagram is actually in 3D-Space. Was my approach correct? Were my angles of depression correctly placed?

Question 2: See the q2 attached picture. I solved all the missing angles and sides a, b, c.
but I have no idea what the question is asking me for. It is very confusing.

my a=8.356 km, b=9.444 km and c=12.61 km.

Thank you.

 

[learningphysics]

In the first problem... it seems like north south east west are in the same plane... and the dips are perpendicular to the plane... Was a picture given in the problem, or are you attaching your own pictures?

How much deeper is B than A? How much deeper is C than B... just add the two numbers to get much deeper C is than A. so it seems to me like, north, south, east and west don't matter... they are all horizontal. unless I'm misunderstanding something.

For the second problem you have a right triangle with an angle of 48 deg 30'. Another right triangle has the same height with an angle of 39 deg 45' (the two triangles have a common side - the height which is opposite the 39 deg 45' and 48 deg 30'). This is all in 2D.

Here's a rough picture of the way I understand it.

x
x
x
x
x
x
x
xxxxxxxxxxxxx First spot xxxxxxxxxxx Second spot

Find the height... then find the horizontal distance from the first point... ie: find the other leg of the 48 deg 30' triangle.

 

[HallsofIvy]

In your picture, you seem to have interpreted "dip" as meaning "south of east". I would interpret it to mean what you initially said, "angle of depression"- downward.

 

[pooface]

Thank you for responding. Both images were drawn/interpreted by me.

learningphysics. Thank you. At least I know that my diagram for the second question was correct. I just didn't know what I was supposed to solve for.

As for the first question, my diagram is kind of misleading.

I am understanding the question in terms of 3D-Space. You walk 220m east on a downwards slope defined by the dip then stop and go south on a downwards slope defined by the other dip then stop after 325m. From this point to where you originally started should be the length of depth between the two points(the origin A and C), is it not?

If that is correct that this forms a triangle in 3D space. I am not sure if it is a right angle triangle. I may have determined angle B incorrectly.

 

[learningphysics]

The depth isn't the distance between the initial and final point... it's the vertical distance...

What is the depth from A to B?

 

[pooface]

ahh yes. Now i fully understand.

Just to make sure, ok I got the depth from A to B as 25.86m determined from sin6.75(220m).

Then depth from B to C is sin10.25(325m) = 57.832m

Adding the two depths= approx. 83.692m

Have I done it right this time?

Thank you very much.

 

[learningphysics]

Yup. That's right.