Vertical post

[IloveManUtd]

The points P, Q and R are on level ground such that Q is due north of P. The bearing of R from P is 018[sup:mloc695q]o[/sup:mloc695q] and the bearing of R from Q is 063[sup:mloc695q]o[/sup:mloc695q]. Given that the vertical post XQ is 32m, and that the length of PQ is 250m, calculate the angle of elevation of X from P?

I do not understand the question and am wondering how the answer my book gives is 7.3[sup:mloc695q]o[/sup:mloc695q]. Please help. Thx

 

[Denis]

Time for you to start showing your work.
This is NOT a "do your homework" site.
Plus your diagram is not clear; what equals 250 m ?

 

[mmm4444bot]

calculate the angle of elevation of X from P?

I do not understand the question

It seems to me that they're asking for the measure of angle QPX.

But we don't need any of the given bearings to find angle QPX, so I'm not sure. EDIT: Yes, that comes out to 7.3 degrees

All we need is the Pythagorean Theorem and the Law of Cosines, working with the right-triangle QPX.

 

[IloveManUtd]

Re:

It seems to me that they're asking for the measure of angle QPX.
All we need is the Pythagorean Theorem and the Law of Cosines, working with the right-triangle QPX.

Can someone show me how to do it? THANKS!

 

[mmm4444bot]

Do you know how to use the Pythagorean Theorem to find the hypotenuse of a right-triangle?

 

[BigGlenntheHeavy]

\(\displaystyle tan(\theta) \ = \ \frac{32}{250}\)

\(\displaystyle \theta \ \dot= \ 7.29^0\)

 

[Denis]

\(\displaystyle tan(\theta) \ = \ \frac{32}{250}\)
\(\displaystyle \theta \ \dot= \ 7.29^0\)

That won't do much BigG: the poor guy does not even know about Pythagorean theorem!!

 

[mmm4444bot]

Can someone show me how to do it? Of course someone can.

We've done a lot of work for you, at this site. It's time for you to show some work and ask specific questions.

 

[IloveManUtd]

That won't do much BigG: the poor guy does not even know about Pythagorean theorem!!

Says who! I know the Pythagorean Theorem but just that I did not realize that angle Q is 90[sup:gq8knfnm]o[/sup:gq8knfnm] :lol:

 

[mmm4444bot]

\(\displaystyle tan(\theta) \ = \ \frac{32}{250}\)

\(\displaystyle \theta \ \dot= \ 7.29^0\)

Of course. (Oops!)

This poster has me too "focused", heh, heh.

 

[mmm4444bot]

I did not realize that angle Q is 90[sup:se0jsicg]o[/sup:se0jsicg]

I'm guessing that you mean angle PQR, when you write "angle Q".

Q is a point.

The point Q is the vertex point of four different angles, in your diagram.

Therefore, the phrase "angle Q" is ambiguous.

Um, what is my point, again? ... Oh, yeah. I wanted to post that angle PQR is not relevant to the original question.

We can ignore all of the provided bearings. It's the legs of the given right-triangle QPX that are relevant because we're asked for the tangent of angle QPX (i.e., the ratio of the legs posted by Glenn).

I was really overthinking it, before.