Trigonometry

[Mohit Pandey]

I request you to see the following trigonometric question and see if I have formed correct diagram of the given problem.

A pole projected outwards from a window 10 m above the ground of a building makes an angle 0f 30 with the wall. The angles of elevation of the bottom and top of the pole from a point on the ground are 30 and 60 respectively. Find the length of the pole.

 

In the attached picture, BQ is the pole. Angle BQC =30, angle BAD= 60 and angle QAD=45.

[sanctus]

I would say that the image is right but the angles are better named if you say:

QBC=30

QAP=30 your BQC

BAD=60

 

I think this is better because then you show really that it is an angle of elevation...anyway probably it is just a matter of preference...

[sanctus]

Was it right?

[Mohit Pandey]

yes. But tell me how to write root 3 here as we usually write in our copy.

[sanctus]

Just quote this post to see:

 

to write it on the same line [imath]\sqrt{3}[/imath] and to write on its own line

[math]

\sqrt{3}

[/math]

once you type "imath" in brackets and once "math".

 

To write formulas like:

[math]

\frac{x_i^2}{2\cdot \sqrt{x_{i-1}^3}}=y_k \quad \forall k[/math]

you have to learn latex, which is not wasted time, since you need it (or at least it is very advantageous) if you wanna do science. Try to search the forums for latex for useful links.

[Mohit Pandey]

you have to learn latex, which is not wasted time, since you need it (or at least it is very advantageous) if you wanna do science. Try to search the forums for latex for useful links.

 

Sometimes I wonder you can read my mind. ;)

Yes , I was thinking on similar lines. But latex appears to me complicated at the moment and I am preoccupied with my Board Exams which in India affects a student's life. So, as soon as I am free , I'll devote time for latex.

 

Meanwhile, please cooperate with me as I am posting solution of the question. Here[imath]\sqrt{3}[/imath]= root 3. Please check it.

 

Let BC be of length x m. Thus, CD=PQ=10-x.

 

In triangle AQP

Tan30=QP/AP

=1/root 3=CD/AP [QP=CD]

Or AP=root 3(10-x)………………………………………….equation 1.

 

In triangle BAD

Tan 60=BD/AD

Or root 3=10/AD

Or AD =10/root 3…………………………………………….. equation 2

 

In triangle BQC

Tan 60=BC/PD

Or x = PD * root 3

Or x= (AD-AP) root 3

Or x= (10/root 3-AP)*root 3…………………………………………….. equation 3

 

Also, sec 30 = BQ/BC

Or 2/ root 3=BQ/x

Or BQ=x*2/root 3…………………………………………………….. equation 4

 

Now from equation 1 and equation 2,

AP=root 3(10-AP*root 3)

Or AP =10*root3 – 3AP

Or 4AP =10*root3

Or AP =5*root3/2…………………………………………………….. equation 5

 

From equation 3 and equation 5,

x= (10/root 3-5*root3/2)*root 3

=5/2 m. …………………………………………………….. equation 6

 

Now from equation 4,

BQ=x*2/root 3

Or BQ=5/root 3 m

 

:sweat:Whew! it took me time without even using latex. I had completed it in copy in less than half time. Please check it.

[sanctus]

Isn't it:

[math]

\begin{split}

sec(30)&=2\\

sec(60) &=\frac{2}{\sqrt{3}}

\end{split}

[/math]

?

But with this values I always get x=10 so there is something wrong. I can't see what I do wrong though...

[Mohit Pandey]

No, you are wrong. see

[sanctus]

but isn't sec equal 1/sin?

I never use sec and csc so I'm not sure about their definition and find them useless since you can do very well without them.

[Mohit Pandey]

No, 1/sin A= cosec A

1/cos A = sec A

I never use sec and cosec so I'm not sure about their definition and find them useless since you can do very well without them.

Do you meant to say without trigonometry you can do well in maths?

[sanctus]

Not at all!!!! I want to say that without sec and cosec you can do well, just write 1/sin or 1/cos instead.

 

Less superfluos (in my opinion) definition implies smaller risk of definition confusion...

[Mohit Pandey]

OK. then tell me if my solution is correct.

[sanctus]

Didn't you have to hand it in in the meantime?

[CraigD]

Isn't it:

sec(30)=2

…

I want to say that without sec and cosec you can do well, just write 1/sin or 1/cos instead.

I think sanctus is experiencing a confusion of terms I often have myself.

 

[math]\mbox{secant}\, x = \frac1{\mbox{cosine}\, x}[/math], [math]\mbox{cosecant}\, x = \frac1{\mbox{sine}\, x}[/math]. For somewhat obscure reasons stemming from antique geometry, the “co”-secant doesn’t go with the “co”-sine, as my memory frequently misinforms me.

 

So [math]\sec 60^\circ = \frac1{\cos 60^\circ} = \frac1{\frac12} = 2[/math].

 

[math]\mbox{cotangent}\, x = \frac1{\mbox{tangent}\, x}[/math], making the “co” terms of trigonometry even less consistent and more difficult to remember.

 

Personally, like sanctus, I try to avoid using more than the minimum convenient trig terms - sine, cosine, and, because it’s so useful, tangent - in everyday conversation.

[Mohit Pandey]

Didn't you have to hand it in in the meantime?

sorry ,I couldn't understand you?

Hey CraigD! please tell me if my solution is correct.

[sanctus]

It is a homework I guessed, so I thought that you had to hand it in (if you do not understand translate "hand in" into your language)

[CraigD]

Hey CraigD! please tell me if my solution is correct.

Without looking at your work in detail, I can see a problem with your diagram. It’s impossible for a triangle with interior angles totaling 180°

 

You have angle ADB = 90° and BAD = 60°, so angle ABD = 30°. But you also have angle ABC = 30°, so AQB = 0°, point Q must lie on line AB, and angle DAQ = 60°, which contradicts the requirement that it equal 30°.

 

To redraw the diagram as something physically possible, you must interpret “makes an angle 0f 30 with the wall” as a 30° angle sloping upward, not downward, getting something like my attached thumbnail sketch.

 

Considering this diagram, you should be able to answer the question without trig functions, using just simple properties of similarity.