Find the angle BHL using trigonometry

[mathlearn]

Hi,

View attachment 5795

As told in the problem I drew a figure,

View attachment 5794

Next the magnitude of \(\displaystyle \angle BHL\) should be found using trigonometric tables.

Can you help me to and find the angle BHL (Smile)

Many Thanks (Smile)

 

[MarkFL]

I think I would begin by finding $\angle {BAL}=\theta$. Then once we have that, we can find $\overline{BL}$ since:

\(\displaystyle \sin(\theta)=\frac{\overline{BL}}{3\text{ km}}\)

What do you find so far?

 

[mathlearn]

\(\displaystyle sin(θ)= \frac{BL}{3 km}\)

sin(θ) * 3 km = BL , Correct ?

Many Thanks (Smile)

 

[MarkFL]

\(\displaystyle sin(θ)= \frac{BL}{3 km}\)

sin(θ) * 3 km = BL , Correct ?

Many Thanks (Smile)

Yes: \(\displaystyle \overline{BL}=3\sin(\theta)\text{ km}\)

Can you find $\theta$ so that you can get a numeric value for $\overline{BL}$

 

[mathlearn]

Many Thanks :)

How ? Can you help me . I'm unable to find \(\displaystyle \theta\)

Many Thanks :)

 

[MarkFL]

Do you see how the $40^{\circ}$ angle in the diagram is repeated and within the $110^{\circ}$ angle?

 

[mathlearn]

Then \(\displaystyle \angle \)BAL =110-40=70 degrees \(\displaystyle \therefore\)

sin 70 = \(\displaystyle \frac{BL}{3 km}\)

By referring to the sine table sin 70 = 9.9730

9.9730 = \(\displaystyle \frac{BL}{3 km}\)

9.9730 x 3 km = BL

29.9190 km = BL

and thereafter AB should be found should I use the Pythagoras theorem?

Many Thanks (Smile)

 

[MarkFL]

Yes, good! (Yes) (but watch the decimal point!)

I think I would wait until the end though to plug in for the trig. functions (this way we can round one time)...for now I would simply state:

\(\displaystyle \overline{BL}=3\sin\left(70^{\circ}\right)\text{ km}\)

We know this represents a real number...but we don't need to throw a rounded version of it into the mix yet.

Now, can you, in a similar way, find $\overline{AB}$? Notice that $\overline{AB}$ is adjacent to $\theta$, and you know the hypotenuse (it was given)...which trig. function relates the leg in a right triangle adjacent to an angle and the hypotenuse?

 

[mathlearn]

cos 70= \(\displaystyle \frac{AL}{AB}\)

cos 70 = sin(90 -70)
= sin 20

sin 20= \(\displaystyle \frac{AL}{AB}\)

Referring to the trigonometric table,

9.5341= \(\displaystyle \frac{AL}{AB}\)

9.5341 x 3 km = AB

28.6023 km = AB

Found It (Smile). Note: These values seem bigger much when compared to the hypotenuse of 3 km.

Many Thanks (Smile)

 

[MarkFL]

Yes, your values are 10 times too big, because you have the decimal point in the wrong place. :D

I would write:

\(\displaystyle \overline{AB}=3\cos\left(70^{\circ}\right)\text{ km}\)

So, now we have:

\(\displaystyle \overline{HB}=\left(4.5+3\cos\left(70^{\circ}\right)\right)\text{ km}\)

And we earlier found:

\(\displaystyle \overline{BL}=3\sin\left(70^{\circ}\right)\text{ km}\)

Now, let's label:

\(\displaystyle \alpha=\angle {BHL}\)

We know the sides opposite and adjacent to $\alpha$...what trig. function can we use here?

 

[mathlearn]

Many Thanks :),

The trig. function would be tan = \(\displaystyle \frac{adjacent side}{opposite side}\)
= \(\displaystyle \frac{3sin(θ) km }{(4.5+3cos(70∘))}\)


Correct?

Many Thanks (Smile)

 

[MarkFL]

Many Thanks :),

The trig. function would be tan = \(\displaystyle \frac{adjacent side}{opposite side}\)
= \(\displaystyle \frac{3sin(θ) km }{(4.5+3cos(70∘))}\)


Correct?

Many Thanks (Smile)

Well, the definition is:

\(\displaystyle \text{tangent}\equiv\frac{\text{opposite}}{\text{adjacent}}\)

Now, with regards to $\alpha$, the adjacent side is:

\(\displaystyle \overline{HB}=\left(4.5+3\cos\left(70^{\circ}\right)\right)\text{ km}\)

and the opposite side is:

\(\displaystyle \overline{BL}=3\sin\left(70^{\circ}\right)\text{ km}\)

Hence:

\(\displaystyle \tan(\alpha)=\frac{3\sin\left(70^{\circ}\right)\text{ km}}{\left(4.5+3\cos\left(70^{\circ}\right)\right)\text{ km}}=\frac{2\sin\left(70^{\circ}\right)}{3+2\cos\left(70^{\circ}\right)}\)

Okay, now use your table to get the most accurate value for $\tan(\alpha)$, and then you can use your table to find the angle closest to that value for the tangent function. :)

 

[mathlearn]

Can you explain a little on how did \(\displaystyle \frac{(4.5+3cos(70∘)) km}{3sin(70∘) km} \) become \(\displaystyle \frac{3+2cos(70∘)}{2sin(70∘)}\)

Tan \(\displaystyle \theta\) = \(\displaystyle \frac{3+2cos(70∘)}{2sin(70∘)}\)

cos 70 = sin(90-70) = sin 20
= \(\displaystyle \frac{3+2 sin(20∘)}{2 sin(70∘)}\)

= \(\displaystyle \frac{3+2 (9.5341)}{2 (9.9730)}\)
= \(\displaystyle \frac{3+19.0682}{19.946}\)
= \(\displaystyle \frac{22.0682}{19.946}\)

Correct ?

Many Thanks :)

 

[MarkFL]

Can you explain a little on how did \(\displaystyle \frac{(4.5+3cos(70∘)) km}{3sin(70∘) km} \) become \(\displaystyle \frac{3+2cos(70∘)}{2sin(70∘)}\)

Tan \(\displaystyle \theta\) = \(\displaystyle \frac{3+2cos(70∘)}{2sin(70∘)}\)

cos 70 = sin(90-70) = sin 20
= \(\displaystyle \frac{3+2 sin(20∘)}{2 sin(70∘)}\)

= \(\displaystyle \frac{3+2 (9.5341)}{2 (9.9730)}\)
= \(\displaystyle \frac{3+19.0682}{19.946}\)
= \(\displaystyle \frac{22.0682}{19.946}\)

Correct ?

Many Thanks :)

If you multiply all those coefficients by 2/3, you get all integral coefficients...necessary? No...nicer to look at? Yes. :D

You are still using values that are ten times too large for the trig. functions...

 

[mathlearn]

Hi,

Would you be kind enough to demonstrate please.Why are the numbers still too large for the trig? And what should I do now?

Many Thanks (Smile);

 

[MarkFL]

Okay, we have:

\(\displaystyle \tan(\alpha)=\frac{3\sin\left(70^{\circ}\right)\text{ km}}{\left(4.5+3\cos\left(70^{\circ}\right)\right)\text{ km}}=\frac{2\sin\left(70^{\circ}\right)}{3+2\cos\left(70^{\circ}\right)}\)

Using my computer, I find:

\(\displaystyle \sin\left(70^{\circ}\right)\approx0.9397\)

\(\displaystyle \cos\left(70^{\circ}\right)\approx0.3420\)

You see, we should keep in mind that both cosine and sine will never have a magnitude greater than 1...can you image a right triangle in which one of the legs is longer than the hypotenuse?

Can you now (using values from your table) get an approximation for $\tan(\alpha)$?

 

[mathlearn]

Hi,
I think you multiply twice 0.9397 and twice 0.3420 +3 and divide .

then i get 1.960200063850165 but the problem is the value is greater than one.

Many Thanks (Smile)

 

[MarkFL]

Using the values I gave for the sine and cosine of 70 degrees, I get:

\(\displaystyle \tan(\alpha)\approx0.5102\)

Check your division...the denominator is larger than the numerator, so you should get a value less than 1.

However, the tangent function isn't limited in its magnitude like sine and cosine. We just happen to have a value whose magnitude is smaller than 1 in this case.

What do you get when you divide again?

 

[mathlearn]

Hi,

I'm sorry I have divided the wrong way the denominator from the numerator by doing it the correct way I get
tan(α)≈0.5102 and by referring to the trigonometric tables I get the angle as 27 degrees and 07 minutes. Many Thanks for the tremendous effort , dedication and patience of yours. :) (Smile)

Many Thanks (Smile)

 

[MarkFL]

Using the inverse tangent function, I find:

\(\displaystyle \alpha\approx27.0285^{\circ}\), so that sounds about right. (Yes)

 

[mathlearn]

Thank you very much again!