Max observes the zoo and the library from a helicopter flying at a height of 400 times square root of 3 feet above the ground, as shown below:
Here are the choices:
A) 800 Feet
B) 600 Feet
C) 200 Feet
D) 400 Feet
The picture for this problem will be shown below:

- anonymous

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- anonymous

##### 1 Attachment

- ganeshie8

what are you trying to find here ?

- anonymous

Oh, sorry, I forgot to put that in haha.
What is the distance between the zoo and the library?

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## More answers

- anonymous

Hello? @ganeshie8

- anonymous

|dw:1441031827999:dw|

- anonymous

\[ZL=GL-GZ\]
Use \[\triangle GHZ\]
to find GZ
and
\[\triangle GHL\]
to find GL

- anonymous

the tangent function will be your help here

- anonymous

Okay, I'm understanding so far. How would I write that formula out though? Well, not just tell me but, explain how I'd work it out.

- anonymous

In
\[\triangle GHZ\]
\[\tan(60)=\frac{GH}{GZ}\]
see the point now?

- anonymous

Okay, so basically there are 2 problems here? But yes, I'm understanding so far since those are the opposite and adjacent sides.

- anonymous

In a sense, yes, you will have to use 2 separate triangles to find 2 values

- anonymous

for you reference, I shall provide the values of tan30 and tan60
\[\tan(30)=\frac{1}{\sqrt{3}}\]
\[\tan(60)=\sqrt{3}\]

- anonymous

Why is tan(30) = 1/3? and tan(60) = 3?

- anonymous

Well, it's a function, functions take in variables (angles in this case) and return some value depending on the value of the variable
and it's 1/sqrt(3) and sqrt(3)
For example consider
\[x^2+7x\]
What you are asking is something like why will x=1 give us the answer as 8?

- anonymous

One moment please. :)

- anonymous

Sure

- anonymous

Okay, I'm back. I'm still a bit confused. I understand functions need variables. But are we using sqrt because of the 400sqrt{3} ??

- anonymous

Ok let's take my same example
\[x^2+7x\]
When x=1, we get
answer as 8
Similarly consider
\[\tan(x)\]
when
\[x=60\]
degrees
we get the answer as
\[\sqrt{3}\]
It's one of those things you need to remember, I can't help much if you ask "why" is this so?

- anonymous

Infact you are given the height of the tower as PRECISELY as 400sqrt{3} so that your calculations would be easier

- anonymous

sorry height of helicopter

- anonymous

Ok, now I understand what you're trying to stay, just for the sake of curiosity, let me tell you that tan(x) can represented as an infinite series of x,
\[\tan(x)=x+\frac{2x^3}{3!}+\frac{16x^5}{5!}+\frac{272x^7}{7!}+....\]
This is mclaurin series for tan(x), basically mclaurin series are a way of representing functions as an infinite polynomial
where x is in radians
let's convert 60 to radians
\[60 \times \frac{\pi}{180}=\frac{\pi}{3}\]
Now when you put x=pi/3, for each term you add, your value will get closer and closer to \[\sqrt{3}\]
Now don't ask where I pulled that series out from, it's totally out of bounds fornow, I did this for the sake of visualization, and don't go around wasting time putting pi/3 in all those terms, you will get the answer as sqrt{3}
the ! means
\[3!=3 \times 2 \times 1\]
\[5!=5 \times 4 \times 3 \times 2 \times 1\]
It's a way of writing product from that number to 1

- anonymous

I'm sorry if I can't find an easier way to tell you, but for now just remember that
\[\tan(60)=\sqrt{3}\]
and
\[\tan(30)=\frac{1}{\sqrt{3}}\]

- anonymous

It's okay, I don't want you to try and help me understand for too long. I'll ask my teacher about it as well and look over my notes again, but let's just continue.

- anonymous

You don't need to worry about asking my help for long, If I was busy I wouldn't come to the website in the first place, right?

- anonymous

Let's go back to triangle GHZ

- anonymous

\[\tan(60)=\frac{400\sqrt{3}}{GZ}\]
From this we can find GZ
\[\sqrt{3}=\frac{400\sqrt{3}}{GZ}\]
Dividing by root 3 and multiplying by GZ throughout we get
\[\frac{GZ}{\sqrt{3}} \times \sqrt{3}=\frac{GZ}{\sqrt{3}} \times \frac{400\sqrt{3}}{GZ}\]
root 3 cancels on left side, GZ and root 3 cancel on right side
we get
\[GZ=400\]
Similarly find GL using triangle GHL and tan 30

- anonymous

Are you understanding?

- anonymous

Yes, I'm understanding. You're finding the distances right?

- anonymous

Yep, and we're gonna subtract them for ZL, we're indirectly finding this distance ZL

- anonymous

Can you find GL for me now?

- anonymous

Oh okay, so you're putting GH and GZ together to find ZL so we know the distance from the two?

- anonymous

\[GL-GZ=ZL\]
Look at the figure, if you remove the length GZ from GL, you're left with ZL

- anonymous

Right, like you said before we'd subtract the two for ZL.

- anonymous

I'll try figuring out GL now.

- anonymous

\[\tan (30) = \frac{ 400\sqrt{3} }{ GL }\]
\[\sqrt{3}= \frac{ 400\sqrt{3} }{ GL}\]
Is this right so far? Because I remember you saying that tan(30) = 1/sqrt3 ??

- anonymous

\[\tan(30)=\frac{1}{\sqrt{3}}\]
But you've written
\[\sqrt{3}\]
Slight error there

- anonymous

\[\frac{1}{\sqrt{3}}=\frac{400\sqrt{3}}{GL}\]

- anonymous

How would you write the next step then if I'd divide by root?

- anonymous

Multiply by sqrt{3} throughout

- anonymous

Oh okay, just a second. I'll continue working out the problem.

- anonymous

\[\frac{ GL }{ \sqrt{3} } \times \sqrt{3}= \frac{ GL }{ \sqrt{3}} \times \frac{ 400\sqrt{3} }{ GL }\]
Root 3 cancles on the left side, GL and root 3 cancel on right side.
So far is this good?

- anonymous

Your right side doesn't look good, we multiplied by root 3 throughout, so how are you getting a root 3 on the denominator on the right side?

- anonymous

Wouldn't you do that for both?

- anonymous

\[\frac{1}{\sqrt{3}}=\frac{400\sqrt{3}}{GL}\]
If you multiply by GL and root 3 throughout,
\[\frac{1}{\sqrt{3}}\times GL \times \sqrt{3}=\frac{400\sqrt{3}}{GL} \times GL \times \sqrt{3}\]

- anonymous

So it changes because \[\tan \left( 30 \right) = \frac{ 1 }{ \sqrt{3} } \] ?

- anonymous

What changes?

- anonymous

The equation since for the other one it was \[\tan \left( 60 \right)= \sqrt{3}\]
Wouldn't the answer and equation become different?

- anonymous

Of course, both the equations from triangle 1 and triangle 2 are different

- anonymous

Okay.. I'm lost now. I'm reviewing the module where this should've been taught to me, I'm not finding it anywhere which is why I'm very confused since my teacher never broke this down for me in the first place. So, the equations would be different, then how come GL's equation isn't the same as GZ, I know they're supposed to be different but in what way?

- anonymous

Because the angles are different, what we are doing is just applying trigonometry to both the triangles separately, it does not necessarily mean that both will give same equations
\[\sqrt{3}=\frac{400\sqrt{3}}{GZ}\]
and
\[\frac{1}{\sqrt{3}}=\frac{400\sqrt{3}}{GL}\]
Think of it as 2 different triangles
|dw:1441037237930:dw|

- anonymous

Okay, and couldn't you simplify \[400\sqrt{3}\] ?

- anonymous

You cannot further simplify
\[400\sqrt{3}\]
However you can approximate(that is not required here)
\[400\sqrt{3}\approx400 \times 1.732\]

- anonymous

Okay, so this is what we'd have:
\[\frac{ 1 }{ \sqrt{3}} \times GL \times \sqrt{3} = \frac{ 400\sqrt{3} }{ GL } \times GL \times \sqrt{3} \]
Right?
What I'm confused about it how we'd get our answer from there?

- anonymous

For the first problem for GZ, you never told me how you broke down the equation and got your final answer, so that's what I'm stuck on mainly.

- anonymous

ok I'll start with GZ
\[\sqrt{3}=\frac{400\sqrt{3}}{GZ}\]
Let's take it one at a time
Divide by root 3 we get
\[\frac{\sqrt{3}}{\sqrt{3}}=\frac{400\sqrt{3}}{GZ \sqrt{3}}\]
This becomes
\[1=\frac{400}{GZ}\]
Now we multiply by GZ
\[GZ=\frac{400}{GZ} \times GZ\]
\[GZ=400\]

- anonymous

Oh, okay, so GZ would stay 400 since GZ isn't a number in the first place?

- anonymous

It's like your x, I've used GZ
you can also do like
\[\sqrt{3}=\frac{400\sqrt{3}}{x}\]
It's an unknown value, but the rules of maths still apply, a value known or unknown to us, so if we multiply \[\frac{1}{GZ}\]
by \[GZ\]
they cancel out and leave us with 1 just like any other number

- anonymous

\[\frac{1}{5} \times 5=1\]
\[\frac{1}{GZ} \times GZ=1\]

- anonymous

It's a variable

- anonymous

Oh, stupid me. I understand that now. So for GL, how would that problem get broken down? The equation is different so it wouldn't be the exact steps that happened for GZ.

- anonymous

Hm let's have a look
\[\frac{1}{\sqrt{3}}=\frac{400\sqrt{3}}{GL}\]
Let's multiply by root 3
\[\frac{\sqrt{3}}{\sqrt{3}}=\frac{400\sqrt{3}\sqrt{3}}{GL}\]
Now we know that
\[\sqrt{a} \times \sqrt{b}=\sqrt{ab}\]
So on right we have
\[\frac{\sqrt{3}}{\sqrt{3}}=\frac{400\sqrt{3 \times 3}}{GL}\]
Left side cancels out because same quantity
\[1=\frac{400\sqrt{9}}{GL}\]
But we know that square root of 9 is 3
\[1=\frac{400 \times 3}{GL}\]
\[1=\frac{1200}{GL}\]
\[1 \times GL=\frac{1200}{GL} \times GL\]\[GL=1200\]

- anonymous

technically it's not a variable, it's just some unknown number, a variable is a quantity that changes, but as you saw GZ has only one value and that is 400

- anonymous

Oh okay! I'm noticing now the whole problem. It seems quite easy now that I see it broken down. So then you'd subtract the two and get 800?

- anonymous

Yep

- anonymous

Okay, I might leave this up for a bit to take notes off of but I'll be sure to close it when I'm finished. Once again, thank you for helping me throughout this problem for such a long time! I REALLY appreciate it. :)

- anonymous

@ScienceAndMath
Try this question:
|dw:1441039105610:dw|
In the triangle ABC above,
Find the value of x for which sin(x) is equal to cos(x)

- anonymous

Wouldn't you need to use a Trigonometry Function, but there are two: sin and cos..?

- anonymous

Yep, try to find sine and cosine and put them equal to each other, see where that can get you

- anonymous

\[\sin = \frac{ Opposite }{ Hypotenuse }= \cos = \frac{ Adjacent }{ Hypotenuse }\]
\[\sin(x) = \frac{ a }{ c } = \cos(x) = \frac{ b }{ c }\]

- anonymous

Good! Now we want to find x for which sin and cosine are equal, so in a sense what we are saying is
given: sin(x)=cos(x)
find x
So if we equate them, we get
\[\frac{a}{c}=\frac{b}{c}\]
Anything you feel you can cancel out in this equation?

- anonymous

We can cancel out the denominators since they are the same? Which leaves you with a = b.

- anonymous

Good!! Now if in that triangle (look at the figure above), a and b are equal, does that tell you anything about the triangle?

- anonymous

Congruent sides?

- anonymous

Yep, what is a triangle with 2 equal sides known as?

- anonymous

Isosceles triangle. It'd have two equal sides and two equal angles. But an isosceles triangle is shaped differently..

- anonymous

Yep, the angles opposite to the sides are equal,
|dw:1441040422979:dw|

- anonymous

Yep, sum of all 3 angles in a triangle is always 180 degrees

- anonymous

so we get
\[x+x+90=180\]
The two angles are equal so the 2nd angle is infact also x

- anonymous

Oh okay, so would we add or subtract from the problem?

- anonymous

Let's club the x's together first
\[2x+90=180\]
Does that makes sense?

- anonymous

Oh, yes it does lol. So I think afterwards you'd subtract subtract 90 from 180 which gives you 90 and then you take 2x and divide that into 90 which is = 45

- anonymous

Yes!! So what does this tells us??
\[\sin(45)=\cos(45)?\]
Is this true?Yep!!
\[\sin(45)=\cos(45)=\frac{1}{\sqrt{2}}\]
I'll make a table of common trigonometric values for you in degrees, make sure to remember it well
|dw:1441041112353:dw|

- anonymous

Okay, I wrote down the chart. I think I'm understanding it more now. I remember in the last problem we used the tan function and the sqrt3 and the 1/sqrt3 since the degrees were 30 and 60. So for Sin and Cos, since they're 45 degrees we'd use 1/sqrt2?

- anonymous

Yep, If you are ever asked sine of 45 degrees, you'd say it's 1/sqrt(2)

- anonymous

You can also verify tangent for each value,
we know that
\[\tan(x)=\frac{\sin(x)}{\cos(x)}\]
Let's look at 60 degrees
\[\tan(60)=\frac{\sin(60)}{\cos(60)}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2}\times 2=\sqrt{3}\]

- anonymous

Actually, tan 90 would be "not defined" but the closer and closer you get to 90 degrees, like 89.9999 the closer and closer tan gets to infinity

- anonymous

Oh okay, I noticed that thank you haha. Could you possibly help me on a new problem?

- anonymous

Sure

- anonymous

Oh, I actually think this problem is similar to the last one lol.

- anonymous

- anonymous

Not much difference between this and last question, you'd again use tangent function, it's just a vertical figure instead of a horizontal figure now

- anonymous

find AC and BC like we did and subtract BC from AC

- anonymous

If you get stuck, don't hesitate to ask, even if I'm not here, people will be here to help, I'll stay up for an hour or two more

- anonymous

Okay, thank you. I'll see what I can work out now. :)

- anonymous

\[\tan \left( 30 \right)= \frac{ 5\sqrt{3} }{ BC }\]
\[\tan \left( 60 \right)= \frac{ 5\sqrt{3} }{ AC}\]
This right so far?

- anonymous

nope, doesn't look good, check your opposite and adjacent

- anonymous

Okay. On the opposite it's just the flagpole? Or do you mean the degrees starting from D and going out towards the flagpole the other way?

- anonymous

|dw:1441043497380:dw|

- anonymous

Right, so
\[\tan \left( 30 \right)= \frac{ BC }{ 5\sqrt{3} } \]
and
\[\tan \left( 60 \right)= \frac{ AC }{ 5\sqrt{3} }\]

- anonymous

|dw:1441043990553:dw|

- anonymous

Okay, then in the chart I wrote down that you showed me, wouldn't I use sqrt 3 for 60 and 1/sqrt3 for 30?

- anonymous

It's okay, thank you for your help! Bye now :)

- anonymous

yep

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