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:

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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:

Mathematics
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what are you trying to find here ?
Oh, sorry, I forgot to put that in haha. What is the distance between the zoo and the library?

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Hello? @ganeshie8
|dw:1441031827999:dw|
\[ZL=GL-GZ\] Use \[\triangle GHZ\] to find GZ and \[\triangle GHL\] to find GL
the tangent function will be your help here
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.
In \[\triangle GHZ\] \[\tan(60)=\frac{GH}{GZ}\] see the point now?
Okay, so basically there are 2 problems here? But yes, I'm understanding so far since those are the opposite and adjacent sides.
In a sense, yes, you will have to use 2 separate triangles to find 2 values
for you reference, I shall provide the values of tan30 and tan60 \[\tan(30)=\frac{1}{\sqrt{3}}\] \[\tan(60)=\sqrt{3}\]
Why is tan(30) = 1/3? and tan(60) = 3?
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?
One moment please. :)
Sure
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} ??
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?
Infact you are given the height of the tower as PRECISELY as 400sqrt{3} so that your calculations would be easier
sorry height of helicopter
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
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 \[\tan(30)=\frac{1}{\sqrt{3}}\]
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.
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?
Let's go back to triangle GHZ
\[\tan(60)=\frac{400\sqrt{3}}{GZ}\] 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
Are you understanding?
Yes, I'm understanding. You're finding the distances right?
Yep, and we're gonna subtract them for ZL, we're indirectly finding this distance ZL
Can you find GL for me now?
Oh okay, so you're putting GH and GZ together to find ZL so we know the distance from the two?
\[GL-GZ=ZL\] Look at the figure, if you remove the length GZ from GL, you're left with ZL
Right, like you said before we'd subtract the two for ZL.
I'll try figuring out GL now.
\[\tan (30) = \frac{ 400\sqrt{3} }{ GL }\] \[\sqrt{3}= \frac{ 400\sqrt{3} }{ GL}\] Is this right so far? Because I remember you saying that tan(30) = 1/sqrt3 ??
\[\tan(30)=\frac{1}{\sqrt{3}}\] But you've written \[\sqrt{3}\] Slight error there
\[\frac{1}{\sqrt{3}}=\frac{400\sqrt{3}}{GL}\]
How would you write the next step then if I'd divide by root?
Multiply by sqrt{3} throughout
Oh okay, just a second. I'll continue working out the problem.
\[\frac{ GL }{ \sqrt{3} } \times \sqrt{3}= \frac{ GL }{ \sqrt{3}} \times \frac{ 400\sqrt{3} }{ GL }\] Root 3 cancles on the left side, GL and root 3 cancel on right side. So far is this good?
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?
Wouldn't you do that for both?
\[\frac{1}{\sqrt{3}}=\frac{400\sqrt{3}}{GL}\] If you multiply by GL and root 3 throughout, \[\frac{1}{\sqrt{3}}\times GL \times \sqrt{3}=\frac{400\sqrt{3}}{GL} \times GL \times \sqrt{3}\]
So it changes because \[\tan \left( 30 \right) = \frac{ 1 }{ \sqrt{3} } \] ?
What changes?
The equation since for the other one it was \[\tan \left( 60 \right)= \sqrt{3}\] Wouldn't the answer and equation become different?
Of course, both the equations from triangle 1 and triangle 2 are different
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?
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|
Okay, and couldn't you simplify \[400\sqrt{3}\] ?
You cannot further simplify \[400\sqrt{3}\] However you can approximate(that is not required here) \[400\sqrt{3}\approx400 \times 1.732\]
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?
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.
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 \[GZ=\frac{400}{GZ} \times GZ\] \[GZ=400\]
Oh, okay, so GZ would stay 400 since GZ isn't a number in the first place?
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
\[\frac{1}{5} \times 5=1\] \[\frac{1}{GZ} \times GZ=1\]
It's a variable
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.
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 \[1=\frac{400 \times 3}{GL}\] \[1=\frac{1200}{GL}\] \[1 \times GL=\frac{1200}{GL} \times GL\]\[GL=1200\]
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
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?
Yep
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. :)
@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)
Wouldn't you need to use a Trigonometry Function, but there are two: sin and cos..?
Yep, try to find sine and cosine and put them equal to each other, see where that can get you
\[\sin = \frac{ Opposite }{ Hypotenuse }= \cos = \frac{ Adjacent }{ Hypotenuse }\] \[\sin(x) = \frac{ a }{ c } = \cos(x) = \frac{ b }{ c }\]
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?
We can cancel out the denominators since they are the same? Which leaves you with a = b.
Good!! Now if in that triangle (look at the figure above), a and b are equal, does that tell you anything about the triangle?
Congruent sides?
Yep, what is a triangle with 2 equal sides known as?
Isosceles triangle. It'd have two equal sides and two equal angles. But an isosceles triangle is shaped differently..
Yep, the angles opposite to the sides are equal, |dw:1441040422979:dw|
Wouldn't the property be: \[m
Yep, sum of all 3 angles in a triangle is always 180 degrees
so we get \[x+x+90=180\] The two angles are equal so the 2nd angle is infact also x
Oh okay, so would we add or subtract from the problem?
Let's club the x's together first \[2x+90=180\] Does that makes sense?
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
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|
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?
Yep, If you are ever asked sine of 45 degrees, you'd say it's 1/sqrt(2)
You can also verify tangent for each value, we know that \[\tan(x)=\frac{\sin(x)}{\cos(x)}\] Let's look at 60 degrees \[\tan(60)=\frac{\sin(60)}{\cos(60)}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2}\times 2=\sqrt{3}\]
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
Oh okay, I noticed that thank you haha. Could you possibly help me on a new problem?
Sure
Oh, I actually think this problem is similar to the last one lol.
Two ropes, AD and BD, are tied to a peg on the ground at point D. The other ends of the ropes are tied to points A and B on a flagpole, as shown below: Angle ADC measures 60° and angle BDC measures 30°. What is the distance between the points A and B on the flagpole? Choice are: A) 20 Feet B) 15 Feet C) 10 Feet D) 25 Feet
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
find AC and BC like we did and subtract BC from AC
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
Okay, thank you. I'll see what I can work out now. :)
\[\tan \left( 30 \right)= \frac{ 5\sqrt{3} }{ BC }\] \[\tan \left( 60 \right)= \frac{ 5\sqrt{3} }{ AC}\] This right so far?
nope, doesn't look good, check your opposite and adjacent
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?
|dw:1441043497380:dw|
Right, so \[\tan \left( 30 \right)= \frac{ BC }{ 5\sqrt{3} } \] and \[\tan \left( 60 \right)= \frac{ AC }{ 5\sqrt{3} }\]
|dw:1441043990553:dw|
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?
It's okay, thank you for your help! Bye now :)
yep

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