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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?

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

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

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

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

One moment please. :)

Sure

sorry height of helicopter

Let's go back to triangle GHZ

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.

\[\tan(30)=\frac{1}{\sqrt{3}}\]
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.

Wouldn't you do that for both?

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

What changes?

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

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

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

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

It's a variable

Yep

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

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

Congruent sides?

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

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

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?

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

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

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

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

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

|dw:1441043497380:dw|

|dw:1441043990553:dw|

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

yep