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Looking at this, Im not sure if we can use trigonometry.

We know that one of the blue triangles has a 60 degree angle with two sides in a ratio of 2:3.

Composite functions,

First, we solve the inside then, the outside using the given information.

so start with:

f(1+i) ,given f(z)=z^2 if z is not real well it is not real here so we square it.

\((1+i)^2\)=\(2i\)

next f

f(2i) , well it is still imaginary and not real so square it again

\((2i)^2=-4\)

next f

f(-4) now it is real, so -z^2 

\(--4^2=16\) 

2 negatives give positive. 

Now the last f 

f(16) , it is real so 

\(-16^2=-256\) answer.

Well since the yellow is equilateral inside another equilateral, we can confirm that logically all three blue triangles are congruent.

The orange triangle and the red triangle are similar.

Using one of the properties of similar triangles, we know that the ratio of the side length of the orange triangle to the side length of the red triangle is 2:3

Because: (2/3)^2 = 4/9 = 20/45 

Now we found the ratio and that the blue triangles are congruent, we know that the side length of the orange triangle to the side length of the blue triangle is 2: (2 + 3) = 2:5

That means the AREA must be (according to one of the similarity properties) 4:25

So we calculate the area of the blue triangle and find its 125. ( we also can find it using the red triangle.)

Help! Im stuck too! I got the blue triangle as 125, can someone work off from here?

thank you!

This is what I did:

I noticed that segment AO and OD are radii, and since radii are all congruent, center O must also be the midpoint of AD.

I calculated OD because it was a radii.

Then I multiplied the value of OD by 2 to get AD. We know know the height of triangle ABC.

Since ABC Is equilateral, that means angles C and B are 60 degrees.

We find that triangles ACD and ABD are 30-60-90 triangles.

Since we know the height of triangle ABC and we know that triangles ACD and ABD are 30-60-90, can you find the area?

Hint: try using the properties of a 30-60-90 triangle.

The normal vector is (-1,3,2), so (a,b,c) = (-2,6,4).

ok that should have worked.

\(\frac3{a+b+c}=\frac1{a+b} + \frac1{a+c}\\ \frac3{a+b+c}(a+b+c)(a+b)(a+c)=\frac1{a+b}(a+b+c)(a+b)(a+c) + \frac1{a+c}(a+b+c)(a+b)(a+c)\\ 3(a+b)(a+c)=(a+b+c)(a+c) + (a+b+c)(a+b)\\ 3(a^2+ac+ab+bc)=a^2+ac+ab+bc+ca+c^2\quad +\quad a^2+ab+ab+b^2+ca+cb\\ 3a^2+3ac+3ab+3bc=2a^2+3ac+3ab+2bc+c^2+b^2\\ \)

You can take it from there.  I doubt that you will get the same answer as you got before.

When you get the algebra right you can continue to the cosine rule relevance.

And yes, you do want to make it look as much like the cosine rule as possible.

a = 1 and b = 2, so a + b = 3.

They are an insult. Like a slap in the face.

Why would you punish a person who is trying to help you.  EVEN if you think the punishment is inconsequential (or anonymous)?

If

 \(sin\angle{PCB}=\frac{\sqrt3}{\sqrt{19}}\\ then\\ \angle{PCB}=sin^{-1}\frac{\sqrt3}{\sqrt{19}}\\ \angle{PCB}\approx 23.413 ^{\circ}\)

Now, check you know how I did that on a calculator and then move onto step 3

sorry it has not displayed properly, I will see if I can fix it.    It is fixed I hope.

I multiplied everything by (a+b+c)(a+b)(a+c) and simplified it to where there is only one term on one side that is squared because that is like the cosine rule.

Do the upvotes and downvotes really matter that much??

I am sorry if they do.

So BC is \(\sqrt{19}\).

Then, by the law of sines, sin BCP is \(\frac{\sqrt3}{\sqrt{19}}\).

What now?

Why should this person bother to help you more.

You have voted their first attempt to help you down.  

Kick a person for offering help then ask for more help.  Not usually a good tactic!

What is wrtong with you MathCuber  ??

Your getting propfessional teaching help and you vote it down.  

Are you one of these cheating creeps who only is interested in the answer?

I thought you were better than that!

That is not finished and it is not what I got.

Can you show your working so I can see where you went wrong? 

             (mine could be wrong too but it worked out nicely so it probbly isn't)

Did you start it one side at a time or what?

You can find BC using cosine rule.  You have 2 sides and the included angle so you can find the 3rd side.

Sorry, but this answer is incorrect. Can you show how you got that?

I expanded it to \(a^2-2ab-2ac-b^2-c^2-bc=2a^2\).

I don't know how to compare that to the law of cosines: can I simplify further?

1) How would I find BC?

Your title is different from your question.

Please proof your questions properly and get rid of meaningless dollar signs.

Why don't you show me your answer?  I'd like that.

5+   5*5 = 30

thanks melody!

You can take 9z of each of them.