Trigonometry

[brazen hull]

can u plz solve this question

[brazen hull]

Trigonometry

[radiant fox]

What is meant by CO+B ?

[placid belfry]

Think it’s cot(B)

[radiant fox]

ah it could be that

I also assume that the segment starting in A and ending on BC is perpendicular to BC

[radiant fox]

Is that correct @brazen hull

[brazen hull]

that is the right phrase

[radiant fox]

and this? I also assume that the segment starting in A and ending on BC is perpendicular to BC

[brazen hull]

perpendicular?

[radiant fox]

it makes an angle of 90°

and which angle is A?

[brazen hull]

upper one

[radiant fox]

yeah, but it's divided by that line

is the whole or just one of the two?

[placid belfry]

Probably the whole one

[radiant fox]

yeah

If we call the angle at B $\beta$, I can calculate $cot(\beta)$ but I don't know how to do the one for A

[silk plover]

zlik

[radiant fox]

Using Pythagoras' theorem we can calculate $AB$ and get $n\sqrt{2}$. To calculate $cot(\beta)$ we write it as $\frac{cos(\beta)}{sin\beta}$. $cos(\beta) = \frac{n}{n\sqrt{2}}$ and $sin(\beta) = \frac{n}{n\sqrt{2}}$. Therefore $cot(\beta) = 1$

[silk plover]

zlik

[radiant fox]

@brazen hull

[brazen hull]

me too

i dont have any ideas about it bro

[radiant fox]

do you know the answer?

Or do you not know what you are supposed to get?

I might have an idea and an answer but I don't know if it's correct

[marsh roost]

-2?

[radiant fox]

I got the same

@brazen hull

[marsh roost]

did u take that line from A as perpendicular?

[radiant fox]

Yeah

[marsh roost]

But I didn't... u can do without that

Coz its not mentioned that its 90 degrees

[radiant fox]

That's why I asked, but I didn't get an answer.

Tell me your way

[marsh roost]

ok so I constructed a perpendicular from C to AB

And triangle ABO (if u consider O to be the point on BC) is isosceles...

[radiant fox]

How do you know it's isosceles?

[marsh roost]

BO=AO=n.. isn't it?

[radiant fox]

I don't think that's always the case

[marsh roost]

A triangle is isosceles if two sides are equal.... isn't it?

here both sides are of length 'n' so its ofc isosceles......

[radiant fox]

how do you know each side is 'n'?

Can you draw it for me?

[marsh roost]

its given

see the diagram....

[radiant fox]

oh wait, I thought O was the point where your line from C to AB was

[marsh roost]

no no....

O is the point on BC....

[radiant fox]

yeah

How did you draw a perpendicular line from C to AB? Doesn't the intersection fall out of the triangle?

[vague galleon]

n/3

[radiant fox]

Why did you assume that A is 90°?

[vague galleon]

Because sometimes you gotta be gangsta and just assume.

[radiant fox]

Well, if A = 90°, then $\cot(A) = 0$. Which means that the ratio $\frac{\cot(B)}{\cot(A)}$ would be
$\frac{\cot(B)}{0}$ ...