Question

Sollutions of triangles

Answer 1

@ganeshie8 , @sidsiddhartha , @perl , @Miracrown

Answer 2

yeah sot have only got formulas , no problem

Answer 3

Wouldn't \(abc = \sqrt R\) though? :|

Answer 4

i made a mistake

Answer 5

\[abc=\sqrt{R}\]

Answer 6

well i used projection formula but i was not able to proceed any further

Answer 7

Also,\[a^2 + b^2 + c^2 = P=( a + b + c)^2 - 2(ab + bc + ca)\]Not really sure if Vieta's Formulas alone would be able to help us. And I have forgotten my trigonometry. :(

Answer 8

Let a,b,c be the sides of a triangle AB.
\[\large a^2 , b^2 ,c^2 \]
are the roots of the equation
\[\huge x^3-Px^2+Qx-R=0\]
where P,Q and R are constants, then find the value of
\[\huge \frac{ cosA }{ a } + \frac{ cosB }{ b }+ \frac{ cosC }{ c }\]
Find the value in terms of P,Q and R

Answer 9

@Miracrown Hey, could you give us a hand here?

Answer 10

@Miracrown

Answer 11

Wondering if cosine rule would be of any help...

Answer 12

I didn't want to take any risk , but i thought of that , good idea thought , i don't if that would work , i am trying ,

Answer 13

though**

Answer 14

I don't immediately see the connection between these questions.
There is a cubic equation in the middle, which has nothing to do with triangle problems.

Answer 15

PERFECT!

Answer 16

It works

Answer 17

Can't sides of a triangle be the roots of a cubic equation , as there are three sides

Answer 18

\[\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\]\[\cos B = \dfrac{a^2 + c^2 - b^2}{2ac}\]\[\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}\]

Answer 19

The denominator turns out to be the same when you plug in...

Answer 20

thank you owlfred

Answer 21

=)

Answer 22

Smart bird! =]

Answer 23

waiting for you @sidsiddhartha , kya type kar rahe ho

Answer 24

What language is this: ''kya type kar rahe ho'' ?

Answer 25

Hindi.

Answer 26

Hindi

Answer 27

I understand the ''type'' bit, but the rest ... no idea lol

Answer 28

It says, "What are you typing?"

Answer 29

Bharath ki rashtra bhasha , hindi

Answer 30

@sidsiddhartha Late lateef!

Answer 31

actually We got the answer hehhe

Answer 32

yeah very much lagging :(

Answer 33

I had no idea \[a\cdot b\cdot c = \sqrt{R}\]

Answer 34

and \[x^3-px^2+qx-r=0\\from~this\\s_1=a+b+c=p\\s_2=\sum ab=q\\s_3=abc=r\]

Answer 35

Theory of equation or vieta's formula