Question

\[In \triangle ABC(A2−B2)/(A2+B2)=\sin(A−B)/\sin(A+B)then the △ is\]

Answer 1

is it exactly like that no brackets missing anywhere right

Answer 2

(a^2-b^2)/(a^2\(\color{Red}{+}\)b^2) =sin(a-b)/sin(a+b)

like this ?

Answer 3

ooo

Answer 4

is he gone xD

Answer 5

@dg2

Answer 6

yes

Answer 7

can u edit the quesiton and fix typo ?

Answer 8

also use parenthesis so that it is less ambiguous

Answer 9

also use capital letters for angles : A,B,C
and small letters for sides : a,b,c

Answer 10

I have modified the question, see if it looks okay now

Answer 11

\[In \triangle ABC (A^2-B^2)/(A^2+B^2)=\sin(A-B)/\sin(A+B) then the \triangle is\]

Answer 12

now okay .

Answer 13

copy paste below :

```
In \(\triangle ABC\) ,
\[
\dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)}
\]

then the \(\triangle \) is ?

```

Answer 14

\[In \(\triangle ABC\) , \[ \dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)} \] then the \(\triangle \) is ?\]

Answer 15

don't use any brackets, copy past exact same stuff

Answer 16

In \(\triangle ABC\) ,
\[
\dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)}
\]

then the \(\triangle \) is ?

Answer 17

finally ! congrats !!

Answer 18

familiar with `componendo and dividendo` ?

Answer 19

yes .i got struck with this site .(a^2-b^2)/(a^+b^2)*(a^+b^2)/(a^+b^2)

Answer 20

oh sorry sin(A-B)/sin(A+B)*sin(A+B)/sin(A+B)

Answer 21

componendo and dividendo :

If
\[\dfrac{p}{q} = \dfrac{r}{s}\]

then :

\[\dfrac{p+q}{p-q} = \dfrac{r+s}{r-s}\]

Answer 22

Since you're given :

\[\dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)}\]

applying componendo and dividendo, you get :

\[\dfrac{(a^2-b^2) + (a^2 + b^2)}{(a^2-b^2) + (a^2 + b^2) } = \dfrac{\sin(A-B) + \sin(A+B)}{\sin(A-B) - \sin(A+B)}\]

Answer 23

latex is not working for me

Answer 24

simplifying a bit, you would get :

\[\large \dfrac{a^2}{b^2 } = \dfrac{\sin A \cos B}{\cos A \sin B}\]

Answer 25

here is a screenshot

http://www.webpagescreenshot.info/img/53f70852d9c8f7-85003353

Answer 26

screenshot also not loaded;;;) i understood a^2/b^2=(sina*cosb)/(cosa*sinb)

Answer 27

so tan a*cot b ?

Answer 28

yes


a^2/b^2 = tanA cotB

Answer 29

what to conclude from here ? hmm

Answer 30

I have posted this question in stackexchange

http://math.stackexchange.com/questions/905844/if-dfraca2-b2a2b2-dfrac-sina-b-sinab-then-what-type-o

Answer 31

Woah ! look at the answer there, we're done !

Answer 32

right angled triangle ahh

Answer 33

isosceles right angled triangle

Answer 34

yes using sine rule, you can replace

a/b with sinA/sinB

Answer 35

a^2/b^2 = tanA cotB


sin^2A / sin^2B = tanA cotB

Answer 36

simplify

Answer 37

(1-cos^2a)/(1-cos^2b)

Answer 38

don't do that

Answer 39

a^2/b^2 = tanA cotB

sin^2A / sin^2B = tanA cotB

sin^2A / sin^2B = (sinAcosB)/(cosAsinB)

sinA/sinB = cosB/cosA

sinAcosA = sinBcosB

sin(2A) = sin(2B)

A = B

Answer 40

so its an `isosceles triangle`

Answer 41

wait :)

Answer 42

ha bolo

Answer 43

yes .thank you so much

Answer 44

np :)