Question

Sollutions of triangles

Answer 1

In a triangle ABC
A:B:C = 3:5:4
Then
\[\huge a+b+c \sqrt{2}\]
is equal to

Answer 2

(A) 2b
(B) 2c
(C) 3b
(D) 3a

Answer 3

let
A=3k
B=5k\C=4k

Answer 4

OK

Answer 5

A+B+C=180

Answer 6

got it

Answer 7

so
3k+4k+5k=180
k=15

Answer 8

ok?

Answer 9

then sine rule

Answer 10

so we have now
A=45
B=75
C=60

Answer 11

yes

Answer 12

Basically I was dong prove that type questions uptillnow so brain was not working
I got it now

Answer 13

\[a+b+c \sqrt{2}=k(sinA+sinB+\sqrt{2} sinC)\]

Answer 14

ok?

Answer 15

yes

Answer 16

now just plug the values

Answer 17

yeah

Answer 18

\[a+b+\sqrt {2} c=k(\frac{ 1 }{ \sqrt{2} }+\frac{ \sqrt{3}+1 }{ 2\sqrt{2} }+\sqrt{2}\frac{ \sqrt{3} }{ 2 })\]

Answer 19

I will do the rest

Answer 20

ok :)

Answer 21

@sidsiddhartha how did you get that equation
a + b + c sqrt(2) = k ( sin A + sin B + sin C )

Answer 22

using sine rule\[\frac{ sinA }{ a }=\frac{ sinB }{ b }=\frac{ sinC }{ c }\]

Answer 23

i dont see how from that step you can get a + b + c = k ( sin A + sin B + sin C )

Answer 24

al that equals a constant ktake the denominator to the right and see

Answer 25

just used that property\[a=k.sinA\\b=k.sinB\\c=k.sinC\]

Answer 26

why @perl ?

Answer 27

oh i thought you were using the same k from befor

Answer 28

so whats your final answer udit?

Answer 29

how to do after that

Answer 30

I got answer in terms of newly defined "k" for sine rule but answer is in terms of b

Answer 31

b ,a ,c,

Answer 32

from where i left--\[a+b+\sqrt{2}c=k(\frac{ 1 }{ \sqrt{2} }+\frac{ \sqrt{3}+1 }{ 2\sqrt{2} }+\frac{ \sqrt{3} }{ \sqrt{2} })\\=k \frac{ 2+\sqrt{3}+1+2\sqrt{3} }{ 2\sqrt{2} }=3k*\frac{ \sqrt{3}+1 }{ 2\sqrt{2} }=3k.\sin75^o=3b\]
thats what i'm getting

Answer 33

ok gotta go now

Answer 34

thanks ok bye!! =:)

Answer 35

is 3b ok as a solution?

Answer 36

so there is no unique real number solution because you are only given the three angles , and you can dilate the triangle to any size

Answer 37

answer check karke dekha