Question

excuse me,
in triangle ABC, AB=5, COSB=-3/5
SINA=?
one more condition, i'll use the tool bar
sorry,cuz i dont know how to describe it in english

Answer 1

can you mark the sides of triangle and is cosb = -3/5?

Answer 2

Is this the diameter?

Answer 3

no picture in the question

Answer 4

i just want to show the circle which the 3 points of triangle would fall on it

Answer 5

and yes cos B =-3/5

Answer 6

Why does there have to be a circle, though? Was it included in the question?

Answer 7

It usually helps if we make a list, or an illustration of what we know.

Answer 8

sorry, it's just a condition
do you call that circumcircle?

Answer 9

Yeah, seems like it :)

Answer 10

cuz the question only has words description, it gives no pic

Answer 11

oh ok then it's just one of the conditions, saying the circumcircle's radius=13/2

Answer 12

Ok, studying about circumcircles now... :P

Answer 13

I'm hoping, sin B = 4/5. :)

Answer 14

i think so , cuz cosB =-3/5 so sinB must be 4/5

Answer 15

Yep. Well, as a result of the law of sines,
\[\huge \frac{\sin B}{AC}=\frac{\sin C}{AB}\]

Answer 16

\[\huge = \frac{\sin A}{BC}\]

Answer 17

And it so happens these are all equal to the diameter of the circumcircle, which is twice 13/2, which is just 13.

Answer 18

sorry, yeah i know that law
sinB/b=13 so b=4/65
sinC/c=13,c=5 so sinC= 65
but what can i do with sinA
thank you

Answer 19

Thinking... this is making me rather dizzy :D

Answer 20

If you can just get sin C, it would be easy :D

Answer 21

Okay, I'm going to be bloody with this
\[\large \frac{AC}{\sin B}=13 \ \ \ \ \ \ ; \ \ \ \ \ \frac{4}{5}\times 13 = AC\]

Answer 22

Since sin B is 4/5

Answer 23

\[\large AC = \frac{52}{5}\]

Answer 24

ah i got a way,
(c^2+a^2-b^2)/2ac=cosB
then i can get a
then cos formula again i can get cosA
then i can get sinA
do you think this is correct?

Answer 25

sorry i didn't notice above, sorry for making you feel bloody hell

Answer 26

\[\large \frac{\sin B}{AC}=\frac{\frac{4}{5}}{\frac{52}{5}}=\frac{4}{52}=\frac{\sin C}{AB}\]
The pieces are all coming together.

@ranxu6j3 I honestly have no idea, this is the first time I worked with triangles with circumcircles :D

There's no need to apologise; this is fun :)

Answer 27

in your calculation, we are still lack of sinA
This was exactly what i was stuck for?

Answer 28

@terenzreignz thanks for discussing with me

Answer 29

Relax :) We know AB = 5, don't we? :P
\[\large \frac{4}{52}=\frac{\sin C}{AB}=\frac{\sin C}{5}\]
\[\large \frac{4 \times 5}{52}=\frac{5}{13}=\sin C \]

Answer 30

And everything is coming together :D

Answer 31

sorry, the question wants sinA

Answer 32

I know. Be patient... and work on it :D

\[\large \frac{\sin C}{AB}=\frac{\sin A}{BC}\]

Answer 33

Might not have been the best way to go about this :D
Your approach may be better at this point :)

Answer 34

yeah, the bloody thing is that it doesn't give us BC
THANKS FOR HELPING , A LOT

Answer 35

Hang on, we can do this :)

Answer 36

sorry for having to be offline now