Question

How do I find a side, when only given 1 angle and 1 side? in an non-right angled triangle

Answer 1

Are the angles and sides opposite each other?

Answer 2

If so, use sin law

Answer 3

Whereas sinA/a=sinB/b

Answer 4

@IsTim , I put a picture up

Answer 5

is the angle at c?

Answer 6

Yes,

Answer 7

you have an iscoseles triangle

Answer 8

Well, one way you could figure this out is Sin Law. Look it up on purplemath, Khan's Academy or wolframalpha if you're new to the concept.

Answer 9

@alekos, yes

Answer 10

Do you understand, @Sulle26 ?

Answer 11

has it clicked yet?

Answer 12

@IsTim, not really....
Can i find the other two angles by using 180-33.10 and then divide it by 2 to get those to angles, since it's an iscoceles triangle? @alekos

Answer 13

You're looking for side lengths though, aren't you?

Answer 14

But yes, Sulle26, you can do that to find the angle, which leads into using Sin Law.

Answer 15

@IsTim, yes. So if i have at least 2 angles, I can find the remaining side, can't i ?

Answer 16

Using Sin Law, yes.

Answer 17

@IsTim, thanks!

Answer 18

you're heading in the right direction
and remember the two unknown sides are also equal

Answer 19

dont find side as simple as that

Answer 20

@alekos Yup! I'll remember!

Answer 21

@thenerdisback What do you mean?

Answer 22

@Sulle26 I think this question is done. You can close it now.

Answer 23

You can also use the Law of Cosines
\[
c^2=a^2-2 a b \cos (c)+b^2=-2
a^2 \cos (c)+a^2+a^2=2 a^2
(1-\cos (c))\\


a^2=\frac{c^2}{2 (1-\cos
(c))}=\frac{6.71^2}{2 (1-\cos (33.1
{}^{\circ}))}=138.72\\


a=\sqrt{138.72}=138.72
\]

Answer 24

How is sqroot of 138.72 the same? but 11.77, right ? :-) @eliassaab

Answer 25

Yes.

Answer 26

I just used sin, and I got the same answer. Thanks everyone! :-)

Answer 27

Sorry that was a misprint
\[
c^2=a^2-2 a b \cos (c)+b^2=-2
a^2 \cos (c)+a^2+a^2=2 a^2
(1-\cos (c))\\


a^2=\frac{c^2}{2 (1-\cos
(c))}=\frac{6.71^2}{2 (1-\cos (33.1
{}^{\circ}))}=138.72\\


a=\sqrt{138.72}=11.78
\]