Question

I need to find the missing parts of the triangle!

Answer 1

I know I need to use SSA, but I'm confused since I don't know a matching side & angle

Answer 2

are you doing trig? have you done yet the Law of Cosines?

Answer 3

Sorry, I didn't see you had typed. & No I'm in pre-calculus. But yes I have. I'm studying for my review for my exam & I hadn't gotten this problem right on my homework so I am trying to figure out how to solve it

Answer 4

so, I gather you haven't covered the law of cosines yet?

Answer 5

I have covered it, but in the very beginning of the school year, so I don't really remember it too well

Answer 6

ok, well, you have 1 angle given, and the two sides "coming out" of that angle, so that'd be the law of cosines :)

Answer 7

that'd give you side c
from there you'd need the law of sines

Answer 8

http://www.mathwarehouse.com/trigonometry/images/law-of-cosines/law-of-cosines-formula-and-picture2.gif

Answer 9

Since i'm finding c, i would use the last equation? & thank you for helping, math is just so darn confusing!

Answer 10

When I did that I got 57.74 so I did something wrong..

Answer 11

so to find side "c" \( \large c = \sqrt{5^2+8^2-(5\times 8)cos(67^o)}\)

Answer 12

Ohh I didn't square root it

Answer 13

So that would give me c = 7.6

Answer 14

So how would I find one of the missing angles? do I use the same equations?

Answer 15

law of sines
https://classroom.peoriaud.k12.az.us/sites/caylward/Classroom%20Images/Precal%20Sem%202/Trenton%20Flores/formula-picture-law-of-sines.gif

Answer 16

$$
\cfrac{c}{sin(67^o)} = \cfrac{8}{sin(B)}
$$

Answer 17

\[\frac{ \sin(67) }{ 7.6 } = \frac{ sinB }{ 8 }\]

then do I cross multiply for..
\[\sin(67) \times 8 = sinB \times 7.6\]

Answer 18

is it me or I got 8.5656 for side "c"

Answer 19

yes, my calculator is in Degrees mode

Answer 20

Oh shoot. So am I wrong about the 7.6?

Answer 21

Mine is in degrees mode too

Answer 22

hmm

sqrt(25+64-(40*cos(67))) = 8.5656

Answer 23

I'll check in radians

Answer 24

even a bigger, so hehe, no that

Answer 25

Well I did it again and got your answer, but 8.5 nor 8.6 was an option but 7.6 was

Answer 26

hold.... I got one mistake

Answer 27

we didn't do 2ab just ab

Answer 28

yes, I just saw that :/

Answer 29

$$
\large c = \sqrt{5^2+8^2-2(5\times 8)cos(67^o)}
$$

Answer 30

Now that is 7.6 :)

Answer 31

hehehe, so it's :)

Answer 32

So now how would I find the angle?

Answer 33

$$
\cfrac{7.6}{sin(67^o)} = \cfrac{8}{sin(B)}\\
\implies sin(B) = \cfrac{sin(67^o)\times 8}{7.6}
$$

Answer 34

so, that gives a number, arcSine it :)

Answer 35

75.68533... or 76

Answer 36

right

Answer 37

So then 67+76 = 143 & 180-143=37 for angle a

Answer 38

so the other angle is 180-(B+C)

Answer 39

Yay!!! Thank you soooooo much. I really wish I could give you more than one medal.

Answer 40

yw