Question

Use the law of cosines to solve the triangle: a=16km B=40 degrees c=10km

Answer 1

Do you know the law of cosine?

Answer 2

Just learned it today

Answer 3

One second I'm going to work it out

Answer 4

\(\large \bf \textit{Law of Cosines}\\ \quad \\
c^2 = a^2+b^2-(2ab)cos(c)\\
c = \sqrt{a^2+b^2-(2ab)cos(c)}\)

Answer 5

Ok

Answer 6

do you have a picture? and what solve, means? find side b?

Answer 7

I have to find the missing sides and angles. How do u insert a picture

Answer 8

[Attache File] blue buttton, to the left-side of [Post] blue button

Answer 9

I don't see it

Answer 10

\[b^2=16^2+10^2-2(10)(16)*\cos(40)\]Square the 16 and the 10. Also multiply the -2 with (10)(16)

Which gives you \[b^2=356-320*\cos(40)\]

Do you know how to solve for b from here?

Answer 11

Yes

Answer 12

@Laurene96 well, you can post it at say http://imgur.com and give us the url to it

Answer 13

I got 5.25

Answer 14

Ok I'm uploading the pic

Answer 15

http://imgur.com/gbOpQ5j It's number 6

Answer 16

Type the whole "356-320*cos(40)" into your calculator. You subtracted first then did the cos

Answer 17

Oh ok. One sec.

Answer 18

10.53

Answer 19

That's right! I'm guessing they want you to round to the nearest tenth? so 10.5

Answer 20

She usually wants us to do hundredths but I'm not sure

Answer 21

\(\bf \textit{Law of Cosines}\\ \quad \\
b^2 = {\color{blue}{ a}}^2+{\color{green}{ c}}^2-(2{\color{blue}{ a}}{\color{green}{ c}})cos(B)\\
b = \sqrt{{\color{blue}{ a}}^2+{\color{green}{ c}}^2-(2{\color{blue}{ a}}{\color{green}{ c}})cos(B)}\\ \quad \\

b = \sqrt{{\color{blue}{ 16}}^2+{\color{green}{ 10}}^2-(2\cdot {\color{blue}{ 16}}\cdot {\color{green}{ 10}})cos(40^o)}\)

Answer 22

anyhow.. as Ryan88 has shown :)

Answer 23

Well it's probably best to do as your teacher likes, so b = 10.53

Answer 24

you have to find the angles A and C, right?

Answer 25

Alright

Answer 26

And yes

Answer 27

Would I use sin-1 to find angle A?

Answer 28

well, you'd use the Law of Cosines as well... lemme .. try instead say angle C \(\bf \textit{Law of Cosines}\\ \quad \\
c^2 = a^2+b^2-(2ab)cos(C)\implies c^2-a^2-b^2=-(2ab)cos(C)\\ \quad \\
a^2+b^2-c^2=(2ab)cos(C)\implies \cfrac{a^2+b^2-c^2}{(2ab)}=cos(C)\\ \quad \\
{\color{red}{ \square }}=cos(C)\implies cos^{-1}({\color{red}{ \square }})=cos^{-1}[cos(C)]\implies cos^{-1}({\color{red}{ \square }})=\measuredangle C\)

Answer 29

Ok

Answer 30

Use the Law of Sin

\[\frac{ 10.53 }{ \sin(40) }=\frac{ 16 }{ Sin(A) }\]

Answer 31

Either or, both do the exact same

Answer 32

I got 77.61

Answer 33

for the angle?

Answer 34

Yeah

Answer 35

\(\bf \cfrac{16^2+10.53^2-10^2}{2(16)(10.53)}=cos(C)\\ \quad \\\implies 37.62480093287110170194 \approx cos(C)\)

Answer 36

wel.... one...hmm ..lemme recheck that

Answer 37

\(\bf \cfrac{16^2+10.53^2-10^2}{2(16)(10.53)}=cos(C)\\ \quad \\\implies 0.79202546296296296296\approx cos(C)\\ \quad \\37.62480093287110170194 \approx \measuredangle C\)

Answer 38

Oh I was looking for angle a instead of c

Answer 39

I got 37.62 for angle C.

Answer 40

And 102.38 for angle A

Answer 41

I got 37.62 for angle C
And 102.38 for angle A \(\Large \checkmark\)

Answer 42

Thanks for the help!

Answer 43

yw