Question

I need help with figuring out angles using Pythagorean theorem. For example, being given a triangle, with all three side's measurements given, and being asked to solve for the missing angles. (No angle's are given)

Answer 1

May I help?

Answer 2

Yes please!

Answer 3

With my pleasure!

What did you get?

Answer 4

Like what is the question?

Answer 5

do you have it or what?

Answer 6

Yeah sorry, The triangles sides are 16km, 19km, and 19km again. And I need to calculate the angles.

Answer 7

But that triangle is not a right one!!

Answer 8

So then it wouldn't be Pythagorean theorum? My options were that, Sine law, Cosine law, and SOHCAHTOA

Answer 9

So we can use the fist two not the third!

Answer 10

How does that work? I thought you needed at least one angle to use those?

Answer 11

Not a conditon!

you can use Cosine law to get the first angle, then Sine law to get the second, the (sum=180) to get the third!


Got that!

Answer 12

Ohh okay that makes sense!

Answer 13

Definitely!

Answer 14

Can you share your works/steps, please?

Answer 15

I just need a minute or two to see if I got this first!

Answer 16

Take your time.

Answer 17

I'm a little rusty with Cosine law as well, so just to make sure I'm setting this up right..
\[a ^{2} = b ^{2} + c ^{2} - 2 (b) (c) cosA \] is the equation, so if I were to plug in the numbers it would be \[19^{2} = 16^{2} + 19^{2} -2 (16) (19) cosA\]

Answer 18

is this right so far?

Answer 19

Let me see

Answer 20

\[\Huge\color{darkred}\checkmark\]

so A=?? degree

Answer 21

yess

Answer 22

Excellent!

you got A=##??

Answer 23

I think I did something wrong, I ended up with 361 = 9cosA

Answer 24

I'm not sure what to do

Answer 25

For me, I get almost A=65

Check again and I am with you

Answer 26

Can I send you my steps and you tell me what I did wrong?

Answer 27

Thank you for the medal!

Answer 28

Of course.

Answer 29

\[19^{2}=16^{2}+19^{2}-2(16)(19)cosA\]
\[19^{2}=16^{2}+19^{2}-2(304)cosA\]
\[361=256+361-2(304)cosA\]
\[361=256+361-608cosA\]
\[361=9cosA\]

Answer 30

Brilliant! Ma Sha' Allah

Answer 31

Thats all right??

Answer 32

but just last one!
\[361=256+361-608cosA~~~~~~~~~~~~~~~-256=-608~\cos(A)\]
\[\Huge \color{LightSalmon }{ \cos(A)=\frac{ -256 }{ -608 } =\frac{ 8 }{ 19 }=0.4211}\]
\[\Huge \color{Tomato }{ A=65}\]

Answer 33

@alexismiichaela

got it?

Answer 34

That's actually confusing me a little bit.. where did the two 361 values go?

Answer 35

cancel each other
what if you subtract 361 from each side????

Answer 36

OH right, okay I'll take another look to make sure I absolutely understand it

Answer 37

Take your look

Answer 38

Okay so I get that, and I also understand how the 256 was transferred onto the right side of the equal sign, but how does Cos(A) get transferred to the right, and the 256 value, back to the left side

Answer 39

\[361=256+361-608cos(A)\]
\[361-361=256+361-361-608cos(A)\]
\[0=256-608cos(A)\]
\[0+608cos(A)=256-608cos(A)+-608cos(A)\]
\[608cos(A)=256\]

\[\Large \cos(A)=\frac{ 256 }{ 608 }\]
got that?

Answer 40

Yes! That looks better