Question

HELP!

Answer 1

Can't see enough of your problem to figure it out.

Answer 2

is it triangle?

Answer 3

then use cosine formula a^2=b^2+c^2-2bc*cosA

Answer 4

just substitute the values, you will get the answer

Answer 5

@SerikMB did you get 18.33?

Answer 6

yes, sth around that, 19.24

Answer 7

which one is correct? lol

Answer 8

they may be both correct. The error can be due to calculation and rounding error.

Answer 9

& can you please set up the equation for this one?

Answer 10

\[a^{2}=b ^{2}+c ^{2}-2bc*\cos(A)\]

Answer 11

use that formula again, it is the same question, just substitute different numbers

Answer 12

yeah but this one only has 3 sides? no angle?

Answer 13

i see, then use sine theorem

Answer 14

but its cos..

Answer 15

that is \[\frac{ a }{ \sin(A) }=\frac{ b }{ \sin(B) }=\frac{ c }{ \sin(C) }\]

Answer 16

here is the trick A+B+C=180

Answer 17

\[34^2=28^2+15^2-2(28)(15)\cos A\]

Answer 18

Solve that for A

Answer 19

Am I right? 99 degrees?

Answer 20

I got 100 degrees.

Answer 21

i got 99?

Answer 22

that's right, both are correct, it's just rounding error

Answer 23

Nope. 100.08 degrees.

Answer 24

am I right on this one? @SerikMB

Answer 25

if i am right, then no, try again

Answer 26

hmm .. im stuck

Answer 27

48 - angle of P
60 angle of Q
36 angle of R

Answer 28

do not get confused

Answer 29

so we use sine? or the cosine ?

Answer 30

Law of cosines

Answer 31

use cosine

Answer 32

as Mertsj has shown you in a previous example

Answer 33

i got 24 degrees