Question

solve each triangle. if there are two triangles solve them both.

A= 9 degrees 52' a= 3761ft b=5293ft

Answer 1

@jim_thompson5910

Answer 2

9 degrees, 52 arcminutes = 9+(52/60) = 9.86667 degrees approx

Answer 3

to find x, you'll need to use the law of cosines. Are you familiar with that formula?

Answer 4

sin9.8667/x = sin a/ 3761

Answer 5

no you're thinking of the law of sines

Answer 6

whats law of cosine then

Answer 7

the law of sines cannot be used yet. We need to figure out x first

actually now that I think about it, we can use the law of sines to find angle B

Answer 8

how do we do that?

Answer 9

let's use the law of sines to find angle B

\[\Large \frac{\sin(B)}{b} = \frac{\sin(A)}{a}\]
\[\Large \frac{\sin(B)}{5293} = \frac{\sin(9.86667)}{3761}\]
does that help at all?

Answer 10

0.645

Answer 11

how are you getting that?

Answer 12

i plugged that equation into the calculator

Answer 13

\[\large \frac{\sin(B)}{5293} = \frac{\sin(9.86667^{\circ})}{3761}\]
\[\large 3761*\sin(B) = 5293*\sin(9.86667^{\circ})\]
\[\large \sin(B) = \frac{5293*\sin(9.86667^{\circ})}{3761}\]
\[\large \sin(B) \approx 0.2411559\]
\[\large B \approx \arcsin(0.2411559) \ \text{ or } \ B \approx 180-\arcsin(0.2411559)\]
\[\large B \approx 13.95477255^{\circ} \ \text{ or } \ B \approx 166.045227^{\circ}\]
Let me know if you have any questions on how I got those values for B

Answer 14

i understand that makes sence

Answer 15

If B is approximately 13.95477255 degrees, then what is the value of angle C?

Answer 16

is it 180-13.955? if it is then i got 166.045

Answer 17

don't forget about angle A

Answer 18

A+B+C = 180 degrees

Answer 19

156.178

Answer 20

180-13.95477255-9.86666666666667 = 156.178560783333

looks good

Answer 21

If B is approximately 166.045227 degrees, then what is the value of angle C?

Answer 22

i thought b was 13.955

Answer 23

B is two possible angles (we're in the SSA case)

Answer 24

ohhh okay sorry i was confused

Answer 25

eg:

sin(30) = 1/2
sin(150) = 1/2

so if you had sin(x) = 1/2 then x = 30 or 150 degrees

Answer 26

oh ok

Answer 27

If B is approximately 166.045227 degrees, then what is the value of angle C?

Answer 28

13.955?

Answer 29

If B is approximately 166.045227 degrees, then what is the value of angle C?


C = 180-A-B
C = 180-9.86666666666667-166.045227
C = 4.08810633333331

Answer 30

agreed?

Answer 31

yes

Answer 32

So we have 2 possible triangles we can form here

Triangle 1
A = 9.86666666666667 degrees
B = 13.95477255 degrees
C = 156.178560783333 degrees


Triangle 2
A = 9.86666666666667 degrees
B = 166.045227 degrees
C = 4.08810633333331 degrees


Angle A stays the same both times
all three angles (for either triangle) are approximate

Answer 33

let me know when you're ready for the next part

Answer 34

im ready

Answer 35

For now, let's just focus on

Triangle 1
A = 9.86666666666667 degrees
B = 13.95477255 degrees
C = 156.178560783333 degrees

Answer 36

okay

Answer 37

using these values, are you able to find x in the drawing?

Answer 38

i think so but im not sure how to

Answer 39

you'll use the law of sines
\[\Large \frac{\sin(C)}{c} = \frac{\sin(A)}{a}\]
\[\Large \frac{\sin(156.17856^{\circ})}{c} = \frac{\sin( 9.8667^{\circ})}{3761}\]
I'll let you solve for c

Answer 40

7301.260

Answer 41

incorrect. Try again

Answer 42

-3549.478

Answer 43

\[\Large \frac{\sin(C)}{c} = \frac{\sin(A)}{a}\]
\[\Large \frac{\sin(156.17856^{\circ})}{c} = \frac{\sin( 9.8667^{\circ})}{3761}\]
\[\Large 3761*\sin(156.17856^{\circ}) = c*\sin( 9.8667^{\circ})\]
\[\Large \frac{3761*\sin(156.17856^{\circ})}{\sin( 9.8667^{\circ})} = c\]
\[\Large c = \ ???\]

Answer 44

8864.263

Answer 45

I'm getting 8,864.68360894793

Answer 46

hmm i wonder why

Answer 47

so...

Triangle 1
A = 9.86666666666667 degrees
B = 13.95477255 degrees
C = 156.178560783333 degrees

a = 3761
b = 5293
c = 8,864.68360894793

Answer 48

there may be rounding error somewhere

Answer 49

probably