Question

Solve the triangle.

B = 73°, b = 15, c = 10

Answer 1

C= 39.61 degrees
A= 67.39 degrees
a= 14.48

Answer 2

Thank you!! @moe_ha

Answer 3

can you help me with another one?...

Answer 4

Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles.

C = 67°, a = 21, c = 20

Answer 5

You're welcome. wait i think theyre is another part to the first one....

Yeah I can help with the other one. hold on.

Answer 6

can you explain what you did on the first one?

Answer 7

yeah sure so on the first one i used the law of sines.
so \[\frac{ \sin73 }{ 15 } =\frac{ sinC }{ 10 } \]
then \[\frac{ 10\sin73 }{ 15 } \]
which equals .637536504 then i did the arcsine of that which gave me C=39.61
then just add 73 and 39.61 and subtract that from 180 and that'll give you your A and that equals 67.39.
to find the side a
\[\frac{ \sin73 }{ 15 } \frac{ \sin67.39 }{ a }\]
that'll give you \[\frac{ 15\sin67.39 }{ \sin73 }\]
it'll equal 14.48
and that's how i did the first question

Answer 8

This is for the first triangle:\[\frac{ \sin67 }{ 20 } \frac{ sinA }{ 21 } = \frac{ 21\sin67 }{ 20 }\]
then use the arcsine to find A
A should equal 75.13 degrees
add 75.13 and 67 the subtract it from 180 it = 37.87 degrees and that is your B
\[\frac{ \sin67 }{ 20 } \frac{ \sin37.87 }{ b } =\frac{ 20\sin37.87 }{ \sin67 } =13.34\]

For the second triangle:
subtract 75.13 from 180
so this will give you a new A = 104.87 degrees
to find your new a you put it back into the law of sine formula
\[\frac{ \sin67 }{ 20 } \frac{ \sin104.87 }{ b } = \frac{ 20\sin104.87 }{ \sin67 } = 21.00\]
side a=21.00

to find your new B you add 67 and 140.87 which gives you 8.13 degrees
\[\frac{ \sin67 }{ 20 } \frac{ \sin8.13 }{ b } = \frac{ 20\sin8.13 }{ \sin67 } = 3.07\]
so your new side b = 3.07

Answer 9

so for your first triangle
A= 75.13 degrees
B= 37.87 degrees
b= 13.34

for the second triangle
A= 104.87 degrees
a= 21.00
B= 8.13 degrees
b= 3.07

hoped I helped :)

Answer 10

Thanks so much! @moe_ha

Answer 11

you're welcome!! if you have anymore questions just ask me :) i'll try to answer them

Answer 12

well there is one more but theres a picture and I can't post the picture.. its just to help.. can you give it a shot? @moe_ha

Answer 13

shoot. I'll try it.

Answer 14

Two weather tracking stations are on the equator 159 miles apart. A weather balloon is located on a bearing of N 38°E from the western station and on a bearing of N 14°E from the eastern station. How far is the balloon from the western station?

Answer 15

oh god...word problem... i cannot do these. i completely suck at them I'm sorry.

Answer 16

no worries! @moe_ha
what about
Solve the triangle.

A = 32°, a = 19, b = 12

A. B = 19.6°, C = 148.4°, c ≈ 22.5

B. B = 19.6°, C = 128.4°, c ≈ 28.1

C. Cannot be solved

D. B = 19.6°, C = 128.4°, c ≈ 16.9

Answer 17

sorry for answering so late @ChandlerT0125 i was working on a paper.
so the answer is b

Answer 18

Thank you! @moe_ha

Answer 19

You're welcome