Need help with algebra 2 triangle prob plz!!!?

Question

okay dudes or dudettes,

Solve the triangle ABC:

A=120 degrees, B = 40 degrees, c=35

I knew how to do this but I forgot. Can someone show me how to do it, or at least gimme the formula so I can figure it out? thanks!

Answer 1

sinA/a=sinB/b=sinC/c

A+B+C=180

first solve for angle C using the second one. turns out to be

120+40+C=180

C=20

sin(20)/35=sin(120)/a

.00977a=.866

a=88.623

sin(20)/35=sin(40)/b

.00977b=.643

b=65.779

make it a good day

Answer 2

Really hate math

But i dont even study :D

Answer 3

there are 180 degrees in a triangle.  therefore, 180-120-40=your third angle.  this is true for all triangles. so, C = 20

to find your other two sides, you have to use the law of sines and cosines.



Law of sines says (sinA/a) = (sin B/b) = (sinC/c)

so, use the equation: sin20/35 = sin 120/a

solve for a.

Answer 4

You have to apply just 2 theorems for triangle:

A + B + C = 180 degrees

a/sinA = b/sinB = c/sinC

Answer 5

what exactly r  u trying to find? u got all 3 angle measures....

Answer 6

You have two angles, so 180 - (120 + 40) = 20 degrees for angle C

Using the sine rule to solve for side a, you have

35/ sin 20 = a/ sin 120

a = 35 * sin 120 / sin 20

a = 88.62

You can use either the sine rule or cosine rule to find side b, but for simplicity's sake we'll use the sine rule using the set C.

35/ sin 20 = b/sin 40

b = 35 * sin 40 / sin 20

b = 65.78

therefore

A = 120 degrees

B = 40 degrees

C = 20 degrees

a = 88.62

b = 65.78

c = 35

Answer 7

C = 180° - A - B

a = cSinA/SinC

b = cSinB/SinC

Answer 8

looks like you have to use the sin, cos & tan formulas

(remember soh cah toa?)

Answer 9

use law of sines and cosines.

Answer 10

I need the whole equation to solve this