Yoo can somebody help me? On each triangle, solve both of them and round to the nearest tenth. Make sure to show your work for both. Thank you in advance.

Step-by-step explanation:

Triangle 1   given:  100° , 17 yds,  30 yds

Triangle 2 given: 30 ,17,18

law of cosines: c^2 = a^2 + b^2 − 2ab cos(C)

we'll also  need the formula for a SSS triangle

Cos(C) = a^2 + b^2 - c^2 / 2ab

keeping the above in mind we can now solve both triangles

c = AC for the first triangle

solve law of cosines for what we know

c^2= 17^2 + 30^2 - 2*17*30*Cos(100)

c = 36.96

now use SSS triangle  formula to find the angle C, but c = 17 now , not 36.96, that's b

C = arcCos(17^2+36.96^2-30^2/ 2*17*36.96)

C = 42.24°

now use 180 = 100 +42.24 +A   to solve the last angle

A= 37.75°

Triangle 1 solved

A= 37.75°

B= 100

C = 42.24°

a= 17

b=30

c=36.96

Triangle 2  

given a=17 , b= 18 , c =30

use SSS formula to find one angle, which ever one you want in this case

Cos(C) = a^2 + b^2 - c^2 / 2ab

C = arcCos(17^2+18^2-30^2 / 2*17*18

C= 117.966°

solve for A now

Cos(A) =b^2+c^2-a^2 / 2bc

A = arcCos(18^2+30^2-17^2/2*18*30)

A = 30.03

use 180 again to solve for last angle

180 = 117.966+30.03+B

32.001 = B

Triangle 2 solved

A = 30.03°

B= 32.001°

C= 117.966°

a=17

b=18

c=30

:P  whew  

answer: what are the following choices

answer: 12x^2+6x

step-by-step explanation:

6x does not equal 24x. hope it works!