Question

Yoo can somebody help me?

Answer 1

Answer:

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