Here it just says this:
Solve triangle ABC. (If an answer does not exist, enter DNE. Round your answers to one decimal place.)
b = 67, c = 33, ∠A = 70°
use these formulea
a^2=b^2+c^2-(2bcCosA)
then to get angle C
sinA/a=sinC/a...(1)
to get angle B
sinB/b=sinC/c....(2)
there you go a big hint
or, once you have A and C, B = 180-(A+C)
To solve triangle ABC, we can apply the Law of Sines or the Law of Cosines. Let's use the Law of Sines in this case.
The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant. This can be written as:
a/sin(A) = b/sin(B) = c/sin(C)
Given:
b = 67
c = 33
∠A = 70°
We need to find side a and angles B and C.
To find side a, we can use the Law of Sines:
a/sin(A) = b/sin(B)
Substituting the known values:
a/sin(70°) = 67/sin(B)
Now we can solve for sin(B):
sin(B) = (67 * sin(70°)) / sin(B)
Using a scientific calculator, we find that sin(B) ≈ 0.932
Now we can solve for angle B using the arcsine function:
B = sin^(-1)(0.932)
B ≈ 69.5°
To find angle C, we can use the fact that the sum of the angles in a triangle is always 180°:
C = 180° - A - B
C = 180° - 70° - 69.5°
C ≈ 40.5°
Therefore, the solution for triangle ABC is:
Side a ≈ 53.9
Angle B ≈ 69.5°
Angle C ≈ 40.5°