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------------------------------------------------ Solve the triangle ABC, if the triangle exists.

Given:
- Angle B = 35°18'
- Side a = 38.2
- Side b = 32.6

Determine the remaining angles and sides of the triangle.

Answer :

The solution is:

a = 38.2

b ≈ 22.5

c = 30.0

A = 90°

B = 35°18'

C ≈ 54°42

'

We are given that B = 35°18', a=38.2 and b=32.6. To solve the triangle ABC, we can use the Law of Sines and Law of Cosines.

First, we can find the measure of angle C:

C = 180° - A - B

C = 180° - 90° - 35°18'

C = 54°42'

Next, we can use the Law of Sines to find the length of side c:

sin(A) sin(C)

----- = -----

a c

sin(90°) sin(54°42')

------- = -------

38.2 c

c = 38.2*sin(54°42') / sin(90°)

c = 30.0 (rounded to one decimal place)

Finally, we can use the Law of Cosines to find the length of side b:

b^2 = a^2 + c^2 - 2*a*c*cos(B)

b^2 = 38.2^2 + 30.0^2 - 2*38.2*30.0*cos(35°18')

b ≈ 22.5

Therefore, the solution is:

a = 38.2

b ≈ 22.5

c = 30.0

A = 90°

B = 35°18'

C ≈ 54°42'

Note that since the value of b is less than the value of a, this triangle is an obtuse-angled triangle.

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