Solve a triangle with a = 32. b = 38, and c = 46.
a. a = 43.5°: b = 54.8°: c= 81.70
b. a = 54.3º: b = 54.8°; c = 81.70
c. a = 43.5°: b = 66.5°; c = 81.7
d. a = 43.50: b = 54.8°; c = 78.4°

Using the law of cosines:

Angle A

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

32^2 = 38^2 + 46^2 - 2 * 38 * 46 * cos(Angle A)

1024 = 1444 + 2116 - 3496 * cos(Angle A)

3496 * cos(Angle A) = 1444 + 2116 - 1024

3496 * cos(Angle A) = 2536

cos(Angle A) = 2536 / 3496

Angle A = arccos(2536 / 3496)

Angle A = 43.5 degrees.

Using the same steps calculate angle b and angle c ( rearrange the formula to have b^2 and c^2 equal to:

Angle b = 54.8 degrees

Angle C = 81.7 degrees.

The answer is A. A = 43.5°: B = 54.8°: C= 81.70

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

i can't see the picture : /