Solve the triangle. A = 46°, a = 31, b = 27

Solution:

The information provided in the problem statement is summarised in the figure below:

By solving the triangle it implies that the values of ∠B, ∠C, and side BA have to be determined.

Therefore to obtain these values a perpendicular is dropped from vertex B onto the side AC as shown in the diagram above.

We know that. cos46° = AD/AC and AC = b = 27

cos46° = 0.6944

Therefore.

AD = ACcos46°

= 27cos46°

= 27(0.6944)

= 18.75

We also determine the perpendicular CD from the following expression

tan46° = CD/AD

= CD/18.75 CD = 18.75tan46°

= 18.75(1.0362)

= 19.43

Sin ∠ACD = AD/AC

= 18.75/27

= 0.6944

Therefore

∠ACD = asin(0.6944)

= 44°

We also know cos∠ACD = CD/BC

= 19.43/31

= 0.6268

∠ACD = acos(0.6268)

= 51.17°

Therefore ∠ACB = ∠ACD + ∠DCB

= 51.17°+ 44°

= 95.17°

Also,

Sin(51.17°) = BD/BC

= BD/31 BD = 31sin(51.17°)

= 31(0.7792)

= 24.16

Now AB = AD + DB

AB = 18.75 + 24.16

= 42.95

Therefore all the parameters of the triangle are obtained as follows:

a = 31; c = 17; b = 27; c = 42.95, ∠ACB = 95.17; ∠ABC = 180 - 46 - 95.17 = 38.83°

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Solve the triangle. A = 46°, a = 31, b = 27

Summary:

Solving the given triangle with ∠A = 46°, a = 31, b = 27 we have ∠B = 95.17; and ∠C = 38.83°