Solve the triangle. A = 19°, C = 102°, c = 6

Solution:

The given information is summarized in the diagram below:

To find the other sides of the triangle we drop a perpendicular (CD) from vertex C to side AB.

tan19° = CD/AD = CD/(6 - x)

CD = (6 - x)tan19°

= (6 - x)(0.3445)

= 2.067 - 0.3445x----------->(1)

Also

tan 59° = CD/DB

= CD/x

CD = x . Tan 59°

= x(1.666)

= 1.666 x----------->(2)

Equating (1) and (2) we have

2.067 - 0.3445x = 1.666x

Therefore

2.067 = 1.666x + 0.3445x

2.067 = 2.0105x

x = 2.067/2.0105

x = 1.0281

cos59° = BD/BC

Therefore,

BC = BD/cos59°

= x/cos59°

BC = 1.0281/0.5147

= 1.997

Similarly,

cos19° = AD/AC

AC = (6 - x)/cos19°

= (6 - 1.0281)/cos19°

= 5.9729/0.9455

AC = 6.3172

Therefore the triangle parameters are : AC = 6.3172; C = 1.997; AB = 6; ∠B = 59° and ∠C = 102° and ∠A = 19°

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Solve the triangle. A = 19°, C = 102°, c = 6

Summary:

Solving the triangle. A = 19°, C = 102°, c = 6 we obtain AC = 6.3172; C = 1.997; AB = 6; ∠B = 59° and ∠C = 102° and ∠A = 19°