How do you solve the triangle given B=19, a=51, c=61?

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Answer 1 · 2016-11-02T18:05:57

Please see the explanation for a description of the process.

I assume that you mean that angle $B = 19°$

Use the Law of Cosines to find the length of side b:

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

$b = sqrt(51^2 + 61^2 - 2(51)(61)cos(19°))$

$b~~ 20.95$

Use the Law of Sines to find angle A:

$sin(A)/a = sin(B)/b$

$A = sin^-1(sin(B)a/b)$

$A = sin^-1(sin(19°)51/20.95)$

$A ~~ 52°$

Find angle C:

$C = 180° - 19° - 52°$

$C = 109°$

We know the lengths of all 3 sides, $a = 51, b = 20.95, and c = 61$ and we know the measures of all 3 angles, $A = 52°, B = 19° and C = 109°$