Q.360. Solving the triangle given two sides and an unincluded angle.

Question 360.
Find all sides and angles of a triangle given b = 125, c = 162 and  angle B = 40°.

Answer 360.
This is a case of two sides and a non-included angle.
b, c and angle B are given.
(i) c sin B = 162 sin40° = 104 < b ( = 125) => solution exists.
(ii) b > csinB and b < c => there are two distinctsolutions.
 sinC
= (c/b) sinB
= (162/125) sin40°
= 0.8331
=> C = 56.4° or C = 180° - 56.4° = 123.6°
A = 180° - (B + C) = 180° - (40° + 56.4°) = 83.6° OR
A = 180° - (40° + 123.6°) = 16.4°

If A = 83.6°,
a = b * (sinA/sinB) = 125 * (sin83.6°/sin40°) = 193.3

If A = 16.4°,
a = 125 * (sin16.4°/sin40°) = 54.9

Answers:
1) a = 193.3, A = 83.6°, C = 56.4°
OR
2) a = 54.9, A = 16.4°, C = 123.6°.

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