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fugativelover fugativelover
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10 years ago
One of the triangles has 2 angles and 1 side length, the remaining unknown side needs solving.

The other has 2 side lengths and one angle, with the remaining side length needing solving.
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wrote...
10 years ago
If you have two sides and the angle opposite the third, to find the third side
LAW OF COSINES:
C² = A² + B² - (2AB)cos(c)
C is the length of the side opposite the angle whose measure is c, and A and B are the lengths of the other two sides

If you have a side, and angle opposite, and either the angle to find the side opposite, or the side opposite to find the angle
LAW OF SINES:
A/sin(a) = B/sin(b) = C/sin(c)
A is the length of the side opposite the angle whose measure is a, B is the length of the side opposite the angle whose measure is b, C is the length of the side opposite the angle whose measure is c. By the way, the ratios all = 2r where r is the radius of the circumscribed circle for the triangle.
wrote...
10 years ago
Law of sines: Sin A/a = sin B/b = sin C/c, where side a is the side opposite angle A and so forth.
Law of cosines: c^2 = a^2 + b^2 - 2ab*cos(C).

Use law of sines for the two angles one side.
Use law of cosines for the two sides one angle.
wrote...
10 years ago
Use Laws of
Cosines  c2 = a2 + b2 - 2ab·cos(C)  
 
or

 Sines  a/sin(A) = b/sin(B) = c/sin(C).
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