Use the cosine rule for this. You have the three sides of the triangle andOriginally Posted by cinbad
you need to find the angle between two of the sides.
RonL
Here is one way.Originally Posted by cinbad
Imagine, or draw the figure on paper.
It is a triangle whose 3 sides are 1.83, 13, and 11.5----all in meters.
You are asked to find the angle included by the 13m and 11.5m sides. Call that angle A.
Use the Law of Cosines.
a^2 = b^2 +c^2 -2bc*cosA
(1.83)^2 = (13)^2 +(11.5)^2 -2(13)(11.5)cosA
2(13)(11.5)cosA = (13)^2 +(11.5)^2 -(1.83)^2
cosA = [(13)^2 +(11.5)^2 -(1.83)^2] / [2(13)(11.5)]
cosA = 297.9011 /299
cosA = 0.996324749
A = arccos(0.996324749)
A = 4.914 degrees ----------------answer.