Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 11 - Review Exercises - Page 525: 27

Answer

The base angle is $\approx 68^{\circ}$.

Work Step by Step

1. Use the Law of Cosines $c^{2} = a^{2} + b^{2} - 2abcos(C)$ $(30)^{2} = (40)^{2} + (40)^{2} - 2(40)(40)cos(C)$ $900 = 1600 + 1600 - 3200cos(C)$ $900 = 3200 - 3200cos(C)$ $-2300 = -3200cos(C)$ $cos(C) = 0.71875$ by GDC / calculator $C = 44.0486...^{\circ}$ 2. Find one of the base angles knowing that angles in a $\triangle$ add to $180^{\circ}$ $= \frac{180-(44.0486...)}{2}$ $= 67.975...^{\circ}$ $\approx 68^{\circ}$
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