Final Exam Review Problems

[warwick]

The solutions to these are not standing out to me. Thanks for any help. There might be an addendum to these as I am almost finished with the review.

15) Find a polynomial equation with real coefficients that has the given roots.

-6i and square root of 3.

69) The angle of elevation of the top of a pole as seen from a point 13 ft away from the base is double its angle of elevation as seen from a point 42 ft farther from the pole. Find the height of the pole above the level of the observer's eyes.

76) Two forces of 437 N and 258 N act at a point. The result force is 517 N. Find the angle between the forces.

 

[skeeter]

15. if -6i is a root, so is +6i ...
\(\displaystyle P(x) = (x - 6i)(x + 6i)(x - \sqrt{3})\)

69. \(\displaystyle tan(2\theta) = \frac{h}{13}\)
\(\displaystyle tan(\theta) = \frac{h}{13+42}\)

note that \(\displaystyle tan(2\theta) = \frac{2tan(\theta)}{1 - tan^2(\theta)}\)

76. let the angle be \(\displaystyle \theta\)
using the law of cosines ...
\(\displaystyle 517^2 = 437^2 + 258^2 - 2(437)(258)cos(180 - \theta)\)

 

[warwick]

15. if -6i is a root, so is +6i ...
\(\displaystyle P(x) = (x - 6i)(x + 6i)(x - \sqrt{3})\)

69. \(\displaystyle tan(2\theta) = \frac{h}{13}\)
\(\displaystyle tan(\theta) = \frac{h}{13+42}\)

note that \(\displaystyle tan(2\theta) = \frac{2tan(\theta)}{1 - tan^2(\theta)}\)

76. let the angle be \(\displaystyle \theta\)
using the law of cosines ...
\(\displaystyle 517^2 = 437^2 + 258^2 - 2(437)(258)cos(180 - \theta)\)

15 is a no-duh one. Haha.

You'll have to explain 69 and 76. Thanks.