Refer to 7(b):

http://i45.tinypic.com/11smtr7.jpg

Help, please? The answer given in the textbook states 57.5 degrees, but I can't seem to get it.

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- Jul 22nd 2012, 02:03 AMfActorThree-Dimensional Problem?
Refer to 7(b):

http://i45.tinypic.com/11smtr7.jpg

Help, please? The answer given in the textbook states 57.5 degrees, but I can't seem to get it. - Jul 22nd 2012, 04:50 AMearbothRe: Three-Dimensional Problem?
1. Triangle $\displaystyle \Delta(ABP)$ is an isoscles triangle because |AP| = |BP|.

2. Use Pythagorean theorem to determine the length of AP:

The diagonal d of the base rectangle has the length 20, thus half of the diagonal $\displaystyle (\frac12 d)$ has the length 10.

$\displaystyle |\overline{AP}|=\sqrt{5^2+10^2}=5\sqrt{5}$

3. Use Cosine rule to evaluate

$\displaystyle \cos(\angle(BAP))=\frac{(BP)^2-(AP)^2-(AB)^2}{-2 \cdot (AB) \cdot (BP)}=\frac 6{5\sqrt{5}}$

You'll get: $\displaystyle \angle(BAP) = 57.544^\circ$ - Jul 22nd 2012, 09:05 AMHallsofIvyRe: Three-Dimensional Problem?
A slightly different method: Set up a coordinate system with origin at A, positive x-axis from A to B, positive y-axis from A to D, and positive z-axis upward. Then the vector AB is <12, 0, 0> and AP is <6, 8, 5>. The magnitude of AB is 12, the magnitude of AP is $\displaystyle 5\sqrt{5}$, and the dot product of A and B is 6(12)= 72. We must have $\displaystyle 6(12)= (12)(5\sqrt{5})cos(\theta)$ so that $\displaystyle cos(\theta)= \frac{6}{5\sqrt{5}}$.