Solved Problem 13 | Rectangular Parallelepiped

(a) For angle AFB
$\tan \angle AFB = \dfrac{AB}{BF}$

$\tan \angle AFB = \dfrac{10}{15}$

$\angle AFB = 33^\circ 41'$           answer
 

For angle BFO
$BD = \sqrt{AB^2 + AD^2}$

$BD = \sqrt{20^2 + 10^2}$

$BD = 10\sqrt{5} \, \text{ in}$
 

$OB = \frac{1}{2}BD$

$OB = \frac{1}{2}(10\sqrt{5})$

$OB = 5\sqrt{5} \, \text{ in}$
 

$\tan \angle BFO = \dfrac{OB}{BF}$

$\tan \angle BFO = \dfrac{5\sqrt{5}}{15}$

$\angle BFO = 36^\circ 42'$           answer
 

For angle AFO
$OA = OB$

$OA = 5\sqrt{5} \, \text{ in}$

$OA^2 = 125$
 

$AF^2 = AB^2 + BF^2$

$AF^2 = 10^2 + 15^2$

$AF^2 = 325$

$AF = 5\sqrt{13} \, \text{ in}$
 

$OF^2 = OB^2 + BF^2$

$OF^2 = (5\sqrt{5})^2 + 15^2$

$OF^2 = 350$

$OF = 5\sqrt{14} \, \text{ in}$
 

Apply cosine law to triangle AFO
$OA^2 = AF^2 + OF^2 - 2(AF)(OF)\cos \angle AFO$

$125 = 325 + 350 - 2(5\sqrt{13})(5\sqrt{14})\cos \angle AFO$

$\cos \angle AFO = \dfrac{325 + 350 - 125}{2(5\sqrt{13})(5\sqrt{14})}$

$\angle AFO = 35^\circ 23'$           answer
 

For angle BOF
$\tan \angle BOF = \dfrac{BF}{OB}$

$\tan \angle BOF = \dfrac{15}{5\sqrt{5}}$

$\angle BOF = 53^\circ 18'$           answer
 

Another solution
$\angle BOF = 90^\circ - \angle BFO$

$\angle BOF = 90^\circ - 36^\circ 42'$

$\angle BOF = 53^\circ 18'$       okay
 

For angle AOF
Cosine law for triangle AOF
$AF^2 = OA^2 + OF^2 - 2(OA)(OF)\cos \angle AOF$

$325 = 125 + 350 - 2(5\sqrt{5})(5\sqrt{14})\cos \angle AOF$

$\cos \angle AOF = \dfrac{125 + 350 - 325}{2(5\sqrt{5})(5\sqrt{14})}$

$\angle AOF = 68^\circ 59'$           answer
 

For angle OFC
$CF^2 = BC^2 + BF^2$

$CF^2 = 20^2 + 15^2$

$CF^2 = 625$

$CF = 25 \, \text{ in}$
 

$OC = OA$

$OC = 5\sqrt{5} \, \text{ in}$

$OC^2 = 125$
 

Cosine law for triangle OFC
$OC^2 = OF^2 + CF^2 - 2(OF)(CF)\cos \angle OFC$

$125 = 350 + 625 - 2(5\sqrt{14})(25)\cos \angle OFC$

$\cos \angle OFC = \dfrac{350 + 625 - 125}{2(5\sqrt{14})(25)}$

$\angle OFC = 24^\circ 40'$           answer
 

(b) For triangle ABO
$A_{ABO} = \frac{1}{2}(10)(10)$

$A_{ABO} = 50 \, \text{ in}^2$           answer
 

For triangle BOF
$A_{BOF} = \frac{1}{2}(OB)(BF)$

$A_{BOF} = \frac{1}{2}(5\sqrt{5})(15)$

$ABOF = 83.85 \, \text{ in}^2$           answer

For triangle AOF
$A_{AOF} = \frac{1}{2}(OA)(OF)\sin \angle AOF$

$A_{AOF} = \frac{1}{2}(5\sqrt{5})(5\sqrt{14})\sin 68^\circ 59'$

$A_{AOF} = 97.62 \, \text{ in}^2$
 

(c) Distance from B to the plane AOF.
Triangles ABC and AGB are similar
$sin \alpha = \dfrac{BC}{AC} = \dfrac{BG}{AB}$

$\dfrac{20}{\sqrt{10^2 + 20^2}} = \dfrac{x}{10}$

$x = 4\sqrt{5}$
 

$\tan \theta = \dfrac{x}{BF}$

$\tan \theta = \dfrac{4\sqrt{5}}{15}$

$\theta = 30.81^\circ$
 

$\sin \theta = \dfrac{d}{BF}$

$d = BF \sin \theta$

$d = 15 \sin 30.81^\circ$

$d = 7.682 \, \text{ in}$           answer