3D Trigonometry Maths Tuition

Solution:

(a) Draw a line to form a small right-angled triangle next to the angle $18^\circ$

Then, you will see that

$\angle ACD=90^\circ-18^\circ=72^\circ$ (vert opp. angles)

$\angle BAC=180^\circ-72^\circ=108^\circ$ (supplementary angles in trapezium)

By sine rule,

$\displaystyle \frac{\sin\angle ABC}{30}=\frac{\sin 108^\circ}{40.9}$

$\sin\angle ABC=0.697596$

$\angle ABC=44.23^\circ$

$\angle ACB=180^\circ-44.23^\circ-108^\circ=27.77^\circ=27.8^\circ$ (1 d.p.)

(b) By Sine Rule,

$\displaystyle\frac{AB}{\sin\angle ACB}=\frac{30}{\sin 44.23^\circ}$

$AB=\frac{30}{\sin 44.23^\circ}\times\sin 27.77^\circ=20.04=20.0 m$ (shown)

(c)

$\angle BCD=\angle ACD-\angle ACB=72^\circ-27.77^\circ=44.23^\circ$

By Cosine Rule,

$BD^2=40.9^2+50^2-2(40.9)(50)\cos 44.23^\circ=1242.139$

$BD=35.24=35.2 m$

(d)

$\displaystyle\frac{\sin\angle BDC}{40.9}=\frac{\sin 44.23^\circ}{35.24}$

$\sin\angle BDC=0.80957$

$\angle BDC=54.05^\circ$

angle of depression = $90^\circ-54.05^\circ=35.95^\circ=36.0^\circ$ (1 d.p.)

(e)

Let X be the point where the man is at the shortest distance from D. Draw a right-angle triangle XDC.

$\displaystyle\cos 72^\circ=\frac{XC}{50}$

$XC=50\cos 72^\circ=15.5 m$

Author: mathtuition88

http://mathtuition88.com View all posts by mathtuition88

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