3D Trigonometry Maths Tuition

angle-3d-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

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