3D Trigonometry Maths Tuition



(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)


\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


\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.)


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

Parallelogram Maths Tuition: Solution



(a) We have \angle APQ=\angle ARQ (opp. angles of parallelogram)

AP=RQ (opp. sides of parallelogram)

AR=PQ (opp. sides of parallelogram)

Thus, \triangle APQ\equiv\triangle QRA (SAS)

Similarly, \triangle ABC\equiv\triangle CDA (SAS)

\triangle CHQ\equiv\triangle QKC (SAS)

Thus, \begin{array}{rcl}\text{area of BPHC}&=&\triangle APQ-\triangle ABC-\triangle CHQ\\    &=&\triangle QRA-\triangle CDA-\triangle QKC\\    &=& \text{area of DCKR}    \end{array}



\angle ACD=\angle HCQ (vert. opp. angles)

\angle ADC=\angle CHQ (alt. angles)

\angle DAC=\angle CQH (alt. angles)

Thus, \triangle ADC is similar to \triangle QHC (AAA)

Hence, \displaystyle\frac{AC}{DC}=\frac{QC}{HC}

Thus, AC\cdot HC=DC\cdot QC


Congruent Triangles Maths Tuition: Solution




\angle DES=\angle ESA=90^\circ

BA=DT (given)

Thus, \triangle ASB is congruent to \triangle DET (RHS)

Hence \angle DTE=\angle SBA

Thus DT//BA (alt. angles)


By Pythagoras’ Theorem, we have

\begin{array}{rcl}DB&=&\sqrt{DE^2+BE^2}\\    &=&\sqrt{SA^2+ST^2}\\    &=&TA    \end{array}

Hence \triangle DEB and \triangle AST are congruent (SSS).

Hence \angle DBE=\angle STA

Thus DB//TA (alt. angles)

Therefore, ABDT is a parallelogram since it has two pairs of parallel sides.