## 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. $\displaystyle\cos 72^\circ=\frac{XC}{50}$ $XC=50\cos 72^\circ=15.5 m$

## Parallelogram Maths Tuition: Solution 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, $\begin{array}{rcl}\text{area of BPHC}&=&\triangle APQ-\triangle ABC-\triangle CHQ\\ &=&\triangle QRA-\triangle CDA-\triangle QKC\\ &=& \text{area of DCKR} \end{array}$

(proved)

(b) $\angle ACD=\angle HCQ$ (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, $AC\cdot HC=DC\cdot QC$

(proved)

## Congruent Triangles Geometry Maths Tuition: Solution Solution:

(a) $\angle EBD=\angle BEC$ (given)

BE is a common side for both triangles $\triangle BCE$ and $\triangle EDB$

BD=EC (given)

Therefore, $\triangle BCE \equiv \triangle EDB$ (SAS)

(proved)

(b)

Since $\triangle BCE\equiv\triangle EDB$ we have $\angle CBE=\triangle DEB$

Thus, $\begin{array}{rcl}\angle ABE&=&180^\circ - \angle CBE\\ &=&180^\circ -\angle DEB\\ &=& \angle AEB \end{array}$

Therefore, $\triangle ABE$ is an isosceles triangle.

Thus, $AB=AE$

(proved)

## Congruent Triangles Maths Tuition: Solution Solution:

BS=BE+ES=ST+ES=ET $\angle DES=\angle ESA=90^\circ$

BA=DT (given)

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

Hence $\angle DTE=\angle SBA$

Thus DT//BA (alt. angles)

(Proved)

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.

(shown)