Challenging Trigonometry Question (ACS(I) Sec 3)

maths tuition trigonometry


(a) \tan (\alpha + \beta )=\frac{10}{x}

Using the formula,

\displaystyle \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}=\frac{10}{x}

\displaystyle \frac{\frac{5}{x}+\tan \beta}{1-(\frac{5}{x})\tan \beta}=\frac{10}{x}


5+x\tan\beta =10-\frac{50}{x}\tan \beta

(x+\frac{50}{x})\tan\beta =5

\displaystyle \tan \beta =\frac{5}{x+\frac{50}{x}}=\frac{5x}{x^2+50}

(b) The trick here is to break up 2\alpha +\beta into \alpha + (\alpha +\beta )

\displaystyle \begin{array}{rcl}    \tan (2\alpha +\beta )&=& \tan (\alpha + (\alpha + \beta ))\\    &=& \frac{\tan \alpha + \tan (\alpha + \beta )}{1-\tan \alpha \tan (\alpha + \beta )}\\    &=& \frac{ \frac{5}{x}+\frac{10}{x} }{ 1-(\frac{5}{x})(\frac{10}{x}) }\\    &=& \frac{\frac{15}{x}}{\frac{x^2-50}{x^2}}\\    &=& \frac{15x}{x^2-50}    \end{array}


Since 2\alpha +\beta is acute (1st quadrant), \tan (2\alpha +\beta ) is positive.

x^2-50 >0



Logarithm and Exponential Question: A Maths Question


Solve (4x)^{\lg 5} = (5x)^{\lg 7}


4^{\lg 5}\cdot x^{\lg 5}=5^{\lg 7}\cdot x^{\lg 7}

\displaystyle\frac{4^{\lg 5}}{5^{\lg 7}}=\frac{x^{\lg 7}}{x^{\lg 5}}=x^{\lg 7-\lg 5}

Using calculator, and leaving answers to at least 4 s.f.,


Lg both sides,

\lg 0.6763=0.1461\lg x

\lg x=\frac{\lg 0.6763}{0.1461}=-1.1626

x=10^{-1.1626}=0.0688 (3 s.f.)

Check answer (to prevent careless mistakes):

LHS=(4\times 0.0688)^{\lg 5}=0.406

RHS=(5\times 0.0688)^{\lg 7}=0.406

Since LHS=RHS, we have checked that our answer is valid.

Challenging Geometry E Maths Question — St Andrew’s Sec 3 Maths Tuition Question


ABCD is a rectangle. M and N are points on AB and DC respectively. MC and BN meet at X. M is the midpoint of AB.

recommended maths tuition geometry

(a) Prove that \Delta CXN and \Delta MXB are similar.

(b) Given that area of \triangle CXN: area of \triangle MXB=9:4, find the ratio of,

(i) DN: NC

(ii) area of rectangle ABCD: area of \triangle XBC. (Challenging)

[Answer Key] (b) (i) 1:3

(ii) 20:3

Suggested Solutions:

\angle MXB=\angle NXC (vert. opp. angles)

\angle MBX = \angle XNC (alt. angles)

\angle BMX = \angle XCN (alt. angles)

Therefore, \Delta CXN and \Delta MXB are similar (AAA).

(b) (i) \displaystyle\frac{NC}{BM}=\sqrt{\frac{9}{4}}=\frac{3}{2}

Let BM=2u and NC=3u

Then DC=2\times 2u=4u

So DN=4u-3u=u

Thus, DN:NC=1u:3u=1:3


We now have a shorter solution, thanks to a visitor to our site! (see comments below)

From part (a), since \Delta CXN and \Delta MXB are similar, we have MX:XC=2:3

This means  that MC:XC=5:3

Thus \triangle MBC:\triangle XBC=5:3 (the two triangles share a common height)

Now, note that \displaystyle\frac{\text{area of }ABCD}{\triangle MBC}=\frac{BC\times AB}{0.5 \times BC \times MB}=\frac{AB}{0.5MB}=\frac{2MB}{0.5MB}=4

Hence area of ABCD=4\times\triangle MBC

We conclude that area of rectangle ABCD: area of \triangle XBC=4(5):3=20:3

Here is a longer solution, for those who are interested:

Let area of \triangle XBC =S

Let area of \triangle MXB=4u

Let area of \triangle CXN=9u

We have \displaystyle\frac{S+9u}{S+4u}=\frac{3}{2} since \triangle NCB and \triangle CMB have the same base BC and their heights have ratio 3:2.

Cross-multiplying, we get 2S+18u=3S+12u

So \boxed{S=6u}

\displaystyle\frac{\triangle BCN}{\triangle BDC}=\frac{3}{4} since \triangle BCN and \triangle BDC have the same base BC and their heights have ratio 3:4.


\begin{array}{rcl}    \triangle BDC &=& \frac{4}{3} \triangle BCN\\    &=& \frac{4}{3} (9u+6u)\\    &=& 20u    \end{array}

Thus, area of ABCD=2 \triangle BDC=40u

area of rectangle ABCD: area of \triangle XBC=40:6=20:3

O Level Logarithm Question (Challenging)


Given \displaystyle\log_9{a} = \log_{12}{b} =\log_{16}{(a+b)}, find the value of  \displaystyle\frac{a}{b}.


Working with logarithm is tricky, we try to transform the question to an exponential question.

Let \displaystyle y=\log_9{a} = \log_{12}{b} =\log_{16}{(a+b)}

Then, we have a=9^y=3^{2y}

b=12^y=3^y\cdot 2^{2y}


Here comes the critical observation:

Observe that \boxed{a(a+b)=b^2}.

Divide throughout by b^2, we get \displaystyle (\frac{a}{b})^2+\frac{a}{b}=1.

Hence, \displaystyle (\frac{a}{b})^2+\frac{a}{b}-1=0.

Solving using quadratic formula (and reject the negative value since a and b has to be positive for their logarithm to exist),

We get \displaystyle\frac{a}{b}=\frac{-1+\sqrt{5}}{2}.

If you have any questions, please feel free to ask me by posting a comment, or emailing me.

(I will usually explain in much more detail if I teach in person, than when I type the solution)