Challenging – Singapore Maths Tuition

Challenging Trigonometry Question (ACS(I) Sec 3)

Solution:

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

Using the formula,

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

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

Cross-multiply,

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

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

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

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

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

Range:

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

$x^2-50 >0$

$x>\sqrt{50}$

Logarithm and Exponential Question: A Maths Question

Question:

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

Solution:

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

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

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

$0.6763=x^{0.1461}$

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

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.

(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:

(a)
$\angle MXB=\angle NXC$ (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, $DN:NC=1u:3u=1:3$

(ii)

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.