Solution:

(a)

Using the formula,

Cross-multiply,

(b) The trick here is to break up into

Range:

Since is acute (1st quadrant), is positive.

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# Tag: Challenging

## Challenging Trigonometry Question (ACS(I) Sec 3)

## Logarithm and Exponential Question: A Maths Question

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

## O Level Logarithm Question (Challenging)

**Question:**

Solve

**Solution:**

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

Lg both sides,

(3 s.f.)

**Check answer (to prevent careless mistakes):**

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

**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 and are similar.

(b) Given that area of : area of =9:4, find the ratio of,

(i) DN: NC

(ii) area of rectangle ABCD: area of . **(Challenging)**

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

(ii) 20:3

**Suggested Solutions:**

(a)

(vert. opp. angles)

(alt. angles)

(alt. angles)

Therefore, and are similar (AAA).

(b) (i)

Let and

Then

So

Thus,

(ii)

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

From part (a), since and are similar, we have

This meansÂ that

Thus (the two triangles share a common height)

Now, note that

Hence area of

We conclude that area of rectangle ABCD: area of

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

Let area of

Let area of

Let area of

We have since and have the same base BC and their heights have ratio 3:2.

Cross-multiplying, we get

So

since and have the same base BC and their heights have ratio 3:4.

Hence,

Thus, area of

area of rectangle ABCD: area of =40:6=20:3

**Question:**

Given , find the value ofÂ .

**Solution:**

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

Let

Then, we have

.

Here comes the** critical observation**:

Observe that .

Divide throughout by , we get .

Hence, .

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

We get .

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)