## Appeal from RGS to NYGH (Success)

Just read that appealing to transfer from RGS (Raffles Girls School) to NYGH (Nanyang Girls High) is possible:

dd moved out of rgs to nygh. she got 262+2. she appealed to nygh and was granted interview on Thursday. was given the good news after her interview.

nygh her first choice. she probably missed by decimal points. thus tried to appeal. was telling her both schools are equally good thus if not successful for nygh in her appeal, just move on 🙂

https://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=48&t=7127&start=3940

NYGH cut off point has been higher than RGS for the recent past years. It is the opposite situation of their boy school counterparts: RI cut off point is usually higher than Hwa Chong (Chinese High).

## Kiasuparents PSLE

Basically to summarize the article above, the co-founder of Kiasuparents’ son scored a respectable 4As and 229 T-score for PSLE. However, as their set target was 250, he did not get the Nintendo DS that was part of the deal for achieving the target of 250. Probably the Nintendo is to play the most recently released Pokemon Sun/Moon. Poor kid! I remember that my highlight of finishing PSLE was to play Pokemon (close to 20 years ago). I still remembered I was playing the Blue version, starting as Bulbasaur.

PSLE can be highly unpredictable (variance of 20-30 marks from usual expected mark is common and expected). This is particularly due to language exams, composition, and also the famous rigid marking scheme of PSLE science, where all the “keywords” must be mentioned in order to get the mark. Mathematics is the more reliable subject here as it is more objective, so try to score as high as possible in it.

Hence DSA becomes increasingly important as a backup plan to act as insurance in the event that something goes wrong in the PSLE. Check out some DSA/GAT/HAST posts here. It is always good to have a “Plan B”.

Also, if you suspect that the child’s school teacher is not that excellent in teaching, e.g. don’t know/emphasize the “keywords” which are necessary to get any marks at all in PSLE science, you may consider engaging a tutor as soon as possible. Check out the most recommended tuition agency in Singapore.

# Right brain training

This post is a review on Right brain training, and also a list of resources that one can research on regarding to the popular method of Right brain training.

When it comes to New Year’s resolutions, getting your body in shape often tops the list. But what about your brain?

If your left or right brain is feeling a little flabby, there’s a wide range of books, teaching programs, and even a Nintendo DS game, purporting to train your left and/or right brain. Indeed, if you Google “right brain training”, you’ll score 53,900,000 hits.

These products are based on the belief that the left and right hemispheres are polar opposites. The left brain is often characterised as your intelligent side: rational, logical and analytic. In contrast the right brain is stereotyped as the “touchy-feely” hemisphere, viewed as artistic, creative, and emotive.

Such left and right brain stereotypes have led theorists to suggest that people can be classified according to their “hemisphericity”. If you’re a logical, rational scientist, for instance, you’re left-brained. But creative types, from artists to writers, are right-brained.

Based on my teaching experience, I do find that left-handers (right-brained) students tend to be very creative and usually excel at arts and humanities. However, their math skills can be good too, especially with practice. This shows that the human brain is like a muscle, it gets better with practice and use.

Did you know our Prime Minister Lee Hsien Loong is left-handed too? Barack Obama is also left-handed. Prime Minister Lee Hsien Loong is very good at math, so this should dispel any myths that left-handed students are not good at math.

# Left Brain vs Right Brain

## Understanding the Myth of Left Brain and Right Brain Dominance

### The Right Brain

According to the left-brain, right-brain dominance theory, the right side of the brain is best at expressive and creative tasks. Some of the abilities that are popularly associated with the right side of the brain include:

• Recognizing faces
• Expressing emotions
• Music
• Color
• Images
• Intuition
• Creativity

### The Left Brain

The left-side of the brain is considered to be adept at tasks that involve logic, language and analytical thinking. The left-brain is often described as being better at:

• Language
• Logic
• Critical thinking
• Numbers
• Reasoning

Also, check out the above Youtube video to check if you are a right-brained or left-brained person!

The Right Brain vs Left Brain test … do you see the dancer turning clockwise or anti-clockwise?

If clockwise, then you use more of the right side of the brain and vice versa.

Most of us would see the dancer turning anti-clockwise though you can try to focus and change the direction; see if you can do it.

LEFT BRAIN FUNCTIONS
uses logic
detail oriented
facts rule
words and language
present and past
math and science
can comprehend
knowing
acknowledges
order/pattern perception
knows object name
reality based
forms strategies
practical
safe

RIGHT BRAIN FUNCTIONS
uses feeling
“big picture” oriented
imagination rules
symbols and images
present and future
philosophy & religion
can “get it” (i.e. meaning)
believes
appreciates
spatial perception
knows object function
fantasy based
presents possibilities
impetuous
risk taking

## Right Brain Training Test

Take the test to see if you are right-brained or left-brained!

## Right Brain Training Video

Watch this free brain training video and follow the instructions to increase your brain power. This is an online “game” that really works to improve brain function. You can actually feel it work!

Brain Training can increase your brain power just like weight training can increase your strength. Use this exercise to work out your brain. Bookmark this video and come back and practice with variations on the basics as discussed in the video.

## 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.

$\displaystyle\cos 72^\circ=\frac{XC}{50}$

$XC=50\cos 72^\circ=15.5 m$

## Tips on attempting Geometrical Proof questions (E Maths Tuition)

Tips on attempting Geometrical Proof questions (O Levels E Maths/A Maths)

1) Draw extended lines and additional lines. (using pencil)

Drawing extended lines, especially parallel lines, will enable you to see alternate angles much easier (look for the “Z” shape). Also, some of the more challenging questions can only be solved if you draw an extra line.

2) Use pencil to draw lines, not pen

Many students draw lines with pen on the diagram. If there is any error, it will be hard to remove it.

3) Rotate the page.

Sometimes, rotating the page around will give you a fresh impression of the question. This may help you “see” the way to answer the question.

4) Do not assume angles are right angles, or lines are straight, or lines are parallel unless the question says so, or you have proved it.

For a rigorous proof, we are not allowed to assume anything unless the question explicitly says so. Often, exam setters may set a trap regarding this, making the angle look like a right angle when it is not.

5) Look at the marks of the question

If it is a 1 mark question, look for a short way to solve the problem. If the method is too long, you may be on the wrong track.

6) Be familiar with the basic theorems

The basic theorems are your tools to solve the question! Being familiar with them will help you a lot in solving the problems.

Hope it helps! And all the best for your journey in learning Geometry! Hope you have fun.

“There is no royal road to Geometry.” – Euclid

## 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)