Monday, 9 September 2013

THIS TIME GO FEARLESS TO CRACK YOUR MATHS EXAM!!

Many students, even some really intelligent and talented ones, have a strange enemy. They often find it difficult to finish the paper within the allotted time. They are forced to leave a few questions just because they run out of time and often it has been found that the questions they leave are those which they otherwise could do very easily. It can be very disappointing if you are forced to skip such easy questions.
But how can you avoid a situation like this? Many people suggest a single tablet for this "Time management". But how to manage time and how to stop it from running out is a difficult proposition, especially for an average 14-15-year-old tenth grader.
Here are certain tips by experts from to help you out to finish your paper well before time.

1. Understand your examThe most important thing is to understand the examination you are about to take. In the class X mathematics paper, there are 30 questions in four sections A, B, C and D and we have 180 minutes to answer these questions. Here, a rough calculation is that we get about six minutes to answer a question. But that is not the fact.
The question paper contains 'very short answer' type, 'short answer' type and 'long answer' type questions and the time requirement for each type is different.  An ideal allotment for the four sections is as shown below:


2. Use the first 15 minutes effectively
You get a good 15 minutes in the beginning to read the question paper -- use this time to do just that, READ. Read all the 30 questions in 15 minutes. While reading, mark the questions into categories viz easy, manageable and tough. This is done to have an overall idea about the questions and make a rough plan.





  



 3. Don't worry about the tough ones
The moment you find that there are a few tough questions; it is natural that you start worrying about them. This is not required and will only harm your performance.

The fact is that they may look a bit tough on the surface, but when you actually work on them you will find most of them to be much easier than they seemed. So be happy about the easy ones and don't get unduly worried about the tougher lot.

4. Prioritise your attempt
Always attempt the easy questions first and then move on to the manageable ones and ensure that you complete them before taking on the difficult ones. This will ensure that you are not leaving any question that you know.

Once you successfully attempt all the easy and manageable questions, your confidence will grow and you will be
mentally ready to take on the more challenging questions.

5. Ensure speed and accuracy
Use quicker methods in calculations to ensure that you are not wasting time and your answers are correct. Mostly, we take a lot of time to solve a problem if we happen to make some error in the process.

For example, if you make an error in the sign of a term (+/-), you may not be able to solve questions involving quadratic equations or linear equations. Therefore avoiding silly mistakes is very important to save time.

6. Keep an eye on your watch
Keeping an eye on your watch is of course not to increase your stress. This is just to see that you are broadly adhering to the time allocation we discussed in the beginning. A minor variation is not at all a reason to worry. 

7. Avoid thinking too much about a questionThinking about the questions before you attempt them is essential; but not to such an extent that you waste a lot of time on one question. 

Also you need not write a very lengthy answer to a question just because the question is easy and you know it very well. Remember, you need to just answer the question and nothing more. Any over-attempt will be a mere waste of time.

Additionally, you must practice the habit of finishing samples papers in 140-150 minutes. This will help you simulate and exercise examination pressures better.

Sunday, 8 September 2013

TRIGNOMETRY-solve these.

QUESTIONS


  •  The area of a right triangle is 50. One of its angles is 45o. Find the lengths of the sides and hypotenuse of the triangle.
  • In a right triangle ABC, tan(A) = 3/4. Find sin(A) and cos(A).
  • In a right triangle ABC with angle A equal to 90o, find angle B and C so that sin(B) = cos(B).
  • A rectangle has dimensions 10 cm by 5 cm. Determine the measures of the angles at the point where the diagonals intersect.

  • The lengths of side AB and side BC of a scalene triangle ABC are 12 cm and 8 cm respectively. The size of angle C is 59o. Find the length of side AC.

  • From the top of a 200 meters high building, the angle of depression to the bottom of a second building is 20 degrees. From the same point, the angle of elevation to the top of the second building is 10 degrees. Calculate the height of the second building.

  • Karla is riding vertically in a hot air balloon, directly over a point P on the ground. Karla spots a parked car on the ground at an angle of depression of 30o. The balloon rises 50 meters. Now the angle of depression to the car is 35 degrees. How far is the car from point P?



  • If the shadow of a building increases by 10 meters when the angle of elevation of the sun rays decreases from 70o to 60o, what is the height of the building?  




  •     ANSWERS




  • The triangle is right and the size one of its angles is 45o; the third angle has a size 45o and therefore the triangle is right and isosceles. Let x be the length of one of the sides and H be the length of the hypotenuse.

    Area = (1/2)x2 = 50 , solve for x: x = 10

    We now use Pythagoras  to find H: x2 + x2 = H2

    Solve for H: H = 10 sqrt(2)

  • Let a be the length of the side opposite angle A, b the length of the side adjacent to angle A and h be the length of the hypotenuse.

    tan(A) = opposite side / adjacent side = a/b = 3/4

    We can say that: a = 3k and b = 4k , where k is a coefficient of proportionality. Let us find h.

    Pythagoras theorem: h2 = (3k)2 + (5k)2

    Solve for h: h = 5k

    sin(A) = a / h = 3k / 5k = 3/5 and cos(A) = 4k / 5k = 4/5

  • Let b be the length of the side opposite angle B and c the length of the side opposite angle C and h the length of the hypotenuse.

    sin(B) = b/h and cos(B) = c/h

    sin(B) = cos(B) means b/h = c/h which gives c = b

    The two sides are equal in length means that the triangle is isosceles and angles B and C are equal in size of 45o.




  • The diagram below shows the rectangle with the diagonals and half one of the angles with size x.

    tan(x) = 5/2.5 = 2 , x = arctan(2)

    larger angle made by diagonals 2x = 2 arctan(2) = 127o (3 significant digits)

    Smaller angle made by diagonals 180 - 2x = 53o.
    solution problem diagonals of rectangle

  • Let x be the length of side AC. Use the cosine law

    122 = 82 + x2 - 2*8*x*cos(59o)

    Solve the quadratic equation for x: x = 14.0 and x = -5.7

    x cannot be negative and therefore the solution is x = 14.0 (rounded to one decimal place).




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