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Mathematics | Trigonometry 2 | English Medium

From the previous lesson we learnt about the three main trigonometrical ratios. Lets see how we can find the values of trigonometrical ratios of some commonly used angles.



The triangle above is an equlateral triangle which means all three sides are equal in length. So AB = AC, hence ABC angle is equal to ACB angle. AC = CB, hence CAB angle is equal to CBA angle thus all three angles are equal. Let the value of an angle be x, 

then x + x + x = 180°

So the ABC angle is 60 degrees whilst the AN perpendicular drawn to BC from A eually divides the BAC angle so that, BAN angle is 30 degrees.
Let AB = BC = CA = 2a then,
BN = a and
AN =  3a , according to the pythagoras theorem (apply pythagoras theorem to the triangle ABN) .

Now from BAN triangle  we can see than sin BAN is equal to the ratio, BN/BA.

Thus sin30 °  = a/2a = 1/2
Likewise we can develop the sine, cosine and tan values of 30° and 60°.
Now lets consider a right angled isoceles triangle.

Let the two equal sides be of length a then according to the pythagoras theorem the hypotaneuse of the triangle will be square root(2)a.

So easily we can see that 
sin45°  = cos 45° = 1/
and 
tan 45 °  = 1.


The table below will sum up all the trignometrical ratios we have considered above and you are expected to memorize them for the O/L exam.



From above table we can roughly deduce that from 0 - 90 °  the sine and tangent values of an angle increases with the angle while the cosine value of an angle decreases with the angle.
Moreover,

sin 0 = 0   sin 90 °  = 1 

cos 0 = 1  cos 90 ° = 0
tan 0 = 0  tan 90 ° = infinity

therefor for any given angle x < 90(degrees) ,    
0 < sinx < 1
0 < cosx < 1
tanx > 0



Written by : Uncle

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2 දෙනෙක් අදහස් කිවුවා:

වකා - WAKA said...

(y)

sheshan narada said...

නියමයි...

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