Right angled triangle and Application of right angled triangle

Right Angled Triangle:

In a

right angled triangle,

ABC with sides

BC = a

CA = b and

AB = c ;

right angled triangle

We know that:

$\sin A = \dfrac{a}{c}$ or , $a = c \sin A$

And $\cos A = \dfrac{b}{c}$ or , $b = c \cos A$

And $\tan A = \dfrac{a}{b}$ or , $a = b \tan A$

It is obvious that SIN A = COS B , COS A = SIN B because:

A+B = 90 , So A is the complement angle of B , this may be stated as:

Sine of the angle A = Sine of the complement of B

So , Sine of the angle A = CoSine of angle B

Or , $\sin A = \sin (90-B) = \cos B$

And similarly:

$\csc A = \sec B$ and $\tan A = \cot B$

In a triangle , there are three angles and three sides. They are known as the six components or elements of a triangle.

To solve a triangle means to find unknown elements from the given parts.

It is always possible to solve a triangle if three of it’s parts are given (Except for the case that all three parts given are angles)

In solving problems of practical interest in which right-angled triangles appear , we shall use some new terms. They are “The Point of observation” , “Horizontal” , “Line of Sight” , “Angle of Elevation” , “Angle of Depression”.

These terms are diagrammatically illustrated illustrated below:

application of right angled triangle

Application Of Right Angled Triangle:

To solve a real life problem involving right angled triangle , we first collect the given information and then solve the triangle and find the unknown parameters.

For example:

Q. A person 30 meters away from the feet of a tower finds that his line of sight of top of the tower is making an angle of 60 degrees with the horizontal , then find the Height of thee tower.

Solution:

First of all let’s make a visualisation of the situation in diagrammatic form as:

application of right angled triangle

Where , AC is the tower and B is the point from where the person is watching the top of thee tower.

So in the Right Angled triangle ABC ,

$\tan 60 = \dfrac{AC}{BC} = \dfrac{AC}{30}$

Or, $AC = \tan 60 \times 30 = \sqrt{3} \times 30 = 51.96$

Thus, height of the tower = AC = 51.96 Meters.

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