The word Trigonometry can be split as Tri - gono - metry, meaning the measure of the three angles of a triangle. The study of trigonometry is very much useful in finding the distance between the stars and planets, height of a building, distance of a ship from the shore, height of a plane at an instant etc. To solve the problems on this we should know the basics of trigonometry. To study about the basics of trigonometry, let us study about the right angled triangle and the trigonometric ratios of each of the acute angles.

## Sine Cosine Tangent Formulas

Sine, Cosine and Tangents are the ratio of the sides of a right angled triangle. In a right angled triangle, let us consider one of the right angled triangle and denote it as $\theta$

Therefore, Sine of the angle
$\theta$ is denoted as sin$\theta$ which is the ratio of the opposite side to the hypotenuse.
which is denoted as
sin $\theta$ = $\frac{Opposite\; side}{Hypotenuse}$
= $\frac{AB}{AC}$

Cosine of the angle $\theta$ is denoted as cos$\theta$ which is the ratio of the adjacent side to the hypotenuse of a triangle.
cos $\theta$ = $\frac{Adjacent\;side}{Hypotenuse}$
= $\frac{CB}{AC}$

Tangent of the angle $\theta$ is denoted as tan$\theta$ which is the ratio of the opposite side to the adjacent side of the triangle.
tan $\theta$ = $\frac{Opposite\; side}{Adjacent\; Side}$

= $\frac{AB}{BC}$

### Sine Cosine and Tangent Ratios:

Let us consider the following diagram containing showing the perpendicular and base of the triangle.

The base of the triangle is b unites, the perpendicular side to the base is a units.
Therefore, According to Pythagoras theorem, the length of the hypotenuse c is given by $\sqrt{(a^{2}+b^{2}}$
sin $\theta$ = $\frac{Perpendicular}{Hypotenuse}$

= $\frac{a}{c}$

cos
$\theta$ = $\frac{Base}{Hypotenuse}$

= $\frac{b}{c}$

tan
$\theta$ = $\frac{Perpendicular}{Base}$

= $\frac{a}{b}$

### Solved Example

Question: Find the three trigonometric ratios of the angle A of the triangle whose base measure 5 cm and the Perpendicular measure 12 cm.
Solution:

The following diagram shows the angle A for which the trigonometric ratios are to be written.

We can find the length of the hypotenuse using the formula,
c =
$\sqrt{(a^{2}+b^{2})}$

= $\sqrt{(12^{2}+5^{2})}$

= $\sqrt{(144 + 25)}$

= $\sqrt{169}$

= 13
Therefore,
sin $\theta$ =  $\frac{perpendicular}{Hypotenuse}$

=
$\frac{12}{13}$

cos $\theta$$\frac{Base}{Hypotenuse} = \frac{5}{13} tan \theta = \frac{Perpendicular}{Base} = \frac{12}{5} ## Sine Cosine Tangent Chart Tables have been constructed giving the numerical values of the trigonometrical ratios of all angles between 0o and 90o at suitable intervals. The following table shows the value of the trigonometric ratios for the special angles, 0o, 30o, 45o, 60o, 90o.  Angles Functions 0o 30o 45o 60o 90o Sin 0 \frac{1}{2} \frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2} 1 Cos 1 \frac{\sqrt{3}}{2} \frac{1}{\sqrt{2}} \frac{1}{2} 0 Tan 0 \frac{1}{\sqrt{3}} 1 \sqrt{3} not defined ### Using the above table evaluate, ### Solved Examples Question 1: (a) sin 30o + cos 60o + tan 45o Solution: We have from the above table, sin 30o = \frac{1}{2} cos 60o = \frac{1}{2} tan 45o = 1 Substituting these values we get, sin 30o + cos 60o + tan 45o = \frac{1}{2} + \frac{1}{2} + 1 = 1 + 1 = 2 Question 2: sin2 60o + cos2 45o + tan60o Solution: We have from the above table, sin 60o = \frac{\sqrt{3}}{2} cos 45o = \frac{1}{\sqrt{2}} tan 60o = \sqrt{3} Substituting these values we get, sin2 60o + cos2 45o + tan60o = \left (\frac{\sqrt{3}}{2}\right )^{2}$$\left ( \frac{1}{\sqrt{2}} \right )^{2}$
+ $\left (\sqrt{3} \right )^{2}$

= $\frac{3}{4}$ + $\frac{1}{2}$ + 3

= $\frac{5}{4}$ + 3

= 1 $\frac{1}{4}$ + 3

= 4 $\frac{1}{4}$

## Sine Cosine Tangent Problems

### Solved Examples

Question 1: In a right angled triangle PQR, sin P = $\frac{24}{25}$ find the trigonometric ratios of Cos P and Tan P. Also find Sin R, Cos R and Tan R.
Solution:

We are given that,
In $\Delta PQR$, we have sin P = $\frac{24}{25}$

From the above diagram, we have PR = 25 k, RQ = 24 k.
Let us use Pythagorean Theorem to find PQ.
We know that,
PQ2 + RQ2     = PR2
=>      PQ2 + ( 24 k )2 = ( 25 k )2
=>     PQ2  + 576 k2   = 625 k2
=>                      PQ2 = 625 k2 - 576 k2
= 49 k2
Therefore, PQ = $\sqrt{49k^{2}}$
= 7 k
Applying the above formulas on ratios, we get,
cos P = $\frac{Base}{Hypotenuse}$

= $\frac{7k}{25k}$

= $\frac{7}{25}$

tan P = $\frac{Perpendicular}{Base}$

=
$\frac{24k}{7k}$

= $\frac{24}{7}$

similarly, for the angle Q, we have,
Sin R = $\frac{Perpendicular}{Hypotenuse}$

= $\frac{PQ}{PR}$

= $\frac{7k}{25k}$

= $\frac{7}{25}$

Cos R = $\frac{Base}{Hypotenuse}$

= $\frac{QR}{PR}$

= $\frac{24k}{25k}$

= $\frac{24}{25}$

Tan R =
$\frac{Perpendicular}{Base}$

= $\frac{PQ}{QR}$

= $\frac{7k}{24k}$

= $\frac{7}{24}$

Question 2: If tan $\theta$ = $\frac{21}{20}$, find the trigonometric ratios of sin $\theta$ and cos $\theta$.
Solution:

We have tan $\theta$ = $\frac{21}{20}$

Since, tan $\theta$ = $\frac{Perpendicular}{Base}$
( or ) $\frac{Opposite\;Side}{Adjacent\;Side}$ =
$\frac{21}{20}$
we can take Opposite side = 21 k and the Adjacent side = 20 k
Applying Pythagoras theorem, we have the Hypotenuse c can be found using the forrmula,
c = $\sqrt{a^{2}+b^{2}}$

=
$\sqrt{\left(21k\right)^{2}+\left(20k\right)^{2}}$

= $\sqrt{441k^{2}+200k^{2}}$

=  $\sqrt{841k^{2}}$

= 29 k
Therefore, the other trigonometric ratios of $\theta$ are,
Cos $\theta$ = $\frac{Base}{Hypotenuse}$

=
$\frac{Adjacent}{Hypotenuse}$

= $\frac{20k}{29k}$

= $\frac{20}{29}$

tan $\theta$ = $\frac{Perpendicular}{Base}$

=
$\frac{21k}{20k}$

= $\frac{21}{20}$

Question 3: A kite is flying at a height of 75 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60o. Find the length of the string, assuming that there is no slack in the string.
Solution:

The above diagram shows the angle made by the string with the ground. The dotted line shows the height of the kite from the ground.
In $\Delta ABC$. $\angle C$ = 60o
The height of the kite = AB
= 75 m
Writing the trigonometric ratio connecting AB and AC for the given angle, 60o ,
we get,                           Sin 60 o = $\frac{AB}{AC}$

= $\frac{75}{AC}$

=>                       $\frac{\sqrt{3}}{2}$ = $\frac{75}{AC}$

=>                      $\sqrt{3}$ AC = 2 x 75

= 150

=>                              AC = $\frac{150}{\sqrt{3}}$

= $\frac{150}{\sqrt{3}}$  x   $\frac{\sqrt{3}}{\sqrt{3}}$ [ by rationalizing the denominator ]

=>                              AC = $\frac{150 \sqrt{3}}{3}$

= 50 $\sqrt{3}$

= 50 x 1.732

= 86.6 m

Therefore the length of the string is 86.6 m