Trigonometry is derived from the Greek words trigonon and
metron, which mean triangle and measure, respectively. Sine, Cosine, Tangent,
Cotangent, Secant, and Cosecant are the six trigonometry ratios. These
trigonometry ratios describe the various combinations of a right-angled
triangle.

#### Trigonometric ratios

Trigonometric ratios are length ratios of right-angled
triangles. These ratios can be used to calculate the ratios of any two sides of
a right-angled triangle out of a total of three sides.

**Sine function:** The sine ratio for the given angle Î¸ in
a right-angled triangle is defined as the ratio of lengths of its opposite side
to its hypotenuse.

i.e., SinÎ¸ = AB/AC

**Cosine function:** The Cosine ratio for the given angle Î¸
in a right-angled triangle is defined as the ratio of lengths of its adjacent
side to its hypotenuse.

i.e, CosÎ¸ = BC/AC

**Tangent Function:** The Tangent ratio for the given angle
Î¸ in a right-angled triangle is defined as the ratio of lengths of its opposite
side to its adjacent.

i.e, TanÎ¸ = AB/BC

**Cotangent Function:** The Cotangent ratio for the given
angle Î¸ in a right-angled triangle is defined as the ratio of lengths of its
adjacent side to its opposite. It's the reciprocal of the tan ratio.

i.e, CotÎ¸ = BC/AB =1/TanÎ¸

**Secant Function:** The Secant ratio for the given angle Î¸
in a right-angled triangle is defined as the ratio of lengths of its Hypotenuse
side to its adjacent.

i.e, SecÎ¸ = AC/BC

**Cosecant Function:** The Cosecant ratio for the given
angle Î¸ in a right-angled triangle is defined as the ratio of lengths of its hypotenuse
side to its opposite.

i.e, CosecÎ¸ = AC/AB

#### Sin Theta Formula

In a Right-angled triangle, the sine function or sine theta
is defined as the ratio of the opposite side to the hypotenuse of the triangle.
In a triangle, the Sine rule helps to relate the sides and angles of the
triangle with its circumradius(R) i.e, a/SinA = b/SinB = c/SinC = 2R. Where a,
b, and c are lengths of the triangle, and A, B, C are angles, and R is
circumradius.

Sin Î¸ = (opposite side / hypotenuse)

From the above figure, sine Î¸ can be written as

SinÎ¸ = AB / AC

According to the Pythagoras theorem, we know that AB2 +
BC2 = AC2. On dividing both sides by AC2

â‡’ (AB/AC)2 + (BC/AC)2

â‡’ Sin2Î¸ + Cos2Î¸ = 1

#### Sample Problems

**Question 1:** If the sides of the right-angled triangle â–³ABC
which is right-angled at B are 7, 25, and 24 respectively. Then find the value
of SinC?

**Solution:**

As we know that SinÎ¸ = (Opposite side/hypotenuse)

SinC = 24/25

**Question 2:** If two sides of a right-angled triangle are 3
and 5 then find the sine of the smallest angle of the triangle?

**Solution:**

By Pythagoras theorem, other side of the triangle is found
to be 4.

As the smaller side lies opposite to the smaller
angle,

Then Sine of smaller angle is equal to 3/5.

**Question 3:** If sinA = 12/13 in the triangle â–³ABC,
then find the least possible lengths of sides of the triangle?

**Solution:**

As we know, SinÎ¸ = opposite/hypotenuse

Here, opposite side = 12 and hypotenuse = 13

Then by the pythagoras theorem, other side of the triangle
is 5 units

**Question 4:** If the lengths of sides of a right-angled â–³PQR
are in A.P. then find the sine values of the smaller angles?

**Solution:**

The only possible Pythagorean triplet for the given
condition is (3, 4, 5).

Therefore, the sine values of the smaller sides are 3/5 and
4/5

**Question 5:** In a triangle â–³XYZ if CosX=1/2 then find the
value of SinY?

**Solution:**

From the given data, the angle X is equal to 60 degrees,
then Y=30 degrees as it's a right angled triangle.

Therefore, SinY = Sin30°

Y = 1/2

**Question 6:** If sinÎ¸.SecÎ¸ = 1/5 then find the value of SinÎ¸?

**Solution:**

As secÎ¸ = 1/cosÎ¸

SecÎ¸ = TanÎ¸ = 1/5.

Therefore, opposite side= k and adjacent side is 5k and
hypotenuse = âˆš26 k.

Then SinÎ¸ = k/âˆš26 k

= 1/âˆš26

**Question 7:** In a right-angled triangle, if the ratio of
smaller angles is 1:2 then find the sum of sines of smaller angles of the
triangle?

**Solution:**

Let the smaller angles be A, B. As A:B = 1:2.

So, A = k and B = 2k. As A + B = 90.

â‡’ k + 2k = 90

â‡’ k = 30.

Therefore, the other angles are 30 and 60

So, their sine values are 1/2 and âˆš3/2