**Trigonometric Ratios
of Standard Angles**

In trigonometry, 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
etc. are considered as the **standard angles**. We can derive the values of **trigonometric ratios of standard angles** geometrically.

We should remember those values for the calculations in trigonometry.
Here is the geometrical procedure for the derivation of trigonometric ratios of
standard angles.

**Trigonometric ratios
of 45°**

Let OX be the initial line and trace out an ∠XOP = 45°. From Q, a point on OP, draw QN⊥OX. Then ∠NQO = 90° – 45° = 45°. Therefore it is an isosceles triangle, i.e. ON = NQ. Let ON = NQ = a.

From the right angled triangle ONQ, using Pythagoras relation,

OQ^{2} = ON^{2} + NQ^{2}

= a^{2} + a^{2}

= 2a^{2}

∴ OQ = a√2

Now for right angled triangle ONQ with angle of reference ∠NOQ = 45°,

**Trigonometric ratios
of 60° and 30°**

Let ABC be a triangle in which AB = BC = CA and hence ∠A = ∠B = ∠C = 60°. From A draw AD⊥BC. Then ∠BAD = ∠CAD = 30° and BD = CD. Let AB = BC = CA = 2a. So, BD = CD = a.

Now, from right angled DABD,

AD^{2} =
AB^{2} – BD^{2}

= 4a^{2} – a^{2}

=
3a^{2}

∴ AD = a√3

Now, in right angled DABD with angle of reference ∠ABD = 60°,

And, with the angle of reference ∠BAD = 30°,

**Trigonometric ratios
of 0°**

Let OX be the initial line and trace out an ∠XOP = θ. Let Q be a point on OP. Draw QM⊥OX.

When θ is decreased and tends to be 0° keeping OM constant, then
QM = 0 and OQ = OM.

Now, from right angled triangle MOQ with angle of reference θ = 0°,

**Trigonometric ratios
of 90°**

Let OX be the initial line and trace out an ∠XOP = θ. Let Q be a point on OP. Draw QM⊥OX.

When θ is increased and tends to be 90° keeping MQ constant,
then OM = 0 and OQ = MQ.

Now, from right angled triangle MOQ with angle of reference θ = 90°,

**Table of
Trigonometric Ratios of Standard Angles**

**Worked
Out Examples**

**Example 1:** Find the value of sin^{2}30° + sin^{2}45° + sin^{2}60°

**Solution:** Here,

**Example 2:** Find the value of:

**Solution:** Here,

**Example 3:** Prove that:

**Solution:** Here,

**Example
4:**
Find the value of x: sin60°cos30° + x sin45°cos45° = sin30°cos60°

**Solution:** Here,

**Example 5:** In the given figure, the distance between house B and House C is
20m and ∠BCA = 30°. Calculate the distance between house
A and B.

**Solution:** Here,

ΔABC is a right angled triangle. BC = 20m, ∠BCA = 30° and ∠BAC = 90°

Sin30° = AB/BC

or, ½ = AB/20m

or, 2AB = 20m

or, AB = 20m/2

or, AB = 10m

Hence, the distance between house A and B is 10m.

If you have any question or problems regarding the **Trigonometric Ratios of Standard Angles**,
you can ask here, in the comment section below.

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