**Trigonometric Ratios of Standard Angles**

We
have to remember the values of **trigonometric ratios of the standard angles** like
0°, 30°, 45°, 60°, 90° and so on. Those values can be derived and understood as
follows:

**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. Now, by 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°,

Sin45° = NQ/OQ = a/a√2 = 1/√2

Cosec45° = OQ/NQ = a√2/a = √2

Cos45° = ON/OQ = a/a√2 = 1/√2

Sec45° = OQ/ON = a√2/a = √2

tan45° = NQ/ON = a/a = 1

cot45° = ON/NQ = a/a = 1

**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°,

Sin60° = AD/AB = a√3/2a = √3/2

Cosec60° = AB/AD = 2a/a√3 = 2/√3

Cos60° = BD/AB = a/2a = 1/2

Sec60° = AB/BD = 2a/a = 2

tan60° = AD/BD = a√3/a = √3

cot60° = BD/AD = a/a√3 = 1/√3

Sin30° = BD/AB = a/2a = 1/2

Cosec30° = AB/BD = 2a/a = 2

Cos30° = AD/AB = a√3/2a = √3/2

Sec30° = AB/AD = 2a/a√3 = 2/√3

tan30° = BD/AD = a/a√3 = 1/√3

cot30° = AD/BD = a√3/a = √3

**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 small enough, QM will be small enough.

And,
when θ = 0°, QM = 0 and OQ = OM. Now, from right angled triangle MOQ with angle
of reference θ = 0°,

Sin0° = QM/OQ = 0/OQ = 0

Cosec0° = OQ/QM = OQ/0 = ∞ (undefined)

Cos0° = OM/OQ = OM/OM = 1

Sec0° = OQ/OM = OM/OM = 1

tan0° = QM/OM = 0/OM = 0

cot0° = OM/QM = OM/0 = ∞ (undefined)

**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
∠MOQ = θ is close to
90°, OM will be close to 0.

And,
when θ = 90°, OM = 0 and OQ = MQ. Now, from right angled triangle MOQ with
angle of reference θ = 90°,

Sin90° = MQ/OQ = MQ/MQ = 1

Cosec90° = OQ/MQ = MQ/MQ = 1

Cos90° = OM/OQ = 0/OM = 0

Sec90° = OQ/OM = OQ/0 = ∞ (undefined)

tan90° = MQ/OM = MQ/0 = ∞ (undefined)

cot90° = OM/MQ = 0/MQ = 0

From
the above relations, the values of various trigonometric ratios of standard
angles can be summarized as follows:

*Worked Out
Examples*

*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:*

* As
triangle ABC is a right angled triangle and 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.*

*You can comment your questions or
problems regarding the values of trigonometric ratios of standard angles here.*

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