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 QNOX. Then NQO = 90° - 45° = 45°. Therefore it is an isosceles triangle, i.e. ON = NQ. Let ON = NQ = a. Now, by Pythagoras relation,

OQ2 = ON2 + NQ2

= a2 + a2

= 2a2

∴  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 ADBC. Then BAD = CAD = 30° and BD = CD. Let AB = BC = CA = 2a. So, BD = CD = a.

Now, from right angled DABD,

= 4a2 – a2

= 3a2

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

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

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 QMOX.

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 QMOX.

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.