Trigonometric Ratios of Standard Angles

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,

          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 AD⊥BC. Then ∠BAD = ∠CAD = 30° and BD = CD. Let AB = BC = CA = 2a. So, BD = CD = a.

Now, from right angled DABD,

          AD² = AB² – BD²

                 = 4a² – a²

                 = 3a²

       ∴  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

 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 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.

 

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