Trigonometric Ratios of Standard Angles

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.

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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, 

OQ² = ON² + NQ²

                  = a² + a²

                  = 2a²

∴        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, 

          AD² = AB² – BD²

                  = 4a² – a²

                  = 3a²

∴         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 sin²30° + sin²45° + sin²60°

 

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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