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Trigonometrical Ratios of 60°

How to find the Trigonometrical Ratios of 60°?

Let a rotating line OX rotates about O in the anti-clockwise sense and starting from its initial position OX traces out ∠XOY = 60° is shown in the above picture.

Take a point P on OY and draw ¯PQ perpendicular to OX.

Trigonometrical Ratios of 60°

Let a rotating line OX rotates about O in the anti-clockwise sense and starting from its initial position OX traces out ∠XOY = 60° is shown in the above picture.

Take a point P on OY and draw ¯PQ perpendicular to OX.

Now, take a point R on OX such that ¯OQ = ¯QR  and join ¯PR.

From △OPQ and △PQR we get,

¯OQ  = ¯QR,

¯PQ common

and ∠PQO = ∠PQR (both are right angles)

Thus, the triangles are congruent.

Therefore,  ∠PRO = ∠POQ = 60°

Therefore, ∠OPR

= 180°  - ∠POQ - ∠PRO

= 180°  - 60° - 60°

=  60°

Therefore, the △POR is equilateral triangle

Let, OP = OR = 2a;

Thus, OQ = a.

Now, from pythagoras theorem we get,

OQ2 + PQ2 = OP2

⇒ a2 + PQ2 = (2a)2

⇒ PQ2 = 4a2 – a2

⇒ PQ2 = 3a2

Taking square roots on both the sides we get,

PQ = √3a (since, PQ > 0)

Therefore, from the right angled triangle POQ we get,
sin 60° = ¯PQ¯OP=3a2a=32;
cos 60° = ¯OQ¯OP=a2a=12
And tan 60° = ¯PQ¯OQ=3aa=3
Therefore, csc 60° = 1sin60°=23=233
sec 60° = 1cos60°= 2
And cot 60° =  1tan60°=13=33


Trigonometrical Ratios of 60° are commonly called standard angles and the trigonometrical ratios of these angles are frequently used to solve particular angles.

 Trigonometric Functions





11 and 12 Grade Math

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