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
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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