Tan 30 degrees

The value of tan30^∘ is 1/√3. If we solved 1√3 we got 0.57735026919. tan 30 degrees is also represented by tan π/6 in terms of radians.

tan30^∘ = 1/√3 = 0.57735026919

How to calculate the value of tan30^∘

There are so many ways of calculating the values of tan30^∘ .

First Method :- Calculate the value of tan30^∘

by using sin∅ & cos∅

We know that tan∅ = Sin∅/Cos∅

Now we discuss about sin∅

Sin∅ = Perpendicular/Hypotenuse

Suppose for a triangle for PQR right angle at Q. ∅ is the angle , PR is the hypotenuse . QR is the base and PQ is the Opposite side or perpendicular.

So we can say that

Sin∅ = Perpendicular/Hypotenuse = PQ/PR

Similarly

Cos∅ = Base /Hypotenuse = QR/PR (∵ ∆ PQR )

Now can say that tan∅ = Sin∅/Cos∅

Now tan∅ = PQ/QR

tan∅ = Perpendicular/Base

2^nd Method

If we know the value of sin30^∘and cos30^∘. We can easily find the value of tan30^∘.

We know that sin30^∘= 1/2 and Cos30^∘= √3/2

So we know tan∅ = Sin∅/Cos∅

= tan∅ = sin30^∘/Cos30^∘

= tan∅ = 1/√3 { ∵ tan30^∘= 1/ √3 }

3^rdMethod

If we know the value of sin30^∘. We can easily find the value of tan30^∘.

We know that sin30^∘= 1/2 { ∵ Sin∅ = Perpendicular/Hypotenuse = 1/2 }

Now we are find base ( QR ) by using the Pythagoras Theorem

= ( Perpendicular )² + (Base)² = (Hypotenuse )²

= ( 1 )² + ( QR )² = (2)²

= 1 + ( QR )² = 4

= ( QR )²=4-1

= ( QR )= √3

So tan∅ = Perpendicular/Base { ∵ tan∅ = PQ/QR }

= tan30^∘= 1/ √3

4^thMethod

If we know the value of Cos30^∘. We can easily find the value of tan30^∘.

We know that Cos30^∘= √3/2 { ∵ Cos∅ = Base /Hypotenuse = √3/2 }

Now we are find Perpendicular ( PQ ) by using the Pythagoras Theorem

= ( Perpendicular )² + (Base)² = (Hypotenuse )²

= ( PQ )² + ( √3 )² = (2)²

= ( PQ )² + 3 = 4

= ( PQ)²=4-3

= ( PQ )= √1

So tan∅ = Perpendicular/Base { ∵ tan∅ = PQ/QR }

= tan30^∘= 1/ √3

^5thMethod

If we know the value of Sec30^∘. We can easily find the value of tan30^∘.

We know that Sec30^∘= 2/√3 { ∵ Sec∅ = Hypotenuse/Base = 2/√3 }

Now we are find Perpendicular ( PQ ) by using the Pythagoras Theorem

= ( Perpendicular )² + (Base)² = (Hypotenuse )²

= ( PQ )² + ( √3 )² = (2)²

= ( PQ )² + 3 = 4

= ( PQ)²=4-3

= ( PQ )= √1

So tan∅ = Perpendicular/Base { ∵ tan∅ = PQ/QR }

= tan30^∘= 1/ √3

6^thMethod

If we know the value of Cosec30 . We can easily find the value of tan30^∘.

We know that Cosec30^∘= 2 { ∵ Cosec∅ = Hypotenuse/Perpendicular = 2/1 }

Now we are find Base ( QR ) by using the Pythagoras Theorem

= ( Perpendicular )² + (Base)² = (Hypotenuse )²

= ( 1 )² + (QR)² = (2 )²

= ( QR )² + ( 1)² = (2)²

= ( QR )² + 1 = 4

= ( QR)²=4-1

= ( QR ) = √3

So tan∅ = Perpendicular/Base { ∵ tan∅ = PQ/QR }

= tan30^∘= 1/ √3

Check the trigonometry table to get the values for all trigonometry ratios.

Angle Radian	^0∘	30∘	45∘	60∘	90∘	180∘	270∘	^360∘
	0	π/6	π/4	π/3	π/2	π	3π/2	2π
Sin∅	0	1/2	1/√2	√3/2	1	0	-1	0
Cos∅	1	√3/2	1/√2	1/2	0	-1	0	1
tan∅	0	1/√3	1	√3	∞	0	∞	0
Cot∅	∞	√3	1	1/√3	0	∞	0	∞
Sec∅	1	2/√3	√2	2	∞	-1	∞	1
Cosec∅	∞	2	√2	2/√3	1	∞	-1	∞