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mathsTrigonometryTrigonometric Ratios for Standard Angles

Trigonometric Ratios for Standard Angles

Students need not memorize a table of trigonometric ratios for standard angles and instead they can quickly calculate the ratios for standard angle. This page explains how to quickly calculate.



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Given the right angled triangle `Delta OPQ`, and `/_POQ = 45^@`.

`bar(OP)= 1`
`bar(OQ)=bar(PQ)=x=y`
`bar(OQ)^2+bar(PQ)^2 = bar(OP)^2 = 1`

`2 x^2 = 1`trigonometric ratios for 45 degree what is point `P`?

  • `(0,0)`
  • `(1/sqrt(2), 1/sqrt(2))`
  • `(1/sqrt(2), 1/sqrt(2))`

The answer is '`(1/sqrt(2), 1/sqrt(2))`'.

It is proven that point `P (1/sqrt(2), 1/sqrt(2))`. That is,

opposite side `bar(PQ) = 1/sqrt(2)`

adjacent side `bar(OQ) = 1/sqrt(2)`

hypotenuse `bar(OP) = 1` trigonometric ratios for 45 degree `sin 45^@ = 1/sqrt(2) `

`cos 45^@ = 1/sqrt(2) `

`tan 45^@ = 1`

Do not memorize the trigonometric ratios, it is easy to work out the lengths of the three sides of a triangle of standard angles.

Once we know the lengths of the three sides, it is easy to work out the trigonometric ratios.

What is `cot 45^@`?trigonometric ratios for 45 degree

  • `1`
  • `1`
  • `sqrt(2)`

The answer is '`1`'.

Given the right angled triangle `Delta OPQ`, and `/_POQ = 30^@`.

`bar(OP)= 1`
Construct an equilateral triangle `Delta OP′P`.
From this, it is derived that `bar(PQ)=1/2`.
`bar(OQ)^2+bar(PQ)^2 = bar(OP)^2 = 1`

`x^2+(1/2)^2 = 1`
trigonometric ratios for 30 degree what is point `P`?

  • `(sqrt(3)/2 , 1/2)`
  • `(sqrt(3)/2 , 1/2)`
  • `(0,1/2)`

The answer is '`(sqrt(3)/2 , 1/2)`'.

It is proven that point `P (sqrt(3)/2 , 1/2)`. That is,

opposite side `bar(PQ) = 1/2`

adjacent side `bar(OQ) = sqrt(3)/2`

hypotenuse `bar(OP) = 1`trigonometric ratios for 30 degree `sin 30^@ = 1/2`

`cos 30^@ = sqrt(3)/2`

`tan 30^@ = 1/sqrt(3)`

It is noted that one does not need to memorize the trigonometric ratios, it is easy to work out the lengths of the three sides of a triangle of standard angles. Once we know the lengths of the three sides, it is easy to work out the trigonometric ratios.

;; What is `sec 30^@`?trigonometric ratios for 30 degree

  • `1`
  • `2/sqrt(3)`
  • `2/sqrt(3)`

The answer is '`2/sqrt(3)`'.

Given the right angled triangle `Delta OPQ`, and `/_POQ = 60^@`.

`bar(OP)= 1`
Construct an equilateral triangle `Delta O Q′P`.
From this, it is derived that `bar(OQ)=1/2`.
`bar(OQ)^2+bar(PQ)^2 = bar(OP)^2 = 1`

`(1/2)^2+ y^2 = 1`trigonometric ratios for 60 degree what is point `P`?

  • `(1/2 , sqrt(3)/2)`
  • `(1/2 , sqrt(3)/2)`
  • `(1/2,0)`

The answer is '`(1/2 , sqrt(3)/2)`'.

It is proven that point `P (1/2 , sqrt(3)/2)`. That is,

opposite side `bar(PQ) = sqrt(3)/2`

adjacent side `bar(OQ) = 1/2`

hypotenuse `bar(OP) = 1`trigonometric ratios for 60 degree `sin 60^@ = sqrt(3)/2`

`cos 60^@ = 1/2`

`tan 60^@ = sqrt(3)`

What is `text(cosec) 60^@`?trigonometric ratios for 60 degree

  • `1`
  • `2/sqrt(3)`
  • `2/sqrt(3)`

The answer is '`2/sqrt(3)`'.

Given the right angled triangle `Delta OPQ`, and `/_POQ = 0^@`.

`bar(OQ)=bar(OP)= 1`trigonometric ratios for 0 degree what is point `P`?

  • `(1,0)`
  • `(1,0)`
  • Trigonometric Ratios cannot be computed for this

The answer is '`(1,0)`'.

It is proven that point `P (1, 0)`. That is,

opposite side `bar(PQ) = 0`

adjacent side `bar(OQ) = 1`

hypotenuse `bar(OP) = 1`trigonometric ratios for 0 degree `sin 0^@ = 0`

`cos 0^@ = 1`

`tan 0^@ = 0`

What is `sec 0^@`?trigonometric ratios for 0 degree

  • `1`
  • `1`
  • `0`

The answer is '`1`'.

Given the right angled triangle `Delta OPQ`, and `/_POQ = 90^@`.

`bar(OP) = bar(QP) = 1`trigonometric ratios for 90 degree what is point `P`?

  • `(0,1)`
  • `(0,1)`
  • Trigonometric Ratios cannot be computed for this

The answer is '`(0,1)`'.

It is proven that point `P (0,1)`. That is,

opposite side `bar(PQ) = 1`

adjacent side `bar(OQ) = 0`

hypotenuse `bar(OP) = 1`trigonometric ratios for 90 degree `sin 90^@ = 1`

`cos 90^@ = 0`

`tan 90^@ = 1/0 `

What is `text(cosec) 90^@`?trigonometric ratios for 90 degree

  • `1`
  • `1`
  • `0`

The answer is '`1`'.

Trigonometric ratios for all standard angles is captured in the figure.trigonometric ratios for all standard angles When a trigonometric ratio is required for an angle, quickly work out the `x,y` for the angle. And sine is chord, cosine is the chord on the complementary angle, tan is the tangent etc. With a little bit of practice, this becomes fast and easy. I hope you do not memorize a table which you will eventually forget. If you can work out from the first principles, the information will last a lot longer. With a little bit of practice, students can recall the ratios as fast as when memorized.

To compute trigonometric ratios for standard angles, use the properties of the triangles having the given angle.

Trigonometric Ratios of Standard Angles:

trigonometric ratios for all standard angles When a trigonometric ratio is required for an angle, quickly work out the `x,y` for the angle.

                            
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