__maths__>__Trigonometry__>__Trigonometric 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` 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` `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^@`?

- `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`

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` `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^@`?

- `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` 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` `sin 60^@ = sqrt(3)/2`

`cos 60^@ = 1/2`

`tan 60^@ = sqrt(3)`

What is `text(cosec) 60^@`?

- `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` 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` `sin 0^@ = 0`

`cos 0^@ = 1`

`tan 0^@ = 0`

What is `sec 0^@`?

- `1`
- `1`
- `0`

The answer is '`1`'.

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

`bar(OP) = bar(QP) = 1` 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` `sin 90^@ = 1`

`cos 90^@ = 0`

`tan 90^@ = 1/0 `

What is `text(cosec) 90^@`?

- `1`
- `1`
- `0`

The answer is '`1`'.

Trigonometric ratios for all standard angles is captured in the figure. 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: **

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

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