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### Trigonometric Ratios for Standard Angles

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Home

Standard Angles

»  Quickly follow the angle description to calculate the ratios. No need to memorize.

### Trigonometric Ratios for Standard Angles

plain and simple summary

nub

plain and simple summary

nub

dummy

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

simple steps to build the foundation

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simple steps to build the foundation

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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.

Keep tapping on the content to continue learning.
Starting on learning "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.

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))

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

It is proven that point P (1/sqrt(2), 1/sqrt(2)). sin 45^@ = 1/sqrt(2)

cos 45^@ = 1/sqrt(2)

tan 45^@ = 1

What is cot 45^@?

• 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)
• (0,1/2)

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

It is proven that point P (sqrt(3)/2 , 1/2). sin 30^@ = 1/2

cos 30^@ = sqrt(3)/2

tan 30^@ = 1/sqrt(3)

What is sec 30^@?

• 1
• 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)
• (1/2,0)

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

It is proven that point P (1/2 , sqrt(3)/2). sin 60^@ = sqrt(3)/2

cos 60^@ = 1/2

tan 60^@ = sqrt(3)

What is text(cosec) 60^@?

• 1
• 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)
• Trigonometric Ratios cannot be computed for this

The answer is '(1,0)'.

It is proven that point P (1, 0). sin 0^@ = 0

cos 0^@ = 1

tan 0^@ = 0

What is sec 0^@?

• 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)
• Trigonometric Ratios cannot be computed for this

The answer is '(0,1)'.

It is proven that point P (0,1). sin 90^@ = 1

cos 90^@ = 0

tan 90^@ = 1/0

What is text(cosec) 90^@?

• 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.

comprehensive information for quick review

Jogger

comprehensive information for quick review

Jogger

dummy

Trigonometric Ratios of Standard Angles:

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

practice questions to master the knowledge

Exercise

practice questions to master the knowledge

Exercise

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Progress

About you

Progress

Given the right angled triangle ; O P Q, and the angle P O Q equals 45 degree. ;; line O P ; equals ; 1 ;; line O Q ; equals ; line PQ which are x and y;; line O Q squared plus line PQ squared ; equals ; line O P squared which equals ; 1 ;; 2 x squared ; equals 1 ;; What is point p?
zero;0
zero comma zero
square;root;one by
one by square root 2 comma one by square root 2
The answer is 'one by square root 2 comma one by square root 2'
It is proven that point P is one by square root 2 comma one by square root 2;; So sine 45 equals one by square root 2;; cos 45 equals one by square root 2;; tan 45 equals one
What is cot 45 degree?
1;one
one
square;root
square root 2
The answer is 'one'
Given the right angled triangle ; O P Q, and the angle P O Q equals 30 degree. ;; line O P ; equals ; 1 ;; Construct an equilateral triangle O P prime P ;; From this it is derived that line P Q equals one by 2 ;; line O Q squared plus line PQ squared ; equals ; line O P squared which equals ; 1 ;; x squared plus one by 2 square ; equals 1 ;; What is point p?
square;root;one by
square root 3 by 2 comma one by 2
zero;0
zero comma one by 2
The answer is 'square root 3 by 2 comma one by 2'
It is proven that point P is square root 3 by 2 comma one by 2. ;; So sine 30 equals one by 2;; cos 30 equals square root 3 by 2;; tan 30 equals one by square root 3
What is secant 30 degree?
1;one
one
square;root;2;two; square root;3;three
two by square root 3
The answer is 'two by square root 3'
Given the right angled triangle ; O P Q, and the angle P O Q equals 60 degree. ;; line O P ; equals ; 1 ;; Construct an equilateral triangle O Q prime P ;; From this it is derived that line O Q equals one by 2 ;; line O Q squared plus line PQ squared ; equals ; line O P squared which equals ; 1 ;; one by 2 square plus y squared; equals 1 ;; What is point p?
square;root;one by
one by 2 comma square root 3 by 2
zero;0
one by 2 comma zero
The answer is 'one by 2 comma square root 3 by 2'
It is proven that point P is one by 2 comma square root 3 by 2. ;; So sine 60 equals square root 3 by 2;; cos 60 equals one by 2;; tan 60 equals square root 3
What is co-secant 60 degree?
1;one
one
square;root;2;two; square root;3;three
two by square root 3
The answer is 'two by square root 3'
Given the right angled triangle ; O P Q, and the angle P O Q equals 0 degree. ;; line O Q ; equals ;line O P ; equals ; 1 ;; What is point p?
one; zero; one comma zero
one comma zero
ratios;computed;for;this
Trigonometric Ratios cannot be computed for this
The answer is 'one comma zero'
It is proved that point p is one comma zero. ;; So sine 0 equals 0 ;; cos 0 equals 1 ;; tan 0 equals 0.
What is secant 0 degree?
1;one
one
zero
zero
The answer is 'one'
Given the right angled triangle ; O P Q, and the angle P O Q equals 90 degree. ;; line O P ; equals ;line Q P ; equals ; 1 ;; What is point p?
zero;one;comma
zero comma one
ratios;computed;for;this
Trigonometric Ratios cannot be computed for this
The answer is 'zero comma one'
It is proved that point p is zero comma one . ;; So sine 90 equals 1 ;; cos 90 equals 0 ;; tan 90 equals 1 by 0.
What is co-secant 90 degree?
1;one
one
zero
zero
The answer is 'one'
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 for standard angles are captured in the image. When a trigonometric ratio is required for an angle, quickly work out the ;x;y; for the angle.

we are not perfect yet...