Two vectors A and B of magnitudes A = 30 units and B = 60 units respectively are inclined to each other at angle of 60 degrees. Find the resultant vector.

First, draw the vectors on any piece of paper. One way to approach this problem is to draw one vector that has an angle of elevation of 0 degrees, which just means that's parallel to the x-axis, and draw the other vector with an angle of elevation of 60 degrees. Let's assume that vector A is horizontal, and vector B is elevated at an angle of 60 degrees. Just connect both vectors with one vector being elevated 60 degrees above the other. If you were to draw the missing side of the triangle, you would see it's not a right triangle and therefore we can't use our basic SOH CAH TOA identities. Fortunately, we can use our beloved Law of Cosines because we know the length of two of the sides and the angle between them. I assume you know how to use it and therefore I will not explain it here, however, if this sounds unfamiliar to you, here is the URL link to the playlist of the Law of Sines and the Law of Cosines https://www.khanacademy.org/math/trigonometry/trig-with-general-triangles Nevertheless, if we were to use the Law of Cosines to solve for the magnitude of the resultant, the answer should be (30)*(3^1/3) or 30 times the square root of 3. ;)

Why kicking a ball with 45 degree will travel further horizontally than 60 degree

the horizontal distance is called the range. and R=(u^2sin(2thetha))/g. and since the largest value of sine is 1. and that only happens if thetha is 45.that means 2thetha would be 90 and the sine of 90 gives u one

should I have studied trigonometry before starting this course? Because I didn't, and I'm kind of confused now.

Definitly. Trig is an integral part of this course.

how do I know when to use the inverse of tan?

When you have both sides ( opposite and adjacent) but no angle. Normally you’re looking for a side or both given an angle. So you’d write it out as Sin 45 = opposite/3 (opposite/hypothenuse). But when you’re give two sides and looking for and angle you’d write it out — > tan 0= 4/3(opposite/adjacent). To solve that you’d write the inverse of tan (tan-1) Which is tan-1(1.33) *divide before using inverse tan or else you’ll get a different answer. You know to use the inverse of tan when given two sides and need to find the angle.

why θ=tan^−1 ?I didn't catch :c

Nice question! First off -1 is not the exponent of tan there. tan is a function, a lot of functions have inverses. tan^-1 denotes the tan inverse function or the _arctan_ function. E.g. If we have tan(y)=x Now if we apply the tan inverse function on both sides, tan^-1(tan(y))=tan^-1(x) This gives: y=tan^-1(x) Hope that helps! And that is a really good question, and its always nice to ask questions! If anything was unclear or needs more explaining feel free to ask a follow up! Cheers!

Main content

Class 11 Physics (India)

Course: Class 11 Physics (India) > Unit 8

Lesson 2: Analyzing vectors using trigonometry

Analyzing vectors using trigonometry review

Google Classroom

Review the skills for analyzing vectors, including how to find horizontal and vertical components of vectors.

Analyzing vectors with trigonometry

To simplify calculations for two-dimensional motion, we analyze the movement in the vertical direction separately from the horizontal direction. Since displacement, velocity, and acceleration are vector quantities, we can analyze the horizontal and vertical components of each using some trigonometry.

Finding horizontal and vertical components

We can find the horizontal component

A_{x}

and vertical component

A_{y}

of a vector using the following relationships for a right triangle (see Figure 1a).

A

is the hypotenuse of the right triangle.

A_{x} = A \cos θ

A_{y} = A \sin θ

[Does cosine always align with the x-axis and sine with the y-axis?]

Determining the magnitude of the resultant

When we know the horizontal and vertical components, we can find the magnitude of their sum using the Pythagorean theorem (Figure 2).

| A | = \sqrt{A_{x}^{2} + A_{y}^{2}}

Finding vector direction

To find the angle

θ

of the vector from the horizontal axis, we can use the horizontal component

A_{x}

and vertical component

A_{y}

in the trigonometric identity:

\tan θ = | \frac{A_{y}}{A_{x}} |

We take the inverse of the

\tan

function to find the angle

θ

θ = \tan^{- 1} | \frac{A_{y}}{A_{x}} |

Common mistakes and misconceptions

Sometimes people forget when to use $\sin$ ‍ or $\cos$ ‍ for calculating vector components. When in doubt, draw a right triangle and remember:

\begin{aligned} \sin θ & = \frac{opposite}{hypoteneuse} \\ \cos θ & = \frac{adjacent}{hypoteneuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}

Learn More

For deeper explanations, see our video visualizing vectors in 2 dimensions.

To check your understanding and work toward mastering these concepts, check out the exercise on adding and decomposing vectors using trigonometry.

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Ah'mir White
Posted a year ago. Direct link to Ah'mir White's post “khan academy is good”
khan academy is good
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(31 votes)
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mr.digvijay1
Posted 5 years ago. Direct link to mr.digvijay1's post “Two vectors A and B of ma...”
Two vectors A and B of magnitudes A = 30 units and B = 60 units respectively
are inclined to each other at angle of 60 degrees. Find the resultant vector.
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(10 votes)
Answer
- Omar
  Posted 5 years ago. Direct link to Omar's post “First, draw the vectors o...”
  First, draw the vectors on any piece of paper. One way to approach this problem is to draw one vector that has an angle of elevation of 0 degrees, which just means that's parallel to the x-axis, and draw the other vector with an angle of elevation of 60 degrees. Let's assume that vector A is horizontal, and vector B is elevated at an angle of 60 degrees. Just connect both vectors with one vector being elevated 60 degrees above the other. If you were to draw the missing side of the triangle, you would see it's not a right triangle and therefore we can't use our basic SOH CAH TOA identities. Fortunately, we can use our beloved Law of Cosines because we know the length of two of the sides and the angle between them. I assume you know how to use it and therefore I will not explain it here, however, if this sounds unfamiliar to you, here is the URL link to the playlist of the Law of Sines and the Law of Cosines https://www.khanacademy.org/math/trigonometry/trig-with-general-triangles
  Nevertheless, if we were to use the Law of Cosines to solve for the magnitude of the resultant, the answer should be (30)*(3^1/3) or 30 times the square root of 3.
  ;)
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  (14 votes)
Jiale Zhang
Posted 4 years ago. Direct link to Jiale Zhang's post “Why kicking a ball with 4...”
Why kicking a ball with 45 degree will travel further horizontally than 60 degree
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(4 votes)
Answer
- physics lover
  Posted 4 years ago. Direct link to physics lover's post “the horizontal distance i...”
  the horizontal distance is called the range. and R=(u^2sin(2thetha))/g. and since the largest value of sine is 1. and that only happens if thetha is 45.that means 2thetha would be 90 and the sine of 90 gives u one
  Button navigates to signup page
  (5 votes)
Ethan
Posted 3 years ago. Direct link to Ethan's post “should I have studied tri...”
should I have studied trigonometry before starting this course? Because I didn't, and I'm kind of confused now.
Button navigates to signup pageComment on Ethan's post “should I have studied tri...”
(4 votes)
Answer
- Aaron Ghosh
  Posted 2 years ago. Direct link to Aaron Ghosh's post “Definitly. Trig is an int...”
  Definitly. Trig is an integral part of this course.
  Button navigates to signup page
  (5 votes)
s632420
Posted 5 years ago. Direct link to s632420's post “how do I know when to use...”
how do I know when to use the inverse of tan?
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(3 votes)
Answer
- nabdul19
  Posted 5 years ago. Direct link to nabdul19's post “When you have both sides ...”
  When you have both sides ( opposite and adjacent) but no angle.
  Normally you’re looking for a side or both given an angle. So you’d write it out as Sin 45 = opposite/3 (opposite/hypothenuse). But when you’re give two sides and looking for and angle you’d write it out — > tan 0= 4/3(opposite/adjacent). To solve that you’d write the inverse of tan (tan-1)
  Which is tan-1(1.33) *divide before using inverse tan or else you’ll get a different answer.
  You know to use the inverse of tan when given two sides and need to find the angle.
  Button navigates to signup page
  (4 votes)
Karina Lvova
Posted 3 years ago. Direct link to Karina Lvova's post “why θ=tan^−1 ?I didn't ca...”
why θ=tan^−1 ?I didn't catch :c
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(0 votes)
Answer
- Swapnal
  Posted 3 years ago. Direct link to Swapnal's post “Nice question! First off ...”
  Nice question!
  First off -1 is not the exponent of tan there. tan is a function, a lot of functions have inverses. tan^-1 denotes the tan inverse function or the arctan function.
  E.g. If we have tan(y)=x
  Now if we apply the tan inverse function on both sides,
  tan^-1(tan(y))=tan^-1(x) This gives:
  
  y=tan^-1(x)
  
  Hope that helps!
  And that is a really good question, and its always nice to ask questions!
  If anything was unclear or needs more explaining feel free to ask a follow up!
  Cheers!
  Comment on Swapnal's post “Nice question! First off ...”
  (6 votes)
divyaparte2021
Posted 2 years ago. Direct link to divyaparte2021's post “= tan-1[Ay\Ax] what will ...”
= tan-1[Ay\Ax] what will further answer ?
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(1 vote)
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Ah'mir White
Posted a year ago. Direct link to Ah'mir White's post “The horizontal distance i...”
The horizontal distance is called the range.
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(1 vote)
Answer
krishi manek
Posted 4 years ago. Direct link to krishi manek's post “why do you take modulus o...”
why do you take modulus of ay/ax while finding theta
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(0 votes)
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- Priyanshu Chauhan
  Posted 3 years ago. Direct link to Priyanshu Chauhan's post “i think it is really a go...”
  i think it is really a good question. Actually it is just for sake of simplicity so as to not confuse those who are not aware of complex trigonometry. Because if ay/ax gets negative then theta comes to the angle where tangent is negative i.e.-2nd and 4th quadrant.
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  (1 vote)
Bronte Tozer
Posted 5 years ago. Direct link to Bronte Tozer's post “How do you tell if the ho...”
How do you tell if the horizontal or vertical components are negative/positive?
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