Solving 2u^2 cos2B/g=0 for B: A Marking Scheme Explained

[markosheehan]

http://thephysicsteacher.ie/Exam Material/AppliedMaths/pdf/Higher Level/Marking Schemes/2000.pdfim trying to understand question 3 on this document. they have given the marking scheme. i understand most of the question but i don't understand how to solve 2u^2 cos2B/g =0 they get B=45 degrees how do they solve this?

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i understand it now.

 

[jvicens]

i was just confused on how they got B=45 degrees.

Hello,

I am happy to help explain how the solution for B=45 degrees was obtained.

In this question, we are given the equation 2u^2 cos2B/g=0 and asked to solve for B. To do this, we first need to understand the properties of cosine and how it relates to the value of an angle.

Cosine is a trigonometric function that relates the ratio of the adjacent side to the hypotenuse of a right triangle. In this case, we can rewrite the given equation as cos2B=0. This means that the cosine of 2B is equal to 0.

Now, we know that the cosine of an angle is equal to 0 when that angle is 90 degrees or π/2 radians. This is because at 90 degrees, the adjacent side of a right triangle is equal to 0, making the ratio 0/hypotenuse=0. Similarly, at 270 degrees or 3π/2 radians, the cosine is also equal to 0.

Since we are looking for the value of B, we need to solve for 2B. To do this, we can divide both sides of the equation by 2, giving us B=45 degrees or π/4 radians. This is because 2B=90 degrees or π/2 radians when B=45 degrees or π/4 radians.

I hope this explanation helps you understand how the solution for B=45 degrees was obtained. Please let me know if you have any further questions. Good luck with your studies!