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hi Max,

Welcome to the forum.

Bob

hi Zeeshan 01

There are two stages to this problem. I think you have done the first but I'll say anyway.

This is a product of two functions so use the product rule:

You can combine the trig functions into a single trig function using the compound angle formulas. You want a cosine so I'll use

To make the derivative look like the right hand side here:

Hope that helps,

Bob

hi yorkmanz

Welcome to the forum.

Looks like you've got these sorted ; I agree with all your answers.

Bob

hi xxStellaxx

Welcome to the forum.

This question has been asked many times. Post 18 of this thread:

http://www.mathisfunforum.com/viewtopic.php?id=18391

has a diagram that may help.

Bob

hi Strangerrr

The area of the triangle will be different as it depends on both the side length and the angle at the top.

n is the number of sides. Here's an example using a regular 12 sided polygon. The centre point is O.

Let's say that A1A2 = side length = 7cm.

First I'll work out the angle A1-O-A2. Calculate 360/12 = 30.

The triangle A1-O-A2 is isosceles. The vertex angle is 30 and the base is 7.

Split it in half; see yellow shaded right angled triangle. This triangle has an angle of 15 and the 'opposite' is 3.5cm.

Use trigonometry to calculate the height.

Now calculate the area of the triangle A1-O-A2.

Now calculate the area of the whole polygon.

Hope this helps,

Bob

That's exactly what I got!

Bob

I agree with all those answers.

Provided the polygon is regular (with n sides) you can draw lines from the centre out to the vertices and make n triangles. Once you have calculated the area of one triangle you can times by n to get the area of the polygon.

These triangles will always be isosceles. The base is the side of the polygon. But you'll also need the height of the triangle.

The angle at the top of each triangle is 360/n in size. With an isosceles triangle you can always split the triangle in two down the middle, making a right angle at the bottom. The top angle will be split into two angles of 180/n each. The base is split into two parts of side/2 each. You can get the height by either

or

Then the area of the polygon is

Bob

**And a Very Happy Christmas and New Year to you!**

Bob

You're welcome!

Bob

hi Strangerrr

I'm not getting that answer. Imagine the prism made from card. If you unfold the net you'd have two triangular ends and three rectangles. The lateral area is the area of those three rectangles. Their measurements are 4 by 8, 6 by 8 and 6 by 8.

To get the volume you need the area of the triangle and then times by the height (8).

The perpendicular height of the triangle requires Pythag: sq rt (6^2 - 2^2). 2 because you make a right angle by dividing the isosceles triangle in half.

The 'slant' height isn't needed for prisms … they don't have one.

Bob

hi Spicca

Welcome to the forum.

I've been hoping that someone would post an answer to this and I could learn too. No such luck

So I'm having a go myself. Please note: I've never done this before, so what follows may be rubbish. Please comment …. ask for clarification … tell me why it's wrong etc. Maybe between us, we can arrive at the correct answer. And oh yes … your English is good.

So let's work with 3D coordinates with the x-y plane horizontal and z going straight up. Further, let's make the sphere have unit radius (cannot see any harm in that) and centred on the origin, O.

If P is one point of the tetrahedron, then we can specify its position using spherical coordinates theta and phi as shown here: https://en.wikipedia.org/wiki/Spherical … ate_system And let P' be the point in the x-y plane below P.

Now, what would be helpful is to have a formula for the volume of a tetrahedron in terms of theta and phi but I cannot find one. Plenty of internet pages giving the formula in vector terms such as https://math.stackexchange.com/question … ot-product

and as a determinant https://stackoverflow.com/questions/986 … n-4-points

I could also expand either into a large algebraic formula but it would take ages to enter all the LaTex so you'll have to ask nicely if you want this.

phi can take any random value from 0 to 2pi and theta any from 0 to pi.

So you can construct your function with 8 variables and there it is. Hhhmmmm.

Bob

hi Zeeshan 01

Sorry, but I do not understand what you are asking. Please post the whole question.

Bob

I would say this means they are the same line and intersect everywhere.

Bob

In algebra if A can be shown to be equal to B by a series of reversible algebraic steps then it follows that B is also equal to A.

Why don't you want to use the method of partial fractions ?

Bob

Why are you still posting 'help Me' questions in the 'exercises' section ?

hi Strangerrr

Wow! That's from a long time ago. Who is still re-cycling these questions?

As far as I can remember post 6 shows the method.

Bob

hi Lauren1415

Thanks for posting back with that result. Well done! It was a pleasure to help you.

Bob

hi Strangerrr

It was a pleasure to help.

Bob

In post 1

Zeeshan 01 wrote:

Integration of (1/(1+sinx+cosx)) dx by letting tanx/2=z

So that's what I did.

The alternative method of post 4 does not use z at all.

You now have the presence of a function and it's derivative, so it is directly integrable as ln(function) .

Bob

Substitute the z expressions:

Then simplify.

Bob

hi Zeeshan 01

Post in 'Help' if you want someone to help you with a problem.

Post in 'Exercises' if you have a problem that you can do and want to challenge others.

I'll assume you want help.

Using

is a way that someone has found out that helps with some integrals. If it works then it shouldn't matter why you choose this route. After all, some integrals can be 'done' just by inspection.

Start by showing that

Substitute and simplify.

Even using this 'trick', it's not an easy integral . You'll find it involves log base e.

Bob

hi Lauren1415

I think Q18 and Q19 are correct, but not Q20. Add up m and 2m and subtract from 180.

Bob

phrontister wrote:

And there I was thinking that you were just testing me to see if I'd notice you'd only done half the job!

Thanks! That's excellent. I shall use this excuse from now on. I've modified my signature.

Bob

hi Strangerrr

It's a property of all circumcircles that have one side as a diameter, that the angle made at the circumference is 90.

http://www.mathisfunforum.com/viewtopic.php?id=17799 post number 6.

So B = 90.

If you sketch the diagram and making BC a mirror line, reflect A and C in the mirror to get A' and C.

A'C = AC so CAA' is isosceles, and angle ACB = A'CB = 30, so ACA' = 60. This makes triangle CAA' an equilateral triangle with sides of 8.

That should help you do get the answers to these.

Bob

hi Lauren1415

Welcome to the forum.

I'm guessing that you were happy with your answers to 1 - 17. Somewhere around about Q15 or Q16 it explained what the problem means. You forgot to tell us that bit

Looking at the possible answers and also Strangerrr's answers I'm further guessing you were told that two out of three angles are given and are supplementary, and you have to work out the third angle.

So, for Q18, add the two together a + a + 80 = 2a + 80, and then subtract from 180, 180 - (2a + 80) and I'll leave you to simplify.

Were the earlier ones numbers with no algebra? Here's a way to make Q18 similar. Make up a value for 'a' , eg a = 30. Then the given angles become 30 and 110 and so the third angle is 180 - (30 + 110) = 40. Substitute a = 30 into each of the possible answers. Only one will give 40. That's the one.

Hope that helps,

Bob