Welcome to the forum.
Yes, this should happen. You resolve horizontally and vertically and you'll have two equations and unknowns T (tension) , Q and alpha.
to eliminate alpha.
You'll have an equation with T squared and a more complicated quadratic expression in Q.
To get the minimum T, make T squared the subject and consider how you can make the right hand side of the equation as small as possible. As the RHS is a quadratic, it's just a case of finding the minimum point on the quadratic ie. what value of Q to make this minimum.
Once you have that you can go back to the two equations with that value for Q, and find the corresponding values for T and alpha.
Hope that helps,
Welcome to the forum.
Call the centre of the circle C. The triangle A1CA2 is isosceles so split it into two identical right angled triangles and
In the same way get expressions for A1A3, A1A4, A1A5, A1A6 and A1A7.
Square these and add up.
This will simplify as
and so on.
Distances from A1 to A8 ... A12 can be found by symmetry.
Then you can repeat for points A2, ...A12.
Every distance will have been calculated twice ( as A1A2 and again as A2A1 for example) so divide the final total by 2.
Hope that helps.
hi Hong Yuan
I see you have become a member.
For the second part, you need to find the growth rate.
The 'cull' rate is 20 in each year so we can model this with a term '20t'. Strictly the cull is not continuous across the year but is all carried out in one go but it is the model you are asked to make.
So you can create a differential equation with this form dP/dt = growth rate minus cull rate. See if you can use this to do part (b). You'll need to use your values of a and b, and also the deer population at year 5 from part (a).
In Q1 you wrote:
3^2 x 6/4 x tan (180/6)
This does lead to the right answer but she cannot understand how you got this calculation (and neither can I). Why did you do this calculation ? Rather than using the simple formulas you have been taught ? That is why she is unhappy with your answers.
With 2, 6 and 7 you need more information. There are many possible answers when you only know one side and a 90 angle. Perhaps you were told something more in question 1.
In a 30-60-90 triangle:
Suppose you start with an equilateral triangle sides all 2s. Cut the triangle in half so that the distance to the middle is s. Rub out one half so all you have is a 30-60-90 triangle.
The hypotenuse is 2s, and the base is s. Use Pythagoras: height = √ (2s.2s - s.s) = √ (4s^2 - s^2) = √ (3s^2) = √ 3.s
So, for example in Q13, you are told the longest leg (height?) = 5 √ 3 so s=5. The base is therefore 5 and the hypotenuse is 2x5.
Post back with more information for the earlier ones and I'll help with those.
(a) Start with AD < AE + ED
Remember how the medians are created. (D, E and F are the midpoints so AE is half AC and DE is parallel to AB and half its length)
(b) Write the two similar inequalities for BE and CF. Add the three together. (If P < Q and R < S then P+Q < R+S)
Construct the three inequalities using ABG, BCG and CAG. (eg. AB < AG + BG)
Add them up.
Remember the centroid is one third of the way up the median. (eg. AG = 2.AD/3)
Hope that's enough for you to complete these.
The first three answers are correct.
As I advised on August 28, these problems need to be solved with tools that you have learned in this course. Those tools include the Pythagorean Theorem, the properties of special right triangles, or basic trigonometry ratios.
You haven't said what tools you have been taught so it is very hard to say what you should be doing.
Here is a diagram which may help.
A REGULAR hexagon can be divided into 6 equilateral triangles.
Each of those can be split in two, to make a right angled triangle with angles 30-60-90.
The area of a triangle is half the base x the height. The key to these problems is that height. It can be calculated using trig., like this
You can also use Pythag like this:
Hope that helps,
My earlier suggestion didn't, work, sorry.
But this does.
Start with the right hand side and differentiate chiefly using the product rule.
Use zetafunc`s suggestion to express sin(3x) in terms of sin(x) and sin cubed.
Now make the left hand side look like the right hand side.
You can quickly determine a to start off the sin cubed term and b to avoid any sin squared (remember sin squared + cos squared).
Then fix c to finalise the sin3x and sin x terms.
Then d and f allow you to tidy up the rest.
I found two cases, x=36 and x=45. I think that's all.
At school I was taught geometry from a variety of books. I don't have a preferred one now. You could always refer to the master:
Your method for calculating AD is exactly the same as mine.
Are you familiar with complex numbers ?
Complex numbers will also fit Pythagoras theorem so the method works whatever.
When I use pythag to calculate AV, AB and AS I get complex numbers not real ones. bobbym has confirmed this. If I use the cosine rule method described in post 4 the arcos is impossible to compute because it is not ≤ 1. I also cannot actually draw the square with those constraints. I conclude that the questioner forgot to check that this square can actually exist in the real world.
arithmetic progression means the angles are x, x + d, x + 2d, x + 3d, x + 4d
The angle sum for a pentagon is 540, so you can add up those angles and set equal to 540.
Also (x + 4d) - x = 44, so you can find 'd'. And hence x, the smallest angle.
Fitting together problem.
Two pentagons and one decagon meeting at a point means 108 + 108 + 144 = 360, which it does, so the fit is good.
An equilateral + an octagon mean the angle total so far is 60 + 135 = 195.
So the other polygon must fit in the space left, which means its internal angle is 360 - 195 = 165.
So the external angle is 180 - 165.
And (external angle) times n = 360. you can solve this for n.
Excellent! Both correct.
The slope intercept equation is one like this:
m is the slope and c is the intercept.
You can read about it here:
In this case m will be negative as the line is sloping downwards and c will be more than 20 as it is the point where the line crosses the y axis.
You can also test yourself here:
I've just tried that and both m and c look like 'nasty' decimals. Actually, its not as bad as it looks because those decimals are fairly straight forward fractions.
To do the next bit look here:
And finally to get when y = 10, just look for the point on the line where y = 10, or solve this 10 = mx + c with your values for m and c.