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# Maths question trig? Watch

1. (Original post by Forevereuphoria)
I still dont understand could you explain what you are talking about i.e why its tan? Put examples in
Okay, right. Let's start from the beginning.

What can you tell me about angles BDA and CAD? And can you give me the size of one of these angles?
2. (Original post by Forevereuphoria)
How are you labelling the sides...
The sides are labelled by their letters, so AC is the side between letters A and C

it can't be 90, the diagonals of a rectangle don't intersect at right angles, it would have to be a square for that to happen.
3. (Original post by Slowbro93)
Okay, right. Let's start from the beginning.

What can you tell me about angles BDA and CAD? And can you give me the size of one of these angles?
They are all the same?
4. Can someone tell me the answer and as its a hw game im doing and i'll speak to my teacher tomorrow
5. (Original post by Forevereuphoria)
They are all the same?
Yes, so now what is the angle? you have two lengths, how can you calculate it?

And no, we will not tell you the answer but we can assist you.
6. (Original post by Slowbro93)
Yes, so now what is the angle? you have two lengths, how can you calculate it?

And no, we will not tell you the answer but we can assist you.
Times 14 and 18 ??? i dONT KNOW
7. Ok look at triangle B-A-D its a right angle triangle that has an area of half the square. Now look at the other information you are given the length C-D = 14cm and A-D = 18cm. What else can we see, we can see that the triangle the angle x is in is and equilateral triangle. Ok we need to use this information to find the the angle x. First you need to do some trigonometry to find the angle at B with in the triangle with the angle x.(its hard to describe without a drawing) . SOHCAHTOA right, so we need TOA because we know the opposite length and the adjacent length. Tan X = 18/14 . then answer divided by Tan and we have the angle B of the triangle B-A-D. Now that we have that and we also know that the angles within a triangle add up to 180. so the triangle with x in it has two angles that equal B and one that equals X. so the next calculation is 180-(B*2) and that will give you the angle X.
8. (Original post by Forevereuphoria)
Times 14 and 18 ??? i dONT KNOW
Have you not done SOH CAH TOA?
9. Haa mymaths
15.4183541cm
(15.4cm to 3sf.)
10. (Original post by Slowbro93)
Have you not done SOH CAH TOA?
could i do toa
and
18/14=0.28
then
tan-1(0.28)=
52.1
11. Image is attached
12. Hi what part are you stuck on with this question. do you know how to start it?
13. Image is attached
14. Have a good read of this it will tell you how to do it,

Ok look at triangle B-A-D its a right angle triangle that has an area of half the square. Now look at the other information you are given the length C-D = 14cm and A-D = 18cm. What else can we see, we can see that the triangle the angle x is in is and equilateral triangle. Ok we need to use this information to find the the angle x. First you need to do some trigonometry to find the angle at B with in the triangle with the angle x.(its hard to describe without a drawing) . SOHCAHTOA right, so we need TOA because we know the opposite length and the adjacent length. Tan X = 18/14 . then answer divided by Tan and we have the angle B of the triangle B-A-D. Now that we have that and we also know that the angles within a triangle add up to 180. so the triangle with x in it has two angles that equal B and one that equals X. so the next calculation is 180-(B*2) and that will give you the angle X.
15. There's no image attached
16. (Original post by Forevereuphoria)
could i do toa
and
18/14=0.28
then
tan-1(0.28)=
52.1
Thats right your half way there. thats the angle at B,

now all you have to do is 180 - (2*52.1) = X

The reason for this is because its an equilateral triangle
17. Pythagoras theorem 18^2+14^2= AC^2
AC= sqroot520
AC/2 gives you DX and XC.

DX, XC = sqroot 130

Cosine Rule a^2 = b^2 + c^2 - 2bc Cos A

Let X be A
Therefore DC (14cm) will be a.
The other two sides are b and c.

Rearrange the cosine rule to make Cos A the subject.

Cos A = (b^2 + c^2 - a^2)/2bc

Enter in the figures we have so far.

Cos A = (sqroot130^2 + sqroot130^2 - 14^2)/2*sqroot130*sqroot130
Cos A = 64/260
A= ????

Therefore X is = ???? degrees.

EDIT: Oh right you guys not giving him the answer.
This is a longer method than the others but I find it easier.
18. (Original post by Forevereuphoria)
Image is attached
Are you sure you attached the images?
19. 2*arctan(7/9)=75.7499673
20. (Original post by spotify95)
Yes there is.

I'll take a look.
There wasn't when I posted my response!

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