Math Is Fun Forum

Karn

Member

Registered: 2009-11-15

Posts: 14

Trigonometry

Hello

I'm having trouble with this question.

In the diagram below, triangle ACD is isosceles with an area of 32m^2 and AD is 11.31m in length whilst angle B has been set 56º. Find the length of BD.

Thank you for any help

cheers

Last edited by Karn (2009-11-18 16:17:56)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Trigonometry

Hi Karn;

I think there is something wrong with your diagram. Please check it for me.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

simplyjasper

Member

Registered: 2009-11-15

Posts: 24

Re: Trigonometry

I think there's something wrong with the diagram too...
Still with the information, it is sufficient to solve it..
Basically since ACD is an isosceles triangle, AC and CD is equidistant hence u should be able to get the angle for CAD and ADC

Then u should be able to find angle BAD

Finally, by using sine rule, you should be able to get BC

Last edited by simplyjasper (2009-11-19 03:52:58)

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,879

Re: Trigonometry

Hi! Here's some figuring that stem from the inconsistencies in the given data:

ACD's area (32m²) yields approx 11.3137 metres for AD, which differs from the text's 11.31 metres and the diagram's 11.3 metres.

There are therefore 3 different answers, and, using the sine rule as per simplyjasper's post:
1. If ACD's area is 32m²: BD = 2 x √32 x sin 79 / sin 56 = approx 13.396 metres
2. If AD is 11.31 metres (text): BD = 11.31 x sin 79 / sin 56 = approx 13.391677 metres
3. If AD is 11.3 metres (diagram): BD = 11.3 x sin 79 / sin 56 = approx 13.3798 metres

I suspect that AD's given length is just an approximation.

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Trigonometry

Hi all;

If AC = CD because the triangle is isosceles. And we call AC = y then by pythagorean theorem.

So CD = AC = 7.99737 m

Now using A = 1 /2 b * h which is A = 1 / 2 * CD * AC we get: A= 31.979025 m^2 not quite his 32 m^2.

Doing the calculations to more places does not bring the Area to 32m^2.

Last edited by bobbym (2009-11-19 12:00:22)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,879

Re: Trigonometry

Doing the calculations to more places does not bring the Area to 32m^2.

True - if using either of the two lengths of AD (11.31 in the text, and 11.3 in the diagram), as they look like being only approximations.

I suspect that the problem involves using the area of ΔACD to find the lengths of AC and CD, and hence AD, rather than using the given length of AD (but which one of the two options?), otherwise there's no point in knowing the area as BD can be determined without it.

I think that the solution may go something like this:

∠BAC = 180 - 90 - 56 = 34
∠CAD = (180 - 90)/2 = 45
∠BAD = 34 + 45 = 79

AC = CD = y

Area of a right-angled isosceles triangle: y² / 2 = 32, from which y=8

Pythagoras: AD = √(y² + y²) = √(8² + 8²) = √128

Sine Rule: BD / sin∠BAD = AD / sin∠ABD; BD = √128 x sin79 / sin56 = approx 13.396

Last edited by phrontister (2009-11-19 13:52:45)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Trigonometry

Hi phrontister;

I think you did it fine.

I used:

For the length of BC. So 8 + 5.39606 = 13.39606 m (same as yours) as the length of BD.

For math latex try this page it practically does it for you.
http://latex.codecogs.com/editor.php

Also just click on any piece of latex in any post and the code will be shown to you. This will help you get the knack of using it.

Last edited by bobbym (2009-11-19 14:25:00)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,879

Re: Trigonometry

Thanks, Bobby.

Yes - I should learn how to present maths equations etc properly...they read so much better that way. On another thread I copied the poster's code and altered it to suit my post, so it looked like I knew what I was doing.

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Trigonometry

Yes - I should learn how to present maths equations etc properly...they read so much better that way. On another thread I copied the poster's code and altered it to suit my post, so it looked like I knew what I was doing.

That's the way.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.