# Trigonometry homework: find area of triangle with known lengths

1. Jul 3, 2009

### Tommzar

1. The problem statement, all variables and given/known data
i need to find angle BDC and side BC when given that

AB = 60m
BAC = 64
AC = 124
BD = 130
CD = 145

and to find the area of the whole plot

2. Relevant equations

3. The attempt at a solution

i have worked out side BC = 111.59 by using the Co-sine rule
a^2 = (60^2 + 124^2) - (2*60*124cos64)
=111.59

and angle BDC = cosC = (145^2) + (130^2) - (111.59^2) = 25472.6719

25472.6719/ (2x 145 x 130)

= 0.675
[PLAIN]http://mathworld.wolfram.com/images/equations/InverseCosine/Inline1.gif0.675[/URL] [Broken] = 47.49

just need help on finding side AD and other internal angles

Last edited by a moderator: May 4, 2017
2. Jul 3, 2009

### HallsofIvy

Staff Emeritus
Re: Triganometry

Strictly speaking, BC is a diagonal, not a side- that puzzled me for a moment. Likewise AD is a diagonal not a side and I don't believe you need to find its length to find the area. Look at triangle BCD now. You know the lengths of the three sides so you can find one angle using the cosine law, then find the other angles using the sine or cosine laws. Once you have the two triangles, you can find the area of each triangle, then add to find the area of the quadrilateral. Use BC as base of each triangle and use trig functions on the appropriate right triangles to find the height.

3. Jul 4, 2009

### Tommzar

Re: Triganometry

i have worked out angle BCD as being 59 degrees

and worked out the area of triangle BCD as having an area of 6946m^2

just trying to find the side AD is puzzling me

4. Jul 4, 2009

### rl.bhat

Re: Triganometry

In triangle ACB, AB/ sinACB = BC/sin BAC. Find angle BCA.
Angle ACD = BCD - BCA. Knowing AC, DC angle ACD you can find AD.

5. Aug 28, 2009

### Tommzar

Re: Triganometry

im still strugling on this i did forget to mention that it is a quadralateral plot of land

6. Aug 28, 2009

### njama

Re: Triganometry

Hello Tommzar!

You need to find the are of the quadrilateral ABCD, right?

Just find the area of the triangles ABC and ACD mate and find the sum ABC+ACD.

Then use, $\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta\,$,

and $\pm\sqrt{1 - \cos^2 \theta}\$

I know it is pretty complicated, but it is worth trying.

Good luck, mate! :tongue:

Edit: Another very good method to find AD, is finding the angle ACD=BCD-BCA and using the cosine law: