1. trig problem

Hi, i have just started a tafe course and have come across this problem. I have not taken trig in along time, any guidance would be appreciated.

question.

6. How do you find the length of the baseline AD when it is interrupted by an obstacle?

Length AB = 436.3m = x
Length BC = ? = y, and
Length CD = 542.8m = z.

Problem: What is length BC?

2. Re: trig problem

Hey mcleod86.

My advice would be to use the sine rule and write out all the relations you can for both the individual triangles and the large triangle as a whole.

Are you aware of the sine rule?

3. Re: trig problem

Hi Chiro, thank you for the prompt reply,

i know the sine rule
a/sinA = b/sinB = c/sinC i tried 542.8/sin54 = x/sin137 but not sure... my answer did not look right.

i have used interpolation to solve , does this look right ?

y - 436 / 38 - 45 = 542 - 436 / 54 - 45

y - 436 / -7 = 11.77

y - 436 = 11.77(-7)

y = -82.39 +436

y = 353.61

what do you think?
thanks

4. Re: trig problem

Can you give me some reasoning behind your solution?

5. Re: trig problem

I thought If I find the ratio between the distance and degree of known values, i can find the unknown distance. yes? but now i think of it, it is not a linear relationship so this isn't correct, is it?

6. Re: trig problem

The first thing you have to do is work out all your angles, from what you have given.

$\angle AED = \alpha + \beta + \gamma = 45^{\circ}12'37'' + 38^{\circ}25'48'' + 54^{\circ}33'28'' = 138^{\circ}11'53''$

$\angle EAD + \angle EDA = 180^{\circ} -138^{\circ}11'53'' = 41^{\circ}48'7''$

Let $\angle EDA = \theta$ and let $\angle EAD = 41^{\circ}48'7'' - \theta$

$\angle ABE = 180^{\circ} - (45^{\circ}12'37'' + 41^{\circ}48'7'' - \theta) = 92^{\circ}59'16'' + \theta$

$\angle DCE = 180^{\circ} - (54^{\circ}33'28'' + \theta) = 125^{\circ}26'32'' - \theta$

$\angle ECB = 54^{\circ}33'28'' + \theta$

$\angle EBC = 87^{\circ}00'44'' - \theta$

The law of sines is now applied to create the three equations needed to find the unknowns $EC, EB$ and $\theta$

$\displaystyle\frac{\sin{54^{\circ}33'28''}}{542.8} = \displaystyle\frac{\sin{\theta}}{EC}$

$\displaystyle\frac{\sin{45^{\circ}12'37''}}{436.3} = \displaystyle\frac{\sin(41^{\circ}48'7'' - \theta)}{EB}$

$\displaystyle\frac{\sin(87^{\circ}00'44'' - \theta)}{EC} = \displaystyle\frac{\sin(54^{\circ}33'28'' + \theta)}{EB}$

Once you have solved for these unknowns then you can apply the law of sines to find $BC$, thus find $AD$