trig help

[storka]

k I tried to do all i can but i just cant get this question, all i can get is angle A and the overall angle C, I dont get how you are supposed to get the other angles and especially side AB.

i made a diagram and here is a link to it:

http://www.freewebs.com/illestkilla/math%20question.bmp

 

[stapel]

I don't know that this forum will serve up bitmaps. Could you reply with a link to a GIF version of the image?

Thank you.

Eliz.

 

[pka]

Note that radii are equal then us the law of cosines.
\(\displaystyle \L
\begin{array}{l}
(10.3)^2 = DA^2 + (15.1)^2 - 2(15.1)(DA)\cos (40^o ) \\
(10.3)^2 = DB^2 + (15.1)^2 - 2(15.1)(DB)\cos (40^o ) \\
\end{array}\)

Solve the syatem.

 

[storka]

im sorry i dont have a converter, and im a bit confused. cant u just click on the link and go to the adress to see the image?

 

[storka]

umm pka, i am in grade 11 and we have not covered those rules, do u think there is another way to solve it? all i have is the sine lawa and cosine law and that case wud be ambiguos wont it?:S

I mean that how can DB equal the same as DA when they are obviously not equal since DB is like half of wat DA is.

 

[jonboy]

To find the link of CB, find the geometric means:

:arrow: 15.1/x=x/10.3=155.53rad

:arrow: use Law of Cosines to find the the other angle measures.

 

[storka]

im sorry i just dont get that. im usually good at math I just dont see how DB can be equal to the same as DA because if CB was equal to AC than wudnt B have to be 70 degrees by doing the sine law, and its obviously not 70 degrees because it is clearly obtuse on the picture. :S

 

[pka]

all i have is the sine lawa and cosine law and that case wud be ambiguos wont it?

All I used is the cosine law!

 

[jonboy]

all i have is the sine lawa and cosine law and that case wud be ambiguos wont it?

All I used is the cosine law!

Yes that would be more precise since it does not say the the triangles are similiar.

 

[jonboy]

I figured out another way to perform this problem (more basic)!!!
Use law of Cosines to find the m<A
SinA/15.1=Sin40/10.3
m<A=approx 70.4
Thus making m<C=69.6
Remember we focused on the big triangle.
Since in Triangle BCA all the radii are all the same, BCA is an
isoceles triangle. Since 70.4 is vertical from the 10.3cm, then the
angle verticle the other 10.3cm must also be 70.4, which makes m<C=39.2
Once again use law of Sine:
Sin39.2/BA=Sin70.4/10.3
You will get an answer of approx. 6.9cm
Glad I could help you