Anybody can tell me how to find y1 , y2 , angle 1

1. Dec 21, 2004

bent al physics

Anybody can tell me how to find y1 , y2 , angle 1 and 2

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2. Dec 21, 2004

dextercioby

The problem cannot be solved because it lacks one equation:there can be written 5 equations for the 6 unknowns.If the angle between Y_{1} and Y_{2} were given,the problem would be solved.
Equations:
$$\sin\alpha_{1}=\frac{150}{Y_{1}}$$
$$\sin\alpha_{2}=\frac{220}{Y_{2}}$$
$$a+b=350$$
$$a^{2}+150^{2}=Y_{1}^{2}$$
$$b^{2}+220^{2}=Y_{2}^{2}$$

As u can see,an equation is missing.There is an option of solving for 5 unknowns wrt une unknown chosen as parameter.But that wouldn't be too convenient,right??

Daniel.

PS.Alpha_{1},alpha_{2} are the two angles unknown (labeled through "angle 1" and "angle 2",respectively),while "a" and "b" are the two remaining sides of the wo trianges which have not been labeled on the picture.

3. Dec 21, 2004

arildno

ahrkon and daniel is absolutely right.
What a dreadful mistake on my part.

Last edited: Dec 21, 2004
4. Dec 21, 2004

ahrkron

Staff Emeritus
I get the impression that Heron's formula is implied by the rest (and hence brings no new information).

I don't see any reason why the vertex in the middle has to be in one position and no other. So far any position between the two segments with known size (150 and 220 m) would satisfy all equations. Isn't that so?