3D trig question

[apple2357]

Can anyone see how you can get length AM=3?
I am thinking there is something not right with the question, i have been using Pythagoras but getting stuck. Something doesn't feel right.
I dont have much to post but be interested in starting points/suggestions rather than complete methods please
Can anyone help here?

 

[Deleted member 4993]

Can anyone see how you can get length AM=3?
I am thinking there is something not right with the question, i have been using Pythagoras but getting stuck. Something doesn't feel right.
I dont have much to post but be interested in starting points/suggestions rather than complete methods please
Can anyone help here?

I am getting AM² = 25/2

 

[apple2357]

I am getting AM² = 25/2

Does that assume BAC is right angled ?

 

[Deleted member 4993]

Does that assume BAC is right angled ?

Yes - I did not realize that that information was not given. The drawing fooled me!!

I think, as given, the problem cannot be solved uniquely.

 

[topsquark]

Yes - I did not realize that that information was not given. The drawing fooled me!!

I think, as given, the problem cannot be solved uniquely.

I agree that there is no unique solution. First note that AD = 12 cm. Then you can write a system of two equations:
[math]\begin{cases} 12^2 + AM^2 = DM^2 \\ DM^2 + x^2 = 13^2 \end{cases}[/math]
Two equations and three unknowns.

-Dan

 

[apple2357]

I agree that there is no unique solution. First note that AD = 12 cm. Then you can write a system of two equations:
[math]\begin{cases} 12^2 + AM^2 = DM^2 \\ DM^2 + x^2 = 13^2 \end{cases}[/math]
Two equations and three unknowns.

-Dan

And you can't use the AB=AC=5 to bring in an additional equation? That was the kind of thing i was trying.

 

[topsquark]

And you can't use the AB=AC=5 to bring in an additional equation? That was the kind of thing i was trying.

That's how I got the AD = 12 cm. You can write [imath]x^2 + AM^2 = 5^2[/imath], but that's equivalent to adding the two equations I gave so it adds nothing new.

(I should mention that I'm defining x = MC.)

-Dan