Rewrite the equation in the form with constants. You know how to solve a quadratic equation for c?
Hello guys,
Here is the question:
"The following three-dimensional diagram shows the four point A, B, C and D
A, B and C are in the same horizontal plane and AD is vertical.
Angle ABC is 45°
BC = 50m
Angle ABD = 30°
and the last on is that Angle ACD is 20°
Using the cosine rule in the trianlge ABC, or otherwise, find AD!"
So the point D is directly above A.
So you can imagine this as a pyramid with different long sides.
So one triangle with D above A as it says.
And now this is as far as I got:
It says that AD is vertical so that we can say that the angles CAD and BAD must be 90 °
Now we have the two triangles on the sides with only 2/3(we donīt need the third angle i think).
To make all algebra easier I named BA as c
and named CA as b
and AD is X what we are looking for
because we have now to rectangular triangles we can use tan twice.
tan 30 = x/c
and tan 20 = x/b
or in the end: c tan 30 = b tan 20
i looked back to the question and saw "use the cosine rule".
The only combination which would make sense is this:
bē=aē+cē-2ac cos (beta)
or
bē=50ē+cē-100c cos 45°
We also could change: c tan 30 = b tan 20 to
b=(c tan 30)/tan20 and square it
bē= (cē x (tan 30)ē)/(tan 20)ē
now we could "eliminate bē) couldnīt we to find c so therefore we have:
(cē x (tan 30)ē)/(tan 20)ē = 50ē+cē-100c cos 45°
Iīve tried several ways but I cant get to c in this equation because I always end up with a cē and a c but cant devide the c.
I hope that is right what Iīve done and that someone can help me please.
Thank you