Finding the length of side in a 3d diagram with a few data only

**ananas1301** · January 7th 2011, 03:28 PM

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 don´t 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:

b²=a²+c²-2ac cos (beta)
or
b²=50²+c²-100c cos 45°

We also could change: c tan 30 = b tan 20 to
b=(c tan 30)/tan20 and square it

b²= (c² x (tan 30)²)/(tan 20)²

now we could "eliminate b²) couldn´t we to find c so therefore we have:

(c² x (tan 30)²)/(tan 20)² = 50²+c²-100c cos 45°

I´ve tried several ways but I cant get to c in this equation because I always end up with a c² and a c but cant devide the c.
I hope that is right what I´ve done and that someone can help me please.

Thank you

**snowtea** · January 7th 2011, 08:25 PM

Rewrite the equation in the form $A_1c^2 + A_2c + A_3 = 0$ with $A_i$ constants. You know how to solve a quadratic equation for c?

**Soroban** · January 7th 2011, 09:56 PM

Hello, ananas1301!

You have a good game plan . . .

$\text{The diagram shows the four points }A, B, C\text{ and }D.$
$A, B\text{ and }C\text{ are in the same horizontal plane.}$
$AD\text{ is vertical; }\,BC = 50\text{ m.}$
$\angle ABC = 45^o;\;\angle ABD = 30^o;\;\angle ACD = 20^o.$ °

$\text{Using the cosine rule in }\Delta ABC\text{, or otherwise, find }AD.$

Let $h = AD.$
Let $a = BC = 50,\;b = AC,\;c = AB$

In right triangle $D\!AC:\;\tan70^o \,=\,\dfrac{b}{h} \quad\Rightarrow\quad b \,=\,h\tan70^o$

In right triangle $D\!AB:\;\tan60^o \,=\,\dfrac{c}{h} \quad\Rightarrow\quad c \,=\,h\tan60^o$

In $\Delta ABC:\;b^2 \:=\:a^2 + c^2 - 2ac\cos B$

We have: . $(h\tan70^o)^2 \;=\;50^2 + (h\tan60^o)^2 - 2(50)(h\tan60^o)\cos45^o$

. . . . . . . . $h^2\tan^270^o \;=\;2500 + h^2\tan^260^o - 100h\tan60^o\cos45^o$

Since $\tan60^o = \sqrt{3}\,\text{ and }\,\cos45^o = \frac{\sqrt{2}}{2}$

. . $h^2\tan^270^o \;=\;2500 + h^2\cdot3 - 100h(\sqrt{3})\left(\frac{\sqrt{2}}{2}\right)$

. . $(\tan^270^o-3)h^2 + 50\sqrt{6}\,h - 2500 \;=\;0$

Now apply the Quadratic Formula . . .

**ananas1301** · January 8th 2011, 02:00 AM

Ohh yeah,

now I saw it.

I just didn´t recognize the quadratic formula.

Now I put it in order and solved it.

The book says 13.6m and I got 13.56.
So its right.

Thank you very much guys.

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