# Thread: HELPP !! 3D tRIG PROBLEM !

1. ## HELPP !! 3D tRIG PROBLEM !

two straight roads PC, PD at right angles to one another, running up a hill ABCD which is inclined at an angle theta to the horizontal. The roads PC, PD are inclined at angles a,b to the horizontal. BC, AD are lines of greatest slope and AB is the horizontal; CE, DF are verticals from C,D to the horizontal plane.

Show that sinsquaredtheta = sinsquareda + sinsquaredb

any help is GREATLLYYY APPRECIATED !!

2. ## 3D Trigonometry

Hello iiharthero
Originally Posted by iiharthero
two straight roads PC, PD at right angles to one another, running up a hill ABCD which is inclined at an angle theta to the horizontal. The roads PC, PD are inclined at angles a,b to the horizontal. BC, AD are lines of greatest slope and AB is the horizontal; CE, DF are verticals from C,D to the horizontal plane.

Show that sinsquaredtheta = sinsquareda + sinsquaredb

any help is GREATLLYYY APPRECIATED !!
Suppose $BC = AD = d$. I assume that you've drawn a diagram. If so, you'll know that:

$CE = d \sin \theta, BE = d\cos\theta, PE = CE\cot a = d\sin\theta\cot a, PC = \frac{CE}{\sin a}=\frac{d\sin\theta}{\sin a}$

So $PB^2 = PE^2-BE^2 = d^2\sin^2\theta \cot^2a -d^2\cos^2\theta$

which can be re-written in terms of $\sin\theta$ and $\sin a$ (which is what we want!) as $\frac{d^2(\sin^2\theta-\sin^2a)}{\sin^2a}$, using $\cos^2\theta = 1 - \sin^2\theta$ and $1 + \cot^2a = \csc^2 a=\frac{1}{\sin^2a}$ (Can you do this bit?)

Similarly, using the other side of the diagram:

$DP = \frac{d\sin\theta}{\sin b}$ and

$PA^2 = \frac{d^2(\sin^2\theta - \sin^2b)}{\sin^2b}$

Now for the bit where you need to keep a clear head!

Using Pythagoras on $\triangle DPC: DC^2 = AB^2= DP^2+PC^2$

$\Rightarrow (PB+PA)^2 = PD^2 + PC^2$

$\Rightarrow \frac{d^2(\sin^2\theta-\sin^2a)}{\sin^2a} + \frac{d^2(\sin^2\theta-\sin^2b)}{\sin^2b} + \frac{2d^2}{\sin a \sin b}\sqrt{(\sin^2\theta-\sin^2a)(\sin^2\theta - \sin^2b)}$ $= \frac{d^2\sin^2\theta}{\sin^2b}+\frac{d^2\sin^2\th eta}{\sin^2a}$

If you now multiply both sides by $\sin^2a\sin^2b$ and divide by $d^2$ this will simplify to

$-2\sin a^2 \sin^2 b + 2\sin a\sin b\sqrt{(\sin^2\theta-\sin^2a)(\sin^2\theta - \sin^2b)}=0$

$\Rightarrow \sin a \sin b = \sqrt{(\sin^2\theta-\sin^2a)(\sin^2\theta - \sin^2b)}$, assuming $\sin a\sin b \ne 0$

$\Rightarrow \sin^2a \sin^2b = \sin^4\theta - \sin^2\theta(\sin^2 a + \sin^2 b) + \sin^2a\sin^2b$

$\Rightarrow \sin^2\theta\Big(\sin^2\theta -(\sin^2a + \sin^2b)\Big) = 0$

$\Rightarrow \sin^2\theta = \sin^2a+\sin^2b$, again assuming $\sin^2\theta \ne 0$

That wasn't so easy, was it?