# HELP!! please and thank you

• Dec 17th 2007, 12:30 AM
Luckegrl_16
A crew is surveying Mr. Gournic's property. From his front door, 10 m away on a bearing S 48deg. E is his mailbox. Seven m from the mailbox on a bearing of S 42 deg. W is a telephone pole. a) How far is it from the door directly to the pole? (exact value) b) What is the bearing from the pole to the door?:confused:
• Dec 17th 2007, 02:01 AM
DivideBy0
Quote:

Originally Posted by Luckegrl_16
A crew is surveying Mr. Gournic's property. From his front door, 10 m away on a bearing S 48deg. E is his mailbox. Seven m from the mailbox on a bearing of S 42 deg. W is a telephone pole. a) How far is it from the door directly to the pole? (exact value) b) What is the bearing from the pole to the door?:confused:

In the diagram, let H be Mr. Gournic's House, L be his letterbox, and P be the telephone pole.

By alternate angles on parallel lines, we can deduce various other angles in the diagram.
After filling in several angles, we note that $\displaystyle \angle HLP = 42 + 48 = 90^{\circ}$, so the paths actually form a right angled triangle $\displaystyle \Delta HLP$.

Then, by Pythagoras' Theorem,

$\displaystyle HP =\sqrt{10^2+7^2}=\sqrt{149}$

So the distance from his door to the pole is $\displaystyle \sqrt{149}$ m.

Next, to find the beaing of H from P, note that

$\displaystyle \tan{(\angle LHP)} = \frac{7}{10}$, so

$\displaystyle \angle LHP = \arctan (\frac{7}{10})$

Now, note that by alternate angles on parallel lines, $\displaystyle x^{\circ} =48- \angle LHP = (48-\arctan (\frac{7}{10}))^{\circ}$

So $\displaystyle x= 13^{\circ}$, approximately.

Therefore the bearing to H from P is $\displaystyle 360 - 13 = 347^{\circ}$, or, if you prefer, $\displaystyle N \ 13^{\circ} W$
• Dec 17th 2007, 03:16 AM
Luckegrl_16
*t*h*a*n*k* *y*o*u*
thank u soooooooooooooo much (Handshake)