# Trigonometry

• Dec 6th 2008, 06:56 PM
helloying
Trigonometry
B is a right angle.DF is the dotted vertical line from D
• Dec 6th 2008, 07:05 PM
Mentia
Do you have any other angles besides C? It seems to me that the distance from D to A could be anything if angle B is not fixed.
• Dec 6th 2008, 07:09 PM
helloying
i forgot to indicate. B is a right angle.
• Dec 6th 2008, 07:18 PM
Skalkaz
"C is 320°"

in the drawing C seems less than 180°
• Dec 6th 2008, 07:19 PM
skeeter
what are you trying to find? your sketch says "DF"
• Dec 6th 2008, 07:21 PM
Skalkaz
Quote:

Originally Posted by helloying
i forgot to indicate. B is a right angle.

You don't forget.
• Dec 6th 2008, 07:26 PM
helloying
C is not 320 degree. sorry if my scan is not clear. What i wrote is The bearing of D from C is 320 degree.The sentence continue to C...

And I am suppose to find the dist DF

:)
• Dec 6th 2008, 07:34 PM
skeeter
Quote:

Originally Posted by helloying
C is not 320 degree. sorry if my scan is not clear. What i wrote is The bearing of D from C is 320 degree.The sentence continue to C...

And I am suppose to find the dist DF

:)

ok ... maybe my eyesight is worse than i thought

where is F?
• Dec 6th 2008, 07:38 PM
helloying
Omg! I didn't indicate. I am so blur. It is the dotted vertical line.
• Dec 6th 2008, 07:43 PM
skeeter
http://www.mathhelpforum.com/math-he...try-img011.jpg

if point F is at the bottom of the dashed line segment, then ...

$\displaystyle DF = 150 + 200\cos(40^{\circ})$
• Dec 6th 2008, 07:46 PM
Mentia
With B a right angle its do-able. Consider the little triangle made if you draw a horizontal line across the rhombus on the right from the intersection at angle C to the dotted line. You have have a right triangle and since you know C you know all if its angles, and you know its hypotenuse is 200, so you can solve for all its sides. Then you know the base length of the rhombus and the height. Then you know the base length of the big triangle on the left and its height, and use pythagorean theorem to solve for the hypotenuse.