To approximate the length of a marsh, a surveyor walks x = 400 meters from point A to point B, then turns 75° and walks 220 meters to point C (see figure). Approximate the length AC of the marsh. (Round your answer to one decimal place.)
(http://imageshack.us/photo/my-images/836/73437200.gif/) is the picture
(http://imageshack.us/photo/my-images/836/73437200.gif/) is the picture
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Hello
You must solve this by the theorem of Al Kashi ( it can have been given another name, never mind)
It is like Pythagoras, but for non right triangles.
It says AC² = AB² + BC² - 2(AB)(BC) Cos B
Here AC² = 400² + 220² -2(400)(220) Cos 105°
AC² = 160000 + 48400 - 176000 (-0.259)
AC² = 208400 + 45552 = 253952
AC = 503.9 m
Here is the sketch
http://img703.imageshack.us/img703/1252/…
Hope it helps
Bye !
You must solve this by the theorem of Al Kashi ( it can have been given another name, never mind)
It is like Pythagoras, but for non right triangles.
It says AC² = AB² + BC² - 2(AB)(BC) Cos B
Here AC² = 400² + 220² -2(400)(220) Cos 105°
AC² = 160000 + 48400 - 176000 (-0.259)
AC² = 208400 + 45552 = 253952
AC = 503.9 m
Here is the sketch
http://img703.imageshack.us/img703/1252/…
Hope it helps
Bye !
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x^2= a^2 + b^2 - 2abcos (q)
501meters
501meters