Asked by Jnr John
Find angle A of a triangle whose vertices are A (3,3) , B(-3,1) and C (-1,-3)
Answers
Answered by
Damon
sketch it first
then find the lengths of the sides
for example
c^2 = 2^2 + 6^2 = 40
c = 2 sqrt 10
then use law of cosines
a^2 = b^2 + c^2 - 2 b c cos A
then find the lengths of the sides
for example
c^2 = 2^2 + 6^2 = 40
c = 2 sqrt 10
then use law of cosines
a^2 = b^2 + c^2 - 2 b c cos A
Answered by
Steve
Or, you can take the slopes of the sides.
AB has slope 1/3
AC has slope 3/2
So, measuring as usual, angle A is
arctan(3/2)-arctan(1/3) = 236.31°-198.43° = 37.88°
AB has slope 1/3
AC has slope 3/2
So, measuring as usual, angle A is
arctan(3/2)-arctan(1/3) = 236.31°-198.43° = 37.88°
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