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°
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.