Triangle ABC is formed by the points A(3, 4), B(2, 6), and C(5, 8). Find angle A to the nearest tenth of a degree?

Triangle ABC is formed by the points A(3, 4), B(2, 6), and C(5, 8). Find angle A to the nearest tenth of a degree.

1 Answer
Mr. Mike
Apr 11, 2018

The measure of angle #A~~53.1^@#

Explanation:

I will solve this using the Law of Cosines which states that for a triangle, with side lengths, #a#, #b#, and #c#, and angle #A# opposite the side with length #a#

#a^2=b^2+c^2-2*b*c*cos(A)#

Solving this for #cosA# we have

#A=cos^-1((b^2+c^2-a^2)/(2bc))#

Original Drawing

Let's first calculate #a#, #b#, and #c# using the distance formula.

#a=sqrt((8-6)^2+(5-2)^2)=sqrt(13)#

#b=sqrt((8-4)^2+(5-3)^2)=2sqrt(5)#

#c=sqrt((6-4)^2+(3-2)^2)=sqrt(5)#

Plugging these values into our formula for A gives

#A=cos^-1[((2sqrt(5))^2+(sqrt(5))^2-(sqrt(13))^2)/(2*2sqrt(5)sqrt(5))]#

#=cos^-1((20+5-13)/20)=cos^-1(3/5)~~53.1^@#

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