How do I find the angle of AHC?

enter image source here

1 Answer
Apr 16, 2018

angle H=74.4 degrees

Explanation:

Not sure if I'm approaching this correctly, but it looks to me like you could make this into some right triangles: DeltaHDA, DeltaHGC, and DeltaABC.
In triangle DeltaHDA, you are looking for HA given HD=3, DA=4, and angleD is right, so using the Pythagorean Theorem: (HD)^2+(DA)^2=(HA)^2
(HA)^2=25
(HA)=5
Using the same process for the other triangles I mentioned (if you need me to expand it feel free to ask), HC=6.708... and AC=7.211....

Therefore, you can use the Law of Cosines to find angleH:
theta=cos^-1((a^2+b^2﹣c^2)/(2ab)) Substituting Values:
angle H =cos^-1(((HA)^2+(HC)^2﹣(AC)^2)/(2(HA)(HC))) Which simplifies to
angle H=~74.44 (in degrees), rounded to the nearest tenth is
angle H=74.4