1. In△ABC, we know that AB=5, BC=9, and CA=7.D is a point on BC such that AD
bisects ∠BAC. What is DC?
2. O is the circumcenter of △ABC. If OB = OC = BC, what is ∠BAC?
3. H is the orthocenter of △ABC. We know AB = AC = BC/sqrt3 What is ∠BHC?
4. E is a point on the side CD of square ABCD. The extensions of AD and BE intersect at F. [DEF] = 8 and [ECB] = 18. Find [ABCD].
5. CD and EF are parallel to AB. They divide the area of △PAB into three equal parts. If PE = 1, what is AC?
Here are the photos * there isn't a photo for #3; the number correspond to the question numbers.
1) Solving SSS triangle ABC [5, 7, 9] using The Law of Cosines, we get:
Angle ∠ C =33.557° = 33°33'26″ = 0.586 rad
Angle ∠ B =50.704° = 50°42'13″ = 0.885 rad
Angle ∠ A =95.739° = 95°44'21″ = 1.671 rad
Angle DAC = 95.739 / 2 = 47.8695 degrees
In triangle ACD, angle C = 33.557 degrees
Solving ASA triangle ACD, we get:
Sides: DC = 5.25 AD = 3.913 AC = 7