The Angle HCA is equal to 52°. This is arrived at using the knowledge of the Total value of Angles in a Triangle and the Total Value of Angles in a Polygon.
What other principles were used to arrive at the answer?
The other principles of mathematics that were used to arrive at the above answer are:
- Total Angles on a straight line;
- Total Angles on a point; and
- Line segments.
What are the Steps to the Solution?
Step 1 - Recall that we have been given Angles AHB and BAH to be 128° and 28° respectively.
We also know that:
- The sum of angles in a triangle is 180°;...................A
- The sum of angles on a straight line is 180°;.........B
- The sum of angles in a polygon is 360°; while.....C
- The total sum of angles at a point is 360°.............D
Since A...therefore:
When ∠AHB (128°) and ∠BAH are taken from 180° we have DBA = ∠28°.
By observation, we can deduce that ∠BDE, ∠CDH, ∠CEH and ∠AEH are all right-angled triangles.
Using the above, we are able to repeat this process of solving for each angle until we have ∠HCA.
To verify that our answer is correct, recall that sum of angles in a polygon is 360°
That means:
∠BDA + ∠DHE + ∠CEH + ∠HCA = 360°
That is, 90+ 128 + 90 + 52 = 360°
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