Question

Altitudes $\overline{ad}$ and $\overline{be}$ of acute triangle $abc$ intersect at point $h$. if $\angle ahb = 128^\circ$ and $\angle bah = 28^\circ$, then what is $\angle hca$ in degrees?

Answer 1

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:

1. The sum of angles in a triangle is 180°;...................A
2. The sum of angles on a straight line is 180°;.........B
3. The sum of angles in a polygon is 360°; while.....C
4. 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°

See attached images and;

Learn more about Angles in a Triangle at:
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