Can You Solve This Trigonometry Problem Involving Equal Sides and Angles?

[Johnny Leong]

In the figure, EA=EB=EC and AB = 200 m. Angle BCA = 30 degrees. Find angle AEB. Anyone can help me?

 

[Diane_]

Start it as a pure geometry problem - sketch the figure, and note that you have three isosceles triangles there. Given that the base angles of an isosceles triangle are congruent, you can find three pairs of congruent angles. Working with these angles (some of which overlap), it turns out to be relatively easy to figure out the measure of angle EBC. From there, the rest of the angles in the bottom triangle can be known. After that, the problem actually gets very easy.

Hope this helps.

 

[Johnny Leong]

I have tried but I am still in confusion. Would Diane give me more hints please?

 

[Diane_]

Ok

Your three isosceles triangles are AEB, BEC, and AEC. Call the base angles for AEB x, the base angles for BEC y, and the base angles for AEC z. We know that the measure of ACB is 30, so from the diagram we can see that 30 + z = y. Again, from the diagram, the measure of CAB is x - z, and the sum of the interior angles of triangle ABC is, of course, 180. This gives us that x-z + x + y + 30 = 180, or 2x + y - z = 150.

Substitute the first relation into the second and you end up with a single equation in one unknown for x. This will tell you something about triangle EAB.

Try it from there. :)

 

[Johnny Leong]

Thank you very much for your help, Diane! I have solved the problem.