Some Right Triangles

Geometry Level 2

Triangle $ABC$ has $\angle ABC = 90 ^\circ$ and $\angle CAB = 30^\circ$ . Triangle $DAC$ has $\angle DAC = 90^\circ$ and $\angle CDA = 30^\circ$ . Similarly, triangle $EDC$ has $\angle EDC = 90^\circ$ and $\angle CED = 30^\circ$ . If $EC = 136$ , what is the value of $BC$ ?

Details and assumptions

For simplicity, you may assume that the triangles do not overlap. However, this assumption isn't necessary.

The answer is 17.

1 solution

Arron Kau Staff
May 13, 2014

Solution 1: Triangles $ABC$ , $DAC$ and $EDC$ are $30^\circ - 60 ^\circ - 90 ^\circ$ triangles, thus $CD:DE:EC = CA:AD:DC = CB:BA:AC = 1:\sqrt{3}:2$ . So, we have $EC = 2 DC = 2 (2 AC) = 4( 2 BC) = 8 BC$ $\Rightarrow BC = \frac{EC}{8} = \frac{136}{8} = 17$ .

Solution 2: Since triangles $ABC$ , $DAC$ and $EDC$ are right triangles, we have $DC = EC \sin 30^\circ = \frac{EC}{2}$ , $AC = DC \sin 30^\circ = \frac{DC}{2}$ and $BC = AC \sin 30^\circ = \frac{AC}{2}$ . Putting these together we have $BC = \frac{EC}{8} = \frac{136}{8} = 17$ .

Please give me the figure of this problem

Aman Real - 6 years, 2 months ago

Some Right Triangles

The answer is 17.

1 solution

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