Exercise 31

The triangle \( \triangle ABC \) is equilateral and it has two of its heights drawn. Then the angle \( \omega \) measures:

  1. \( 30^{\circ} \)

  2. \( 45^{\circ} \)

  3. \( 60^{\circ} \)

  4. \( 90^{\circ} \)

  5. \( 120^{\circ} \)

Try to solve it before checking the answer.

  1. \( 120^{\circ} \)

Recall that the sum of the internal angles of a convex polygon of \( n \) sides is:

\[ S = (n – 2) 180^{\circ} \]

Therefore, the sum of the internal angles of a quadrilateral (\( n = 4)\) is:

\[ S = (4 – 2) 180^{\circ} = 360^{\circ} \]

Now, let’s consider the angles of the quadrilateral \( ADFE \). The angles \( \angle D\) and \( \angle E\) measure \( 90^{\circ} \), whereas the angle \(\angle A \) measures \( 60^{\circ} \) for being an angle of the equilateral triangle \( \triangle ABC\). As a result, we have that:

\[ \begin{aligned} &\omega + 90^{\circ} + 90^{\circ} + 60^{\circ} = 360^{\circ} \\[.5em] &\hspace{10em} \Rightarrow \boldsymbol{ \omega = 120^{\circ} } \end{aligned} \]
\[ \begin{aligned} \omega + 90^{\circ} + 90^{\circ} + 60^{\circ} = 360^{\circ} \Rightarrow \boldsymbol{ \omega = 120^{\circ} } \end{aligned} \]

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