Question

Two altitudes of an isosceles triangle are equal to 20 cm and 30 cm. Determine the possible measures of the base angles of the triangle

Answer 1

Answer on Question #57437 – Math – Geometry

Question

Two altitudes of an isosceles triangle are equal to $20\,\mathrm{cm}$ and $30\,\mathrm{cm}$. Determine the possible measures of the base angles of the triangle

Solution

Let we have isosceles triangle ABC, where AB=AC and two altitudes AK and BM.

Here $\tan \alpha = \frac{AK}{KC}$. So $KC = \frac{AK}{\tan \alpha}$.

Also $BC * \sin \alpha = BM$.

$BC = BK + KC = 2 * KC$, because $BK = KC$.

So $2 * KC * \sin \alpha = BM$, $2 * \frac{AK}{\tan \alpha} * \sin \alpha = BM$.

Since $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$, we have $\frac{2 * AK * \sin \alpha}{\frac{\sin \alpha}{\cos \alpha}} = BM$, hence $2 * AK * \cos \alpha = BM$.

So $\cos \alpha = \frac{BM}{2 * AK}$ and angle $\alpha = \cos^{-1} \frac{BM}{2 * AK}$.

If $AK = 20$, $BM = 30$ then answer will be $\cos^{-1} \frac{3}{4} = 0.72273424$ in radians.

If $AK = 30$, $BM = 20$ then answer will be $\cos^{-1} \frac{1}{3} = 1.23095942$ in radians.

**Answer**: base angle is equal to 0.72273424 rad or 1.23095942 rad.

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