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# halp

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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?

Mar 31, 2023

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Let $F$ be the foot of the altitude from $H$ to $BC$. Since $H$ lies on altitude $AD$, we have $\angle AHB = 90^\circ$, which means $\angle BHF = 90^\circ - \angle AHB/2 = 56^\circ$. Also, since $H$ lies on altitude $BE$, we have $\angle AHB = 90^\circ$, which means $\angle AHB = 180^\circ - \angle BAC$. Thus, $\angle BAC = 180^\circ - 2\angle BAH = 124^\circ$.

Now, we use the fact that the sum of the angles in a triangle is $180^\circ$ to find $\angle HCA$. We have:

\begin{align*}
\angle HCA &= 180^\circ - \angle ACH - \angle AHC \
&= 180^\circ - \angle ACH - (180^\circ - \angle BHA) \
&= \angle BHA - \angle ACH \
&= (180^\circ - \angle BAC) - \angle ACH \
&= 56^\circ - \angle ACH.
\end{align*}

Thus, we need to find $\angle ACH$. Since $AF$ is the altitude of triangle $ABH$, we have $\angle ABF = 90^\circ$. Since $\angle BAH = 28^\circ$, we have $\angle BAF = \angle BAH + \angle HAF = 28^\circ + 90^\circ - \angle AHB/2 = 46^\circ$.

Using the fact that the sum of the angles in a triangle is $180^\circ$ in triangle $ABC$, we have:

\begin{align*}
\angle ACH &= 180^\circ - \angle A - \angle CHA \
&= 180^\circ - \angle A - (180^\circ - \angle BHF - \angle BHA) \
&= \angle BHA - \angle A - \angle BHF \
&= (180^\circ - \angle BAC) - \angle A - \angle BHF \
&= 56^\circ - \angle A - \angle BHF.
\end{align*}

Thus, we need to find $\angle A$ and $\angle BHF$. Since $AF$ is the altitude of triangle $ABH$, we have $\angle AFB = 90^\circ$, so $\angle BAF + \angle ABF = 46^\circ + 90^\circ = 136^\circ$. Since $\angle BAC = 124^\circ$, we have $\angle BCA = \angle BAC/2 = 62^\circ$. Thus, $\angle A = 180^\circ - \angle BAC - \angle BCA = 180^\circ - 124^\circ - 62^\circ = 14^\circ$.

To find $\angle BHF$, note that $\angle BHF + \angle BHA + \angle AHF = 180^\circ$. Since $\angle BHA = 128^\circ$ and $\angle AHF = 90^\circ - \angle BAF = 44^\circ$, we have $\angle BHF = 180^\circ - \angle BHA - \angle AHF = 8^\circ$.

Therefore, $\angle ACH = 56^\circ - \angle A - \angle BHF = 56^\circ - 14^\circ - 8^\circ = 34^\circ$.

Mar 31, 2023