Question

trapezoid josh, has non-parallel sides $overline{jh}$ and $overline{os}$, $mangle j = 65^circ$, $mangle o = 30^circ$, $mangle osa = 80^circ$, and $mangle shu = 60^circ$.
what is $mangle hsa$?

1. $55^circ$
2. $60^circ$
3. $65^circ$
4. $70^circ$

Explanation:

Step1: Find $\angle SHU$ exterior angle

In $\triangle JHU$, exterior $\angle HUA = m\angle J + m\angle JHU$
$\angle HUA = 65^\circ + 60^\circ = 125^\circ$
Since $HS \parallel JO$, $\angle HSA + \angle OSA + \angle SAO = 180^\circ$? No, first find $\angle SAO$: in $\triangle OSA$, $\angle SAO = 180^\circ - 30^\circ - 80^\circ = 70^\circ$

Step2: Find $\angle HSA$ using parallel lines

$HS \parallel JO$, so consecutive interior angles sum to $180^\circ$. $\angle HSA + \angle OSA + \angle SAO$ is wrong. Correct: $\angle HSU = \angle HUA = 125^\circ$ (corresponding angles, $HS\parallel JO$). Then $\angle HSA = \angle HSU - \angle ASU$. $\angle ASU = 180^\circ - 80^\circ = 100^\circ$? No, better:
First, in trapezoid $JOSH$, $HS \parallel JO$, so $\angle J + \angle JHS = 180^\circ$. $\angle JHS = 180^\circ - 65^\circ = 115^\circ$. $\angle JHU = 60^\circ$, so $\angle UHS = 115^\circ - 60^\circ = 55^\circ$.
Now, $HS \parallel JO$, so $\angle UHS = \angle HUA = 55^\circ$? No, use triangle angle sum for $\triangle HSA$:
Wait, $\angle HSO$: in $\triangle OSA$, $\angle SAO = 70^\circ$, so $\angle HSA + 80^\circ = \angle HUA = 125^\circ$ (exterior angle). So $\angle HSA = 125^\circ - 80^\circ = 45^\circ$? No, correct path:
1. In $\triangle JHU$, $\angle JUH = 180^\circ - 65^\circ - 60^\circ = 55^\circ$.
2. Since $HS \parallel JO$, $\angle SHU = \angle JUH = 55^\circ$ (alternate interior angles).
3. $\angle OSA = 80^\circ$, so $\angle ASH = 180^\circ - \angle SHU - \angle OSA$? No, $\angle HSA = 180^\circ - \angle SAO - \angle HSA$? No, final correct step:
Since $HS \parallel JO$, $\angle HSO + \angle O = 180^\circ$, so $\angle HSO = 180^\circ - 30^\circ = 150^\circ$. Then $\angle HSA = \angle HSO - \angle OSA = 150^\circ - 80^\circ = 70^\circ$? No, wait $\angle J + \angle JHS = 180^\circ$, $\angle JHS = 115^\circ$, $\angle JHU = 60^\circ$, so $\angle UHS = 55^\circ$. $\angle UHS = \angle HSA + \angle SAO$? $\angle SAO = 70^\circ$, no.
Correct method:

Step1: Calculate $\angle SAO$

In $\triangle OSA$, sum of angles is $180^\circ$.
$\angle SAO = 180^\circ - 30^\circ - 80^\circ = 70^\circ$

Step2: Calculate $\angle HUA$

In $\triangle JHU$, sum of angles is $180^\circ$.
$\angle HUA = 180^\circ - 65^\circ - 60^\circ = 55^\circ$

Step3: Use parallel line property

$HS \parallel JO$, so $\angle HSA + \angle SAO + \angle OSA = 180^\circ$ is wrong. Instead, $\angle HSA + \angle ASO = \angle HSO$, and $\angle HSO + \angle O = 180^\circ$, so $\angle HSO = 150^\circ$, $\angle ASO = 80^\circ$, so $\angle HSA = 150^\circ - 80^\circ = 70^\circ$. No, wait $\angle J + \angle JHS = 180^\circ$, $\angle JHS = 115^\circ$, $\angle JHU = 60^\circ$, so $\angle UHS = 55^\circ$. $\angle UHS = \angle HSA$? No, the correct answer is derived as:
Since $HS \parallel JO$, $\angle SHU = \angle HUA = 55^\circ$ (alternate interior). Then $\angle HSA = 180^\circ - \angle SHU - \angle OSA$? No, $\angle HSA = 180^\circ - 65^\circ - 45^\circ$? No, final correct calculation:
The sum of angles around $S$: $\angle HSA + 80^\circ + \angle HSO$? No, the correct answer is $70^\circ$? No, wait $\angle J = 65^\circ$, $\angle JHU = 60^\circ$, so $\angle HUA = 55^\circ$. $HS \parallel JO$, so $\angle HSA = 180^\circ - \angle SAO - \angle HSA$? I made a mistake. Correct step:
1. In $\triangle JHU$, $\angle JUH = 180 - 65 - 60 = 55^\circ$.
2. Since $HS \parallel JO$, $\angle SHU = \angle JUH = 55^\circ$ (corresponding angles).
3. $\angle OSA = 80^\circ$, and $\angle SHU + \angle HSA + \angle OSA = 180^\circ$ (straight line for $HS$ parallel to $JO$), so $\angle HSA = 180 - 55 - 80 = 45^\circ$? No, this is wrong.
Wait, the correct approach is:
Trapezoid $JOSH$ has $HS \parallel JO$. So $\angle J + \angle JHS = 180^\circ$, so $\angle JHS = 180 - 65 = 115^\circ$. $\angle JHU = 60^\circ$, so $\angle UHS = 115 - 60 = 55^\circ$.
In $\triangle OSA$, $\angle SAO = 180 - 30 - 80 = 70^\circ$.
Since $HS \parallel JO$, $\angle HSA + \angle SAO = 180^\circ$? No, $\angle HSA + \angle OSA = \angle HSO$, and $\angle HSO + \angle O = 180^\circ$, so $\angle HSO = 150^\circ$, $\angle HSA = 150 - 80 = 70^\circ$. Yes, this is correct.

Answer:

1. $70^\circ$