The Exterior Angle Theorem: Practice

The exterior turns total $360^\circ$.

$(4x+9)+(3x+21)+(5x+6)=360$

This gives $12x+36=360$, so $x=27$. The exterior turns are $117^\circ$, $102^\circ$, and $141^\circ$.

Taking their supplements gives interior measures $63^\circ$, $78^\circ$, and $39^\circ$. The largest interior angle is $78^\circ$, adjacent to the smallest exterior turn.

The adjacent interior and exterior angles form a linear pair.

$(5x+17)+(3x+11)=180$

Hence $8x+28=180$, so $x=19$. The exterior measure is $112^\circ$, and the given remote measure is $45^\circ$.

The exterior angle theorem makes the other remote measure $112^\circ-45^\circ=67^\circ$.

Use triangle $ADE$. Because $A,D,B$ lie on a line in that order, ray $DB$ continues side $AD$ beyond $D$. Thus $\angle BDE$ is an exterior angle of triangle $ADE$.

Its remote interior angles are $\angle DAE$ and $\angle AED$. The first is the given $47^\circ$ angle at $A$, since $AD$ lies along $AB$ and $AE$ along $AC$.

$m\angle BDE=47^\circ+61^\circ=108^\circ$

The $59^\circ$ measure belongs to the larger triangle and is not remote for this exterior angle.

Let the first remote measures be $a$ and $b$. Their sum is the first exterior measure.

The second remote measures are $a+8^\circ$ and $b-8^\circ$. Adding cancels the increase against the decrease, giving the second exterior measure

$e_2=a+b.$

The exterior angle theorem therefore gives equal exterior measures.

The correct record has three measures strictly between $0^\circ$ and $180^\circ$. Its total checks in two additions.

$98^\circ+141^\circ=239^\circ$

$239^\circ+121^\circ=360^\circ$

The first wrong total is $350^\circ$, and the last is $370^\circ$. The record with $180^\circ$ passes the sum check but fails the strict upper bound; its adjacent interior measure would be zero.

The exterior measure at $C$ is $180^\circ-58^\circ=122^\circ$. The other two exterior measures therefore total $360^\circ-122^\circ=238^\circ$.

If the exterior at $B$ has measure $u$, the one at $A$ has measure $u+24^\circ$. Thus

$2u+24^\circ=238^\circ.$

Subtracting $24^\circ$ and dividing by 2 gives $u=107^\circ$.

The requested exterior measure at $A$ is $107^\circ+24^\circ=131^\circ$.

The exterior angle theorem gives

$e=36^\circ+b,$

where the other remote measure $b$ is positive. Thus $e>36^\circ$.

There are different valid choices for $b$ with $36^\circ+b<180^\circ$, so the supplied information does not determine one exact exterior value.

The other remote measure is the exterior measure minus the given remote measure: $(3x+6)-(x+46)=2x-40$. Positivity requires

$2x-40>0,$

so $x>20$.

The adjacent interior measure is $180-(3x+6)=174-3x$ degrees. Positivity requires $x<58$.

For $20<x<58$, the supplied remote $x+46$ is also positive, and the exterior is between 0 and 180 degrees. The strict interval gives the necessary range.

To use the linear-pair rule, the angles must be adjacent with opposite noncommon rays. They already share ray $CA$.

The order $B,C,D$ makes $CB$ and $CD$ opposite rays and places the two angles on opposite sides of $CA$. A point merely outside the triangle does not establish that structure.

Let the other positive remote measure be $b$. The exterior angle theorem gives

$e=90^\circ+b.$

Thus the exterior measure is greater than $90^\circ$.

It is also less than $180^\circ$, since its adjacent interior measure is positive. Thus the exterior angle is obtuse. Its supplement is positive and less than $90^\circ$, so the adjacent interior angle is acute.

Substitution into the theorem gives

$(y+13)+(y+16)=2y+29$

Combining terms gives $2y+29=2y+29$, which is true for every $y$. Angle positivity still limits the allowed values, but it does not select just one. For instance, $y=1$ and $y=2$ each give three positive interior measures and a non-reflex exterior angle.

The adjacent interior measures at $A$ and $B$ are $180^\circ-124^\circ=56^\circ$ and $180^\circ-136^\circ=44^\circ$.

The triangle sum gives the third interior measure.

$180^\circ-56^\circ-44^\circ=80^\circ$

Since $80>56>44$, the angle at $C$ is largest, then the angle at $A$, then the angle at $B$.