Angle Pairs: Practice

The first two sectors form a linear pair, so $3x-4+2x+9=180$. This gives $5x+5=180$, hence $x=35$.

The first and third sectors are vertical, so their measures are equal. Substituting gives $5y+6=101$, hence $y=19$.

Therefore $x+y=35+19=54$.

$5x+5=180$

The vertical pair is congruent. Because its equal measures also total $180^\circ$, each is $90^\circ$. Each remaining sector forms a linear pair with one of those angles, so it too measures $180^\circ-90^\circ=90^\circ$.

$180^\circ\div2=90^\circ$

Complements of congruent angles are congruent, so $8t-11=5t+19$. This gives $3t=30$, hence $t=10$.

Both $E$ and $F$ measure $69^\circ$. Each angle being complemented therefore measures $90^\circ-69^\circ=21^\circ$.

$90^\circ-69^\circ=21^\circ$

The three sectors must sum to $360^\circ$. Since the first two total $253^\circ$, the third must measure $107^\circ$. All three are positive and less than $180^\circ$, so placing the rays successively at these turns gives a possible configuration.

$360^\circ-253^\circ=107^\circ$

Because $OA$ and $OC$ are opposite, $\angle AOD$ and $\angle DOC$ form a linear pair. Thus $m\angle AOD=180^\circ-151^\circ=29^\circ$.

The bisector makes $\angle AOD$ half of $\angle AOB$, so $m\angle AOB=2(29^\circ)=58^\circ$.

$2(180^\circ-151^\circ)=58^\circ$

Let the smaller measure in degrees be $s$. The larger is $s+44$, and their sum gives $2s+44=180$. Subtracting 44 and dividing by 2 gives $s=68$.

The larger measure is $(68+44)^\circ=112^\circ$. Their sum is $180^\circ$ and their difference is $44^\circ$.

$2s+44=180$

The entire interior of $\angle AOB$ lies inside $\angle AOC$. Although they share vertex $O$ and ray $OA$, their interiors overlap. This violates the definition of adjacency.

Angle addition makes the split sector's measure $(2x+24)^\circ$. It equals its vertical opposite, so $2x+24=5x-15$. This gives $3x=39$, hence $x=13$.

The whole sector measures $50^\circ$. Each neighboring sector forms a linear pair with it, so each measures $180^\circ-50^\circ=130^\circ$.

$2x+24=5x-15$

The complements of $C$ are congruent, so $2t+5=3t-14$, giving $t=19$. Each complement measures $43^\circ$, so $m\angle C=90^\circ-43^\circ=47^\circ$.

Since $D$ supplements $C$, $m\angle D=180^\circ-47^\circ=133^\circ$.

$180^\circ-47^\circ=133^\circ$

The complete-turn equation is $(b+40)+b+(2b-12)=360$. Combining terms gives $4b+28=360$, so $b=83$.

The measures are $123^\circ$, $83^\circ$, and $154^\circ$. The largest is $154^\circ$, and their sum is $360^\circ$.

$4b+28=360$

Opposite ray pairs combine into two distinct lines through $O$. The two angles occupy opposite sectors of that crossing, which is exactly the vertical-angle configuration. Equal or supplementary measures alone do not fix the rays' positions.

The supplement is $(180-u)^\circ$ and the complement is $(90-u)^\circ$. The proposed condition gives $180-u=2(90-u)$.

Expanding and adding $2u$ to both sides gives $180+u=180$, hence $u=0$. This is outside the required positive acute range, so no allowed angle works.

$180-u=2(90-u)$