Exercises — Angle measurement — protractor use, angle relationships (complementary, supplementary)

They are complementary. Why: their sum is exactly a right angle, $90^{\circ }$. Memory pairing: Complement → Corner → the shorter word matches the smaller number ($90<180$).

Acute. Why: $65^{\circ }<90^{\circ }$. Anything strictly below $90^{\circ }$ is acute (see Types of Angles); exactly $90^{\circ }$ is right; between $90^{\circ }$ and $180^{\circ }$ is obtuse.

They are supplementary (their measures sum to $180^{\circ }$), and because they sit on a straight line sharing a ray they form a linear pair (Linear Pairs). Why both names? "Supplementary" is about the number ($180^{\circ }$). "Linear pair" is about the picture (two adjacent angles making a straight line). Every linear pair is supplementary, but two supplementary angles need not be side by side.

Complement means "what's left to reach $90^{\circ }$". $\text{complement}=90^{\circ }-37^{\circ }=53^{\circ }$ Check: $37^{\circ }+53^{\circ }=90^{\circ }$ ✓

Supplement means "what's left to reach $180^{\circ }$". $\text{supplement}=180^{\circ }-110^{\circ }=70^{\circ }$ Check: $110^{\circ }+70^{\circ }=180^{\circ }$ ✓ Notice $110^{\circ }$ has no complement — you can't subtract it from $90^{\circ }$ and get a real angle. (More on this in the L2 trap.)

Let the angle be $x$. Its supplement is $180^{\circ }-x$. "Equals its own supplement" means: $x=180^{\circ }-x$ Add $x$ to both sides: $2x=180^{\circ }⟹x=90^{\circ }$ What it means: the only angle that is its own supplement is the right angle $90^{\circ }$ — half of a straight line. Check: $90^{\circ }+90^{\circ }=180^{\circ }$ ✓

Let the angle be $x$; its complement is $90^{\circ }-x$. $x=90^{\circ }-x⟹2x=90^{\circ }⟹x=45^{\circ }$ Check: $45^{\circ }+45^{\circ }=90^{\circ }$ ✓ — the half-of-a-corner angle.

"Ratio $2:3$" means the angles are $2x$ and $3x$ for some unit $x$. Why $x$? It lets both angles grow together while keeping the $2:3$ shape. Supplementary → they add to $180^{\circ }$: $2x+3x=180^{\circ }⟹5x=180^{\circ }⟹x=36^{\circ }$ So the angles are $2(36^{\circ })=72^{\circ }$ and $3(36^{\circ })=108^{\circ }$. Check: $72+108=180$ ✓ and $72:108=2:3$ ✓

Let the angle be $x$; its complement is $90^{\circ }-x$. "The angle is $30^{\circ }$ more than its complement": $x=(90^{\circ }-x)+30^{\circ }$ Combine the constants ($90+30=120$): $x=120^{\circ }-x⟹2x=120^{\circ }⟹x=60^{\circ }$ Check: complement $=90^{\circ }-60^{\circ }=30^{\circ }$; is $60^{\circ }$ thirty more than $30^{\circ }$? Yes ✓

Let the angle be $x$. Its supplement is $180^{\circ }-x$. We're told: $180^{\circ }-x=4x$ Why: "supplement is $4$ times the angle" written straight into symbols. Add $x$ to both sides: $180^{\circ }=5x⟹x=36^{\circ }$ Supplement $=180^{\circ }-36^{\circ }=144^{\circ }$. Check: $144^{\circ }=4\times 36^{\circ }$ ✓ and $36+144=180$ ✓

Label the known angle $a=70^{\circ }$. Go around the crossing:

• The angle next to $a$ on the same line, call it $b$, forms a linear pair with $a$: $a+b=180^{\circ }⟹b=180^{\circ }-70^{\circ }=110^{\circ }(\text{[[Linear Pairs]]})$
• The angle opposite $a$, call it $c$, is a vertical angle: $c=a=70^{\circ }(\text{[[Vertical Angles]]})$
• The angle opposite $b$, call it $d$, is vertical to $b$: $d=b=110^{\circ }$ Check: all four around the point: $70+110+70+110=360^{\circ }$ ✓ (a full turn).

Step 1 — use the complement to get $\angle B$: $\angle B=90^{\circ }-\angle A=90^{\circ }-25^{\circ }=65^{\circ }$ Step 2 — use the supplement to get $\angle C$: $\angle C=180^{\circ }-\angle B=180^{\circ }-65^{\circ }=115^{\circ }$ Why chain them? $\angle B$ is the shared link: it sits in both relationships, so we compute it first, then step outward to $\angle C$. Check: $\angle A+\angle B=25+65=90$ ✓; $\angle B+\angle C=65+115=180$ ✓

Three parts of a straight line must total $180^{\circ }$: $x+2x+(x+40^{\circ })=180^{\circ }$ Why $180^{\circ }$? They tile a straight angle end to end — the whole is a half turn. Combine: $4x+40^{\circ }=180^{\circ }⟹4x=140^{\circ }⟹x=35^{\circ }$. Parts: $x=35^{\circ }$, $2x=70^{\circ }$, $x+40^{\circ }=75^{\circ }$. Check: $35+70+75=180^{\circ }$ ✓

Let the angle be $x$.

• complement $=90^{\circ }-x$
• supplement $=180^{\circ }-x$

Condition: complement is one-third of the supplement: $90^{\circ }-x=\frac{1}{3}(180^{\circ }-x)$ Multiply both sides by $3$ to clear the fraction: $3(90^{\circ }-x)=180^{\circ }-x$ $270^{\circ }-3x=180^{\circ }-x$ Gather $x$ on one side, numbers on the other. Add $3x$ to both sides and subtract $180^{\circ }$: $270^{\circ }-180^{\circ }=3x-x⟹90^{\circ }=2x⟹x=45^{\circ }$ Check: complement $=45^{\circ }$; supplement $=135^{\circ }$; is $45=\frac{1}{3}(135)$? $\frac{135}{3}=45$ ✓

Let the smaller be $s$. "Larger exceeds twice the smaller by $30^{\circ }$" means larger $=2s+30^{\circ }$. They are supplementary: $s+(2s+30^{\circ })=180^{\circ }$ $3s+30^{\circ }=180^{\circ }⟹3s=150^{\circ }⟹s=50^{\circ }$ Larger $=2(50^{\circ })+30^{\circ }=130^{\circ }$. Check: $50+130=180^{\circ }$ ✓ and $130=2(50)+30$ ✓ and indeed $130>50$ (the "larger" label holds).

Angles filling a full turn around a point sum to $360^{\circ }$: $y+2y+3y+90^{\circ }=360^{\circ }$ Why $360^{\circ }$? They wrap completely around the point — one full rotation. $6y+90^{\circ }=360^{\circ }⟹6y=270^{\circ }⟹y=45^{\circ }$ So the angles are $45^{\circ },90^{\circ },135^{\circ },90^{\circ }$. Which pair of variable angles is supplementary? $y+3y=45^{\circ }+135^{\circ }=180^{\circ }$ ✓ — so $\angle y$ and $\angle 3y$ are supplementary. Check: $45+90+135+90=360^{\circ }$ ✓

Let the angle be $x$ (and note we'll need $x<90^{\circ }$ for a complement to exist).

• complement $=90^{\circ }-x$
• supplement $=180^{\circ }-x$

"Supplement is $10^{\circ }$ less than three times the complement": $180^{\circ }-x=3(90^{\circ }-x)-10^{\circ }$ Expand the right side: $180^{\circ }-x=270^{\circ }-3x-10^{\circ }=260^{\circ }-3x$ Add $3x$ to both sides, subtract $180^{\circ }$: $2x=80^{\circ }⟹x=40^{\circ }$ Check: complement $=50^{\circ }$, supplement $=140^{\circ }$; is $140=3(50)-10=150-10=140$? ✓ And $40^{\circ }<90^{\circ }$, so the complement is legal.

Complement of $x$ ::: $(90-x{)}^{\circ }$, and it exists only when $x<90^{\circ }$. Supplement of $x$ ::: $(180-x{)}^{\circ }$, existing whenever $x<180^{\circ }$. Angle equal to its own supplement ::: $90^{\circ }$. Angle equal to its own complement ::: $45^{\circ }$. Angles around a point sum to ::: $360^{\circ }$. "A is $k$ more than B" translates to ::: $A=B+k$.