Eight angles at a pair of parallel lines, and only two different sizes among them.
What is the value of the angle?
In the xy-plane, lines m and n are parallel and are cut by a transversal.
One interior angle at line m measures (3x + 20)°, and the
interior angle on the same side of the transversal at line n measures
(7x)°. What is the measure, in degrees, of the angle
that measures (7x)°?
Two interior angles on the same side of the transversal are co-interior, so they add to 180°: (3x + 20) + 7x = 180 gives 10x = 160 and x = 16. The angle asked for is 7(16) = 112°, and the other is 3(16) + 20 = 68°, which checks: 68 + 112 = 180. A stops at x — the single most common slip on this item type, because x is never the answer unless the question says so. B sets the two expressions equal (3x + 20 = 7x → x = 5, angle 35°), which is the rule for corresponding and alternate angles, not same-side ones. C is the other angle of the pair, 68°.
1
Two numbers carry this whole topic: 90 and 180.
Every angle question here reduces to “equal” or “adds to 180”.
complementary
add to 90°
C comes first — 90 is smaller
supplementary
add to 180°
S comes second — 180 is bigger
C before S in the alphabet, 90 before 180 on the number line. That is the whole
memory trick. The two angles do not have to touch — the definition is only about
the sum, so angles in different corners of a figure can still be complementary.
The SAT often hides the word. “Angle ABD and angle DBC are complementary” may
instead read “ray BD lies in right angle ABC”, and “supplementary” usually
arrives as “points A, B and C lie on a line”. Same two numbers either way.
2
Two crossing lines make four angles and only two sizes.
Across the crossing they are equal; side by side they add to 180°.
vertical angles
opposite through the crossing — equal
linear pair
side by side on one line — 180°
These two rules never need parallel lines. Any two lines that cross, anywhere in a
figure, hand you both of them free.
Equal or supplementary — read the position
Angles that sit next to each other are supplementary, not equal. Only the pair
directly opposite through the crossing point are equal. If the stem gives you
(4x − 6)° and (2x + 30)° as vertical angles you set them equal; as a linear pair
you set the sum to 180. Picking the wrong one is worth a whole question.
3
A transversal across two parallel lines copies that crossing.
Eight angles now, but the bottom crossing is an exact copy of the top one — so still only two sizes.
Say ∠2 is 68°. Then ∠1, ∠4, ∠5 and ∠8 are 112°, and ∠2, ∠3, ∠6 and ∠7 are 68°.
Nothing else is possible.
The arrowheads on m and n are the standard mark for “these are parallel”. Without
them — or without a sentence saying so — none of the rules on the next slide
apply, and the SAT will not give you the marks by accident.
4
Three of the four pair names mean equal; one means 180°.
Pair
Where they sit
Example
Rule
corresponding
same corner at both crossings
∠1 and ∠5
equal
alternate interior
between the lines, opposite sides
∠3 and ∠6
equal
alternate exterior
outside the lines, opposite sides
∠1 and ∠8
equal
co-interior (same-side)
between the lines, same side
∠4 and ∠6
add to 180°
The shortcut that beats the names: if the two angles look alike — both acute or
both obtuse — they are equal. If one is acute and one is obtuse, they add to 180°.
Check it on the numbers from the last slide. ∠3 and ∠6 are both 68°, both acute,
and they are equal. ∠4 is 112° and ∠6 is 68° — one of each — and 112 + 68 = 180.
The shortcut is not a guess; it is what “only two sizes” forces.
5
The SAT gives one angle and asks for one several steps away.
In the figure, m is parallel to n. The marked angle is 68°. What is the value of x?
68° is ∠2; ∠3 = 68°
vertical angles at the top crossing
∠5 = 180 − 68 = 112°
∠3 and ∠5 are co-interior — same side, between the lines
x = 112
vertical to ∠5 at the bottom crossing
x = 112
The one-line version: 68° is acute and x° is drawn obtuse, and both lie outside
the parallel lines on the same side of the transversal, so they are supplementary
— x = 180 − 68 = 112 in a single step. Write the chain out until the shortcut
feels safe; the chain is what proves it.
A calculator is allowed on every SAT Math question, but nothing here needs one.
Desmos will not redraw the figure for you — the work is reading position, then
subtracting from 180.
6
When the angles arrive as expressions, name the relationship first.
Back to the opening problem: (3x + 20)° at line m and (7x)° at line n, same side of the transversal, both between the lines.
same side, interior → co-interior
so the two add to 180, not equal
(3x + 20) + 7x = 180
write the relationship as an equation
10x + 20 = 180
collect
x = 16
− 20, then ÷ 10
7x = 7(16) = 112
the angle the question asked for
112°
x is almost never the answer
Here x = 16, and 16 sits in the choices waiting for anyone who stops one line
early. Underline what the stem asks for — x, one angle, the larger angle, the
sum — before you start, and substitute back at the end. The check is fast:
3(16) + 20 = 68 and 7(16) = 112, and 68 + 112 = 180 ✓.
✓
Check: do the sums close?
the pair68 + 112 = 180co-interior ✓
round a point68 + 112 + 68 + 112 = 360one full turn at each crossing ✓
sensible?112° is the obtuse oneand it is drawn obtuse ✓
Remember the flow
Name the pair · Equal or 180 · Answer what was asked
Same look, equal. Different look, supplementary.
Now you try
Practice problem
What is the value of y?
Parallel lines p and q are cut by a transversal. Angle 1 and angle 2 are alternate
exterior angles, and angle 3 forms a linear pair with angle 2. If the measure of
angle 1 is (2y + 6)° and the measure of angle 3 is
(4y − 12)°, what is the value of y?
Alternate exterior angles are equal, so ∠2 = ∠1 = (2y + 6)°. A linear pair adds to 180°, so (2y + 6) + (4y − 12) = 180, giving 6y − 6 = 180 and y = 31. Substituting back, ∠1 = ∠2 = 68° and ∠3 = 112°, and 68 + 112 = 180 ✓. A sets the two expressions equal (2y + 6 = 4y − 12 → y = 9), treating ∠1 and ∠3 as the equal pair when one of them is the supplement. B uses 90 instead of 180 — complementary when the pair is supplementary. C drops the sign on −12, solving 6y + 6 = 180.
Hint: two deductions, not one. Carry angle 1 across to angle 2 first — that is the
step with the parallel lines in it — and only then use the linear pair. And check
which letter the question wants at the end.
Every lesson in the SAT Math Review has one of these. 10 questions, marked as you answer, with the working shown for each.
Question 1 of 10
In the figure, line m is parallel to line n. What is the value of x?
The 142° angle and the x° angle are corresponding angles (both above and to the left of the transversal at their lines), so x = 142.
Question 2 of 10
In triangle RST, the measure of ∠R is 37° and the measure of ∠S is 58°. What is the measure of ∠T?
The angles of a triangle add to 180°: ∠T = 180 − 37 − 58 = 85°.
Question 3 of 10
Triangle ABC is similar to triangle DEF, where A, B, and C correspond to D, E, and F, respectively. AB = 4, BC = 6, and DE = 10. What is the length of EF?
The scale factor is DE/AB = 10/4 = 2.5, so EF = 2.5 × BC = 2.5 × 6 = 15.
Question 4 of 10
In the figure, lines k and l intersect. What is the measure of each acute angle formed by the two lines?
The marked angles are vertical angles, so 4x + 15 = 6x − 25, giving x = 20 and each marked angle is 4(20) + 15 = 95°. The acute angles are supplementary to these: 180 − 95 = 85°.
Question 5 of 10
In the figure, points A, C, and D lie on a line. The measure of ∠BAC is 61° and the measure of ∠BCD is 133°. What is the value of x?
∠BCD is an exterior angle of triangle ABC, so it equals the sum of the two remote interior angles: 133 = 61 + x, giving x = 72. (Equivalently ∠BCA = 47°, so x = 180 − 61 − 47 = 72.)
Question 6 of 10
In the figure, PQ = PR and the measure of ∠P is 40°. What is the value of x?
The base angles of an isosceles triangle are equal, so each is (180 − 40)/2 = 70°. Then 2x + 14 = 70, so x = 28.
Question 7 of 10
In triangles JKL and MNP, angles J and M each have measure 35° and angles K and N each have measure 80°. Which additional piece of information is sufficient to prove that triangle JKL is congruent to triangle MNP?
Two pairs of equal angles already make the triangles similar; congruence needs one pair of corresponding sides equal. JK and MN are corresponding sides (between the 35° and 80° angles), so JK = MN gives ASA. ∠L and ∠P are already determined (65°), and JK and NP are not corresponding sides.
Question 8 of 10
In the figure, point D lies on side AB, point E lies on side AC, and DE is parallel to BC. AD = 4, DB = 6, and the area of triangle ADE is 12 square units. What is the area, in square units, of triangle ABC?
Triangles ADE and ABC are similar with scale factor AB/AD = 10/4 = 5/2. Areas scale by the square of the scale factor: 12 × (5/2)² = 12 × 25/4 = 75. Multiplying by 5/2 only once gives 30.
Question 9 of 10
In the figure, points D, E, and A lie on a line, point C lies on segment DB, and AC = CD. The measure of ∠ACD is 112° and the measure of ∠DBE is 38°. What is the value of x?
AC = CD makes triangle ACD isosceles, so ∠CDA = ∠CAD = (180 − 112)/2 = 34°. In triangle BDE, ∠BED = 180 − 34 − 38 = 108°, and x° is its supplement along line DA: x = 72.
Question 10 of 10
In the figure, triangle ABC has a right angle at C, and CD is the altitude to hypotenuse AB. AC = 9 and BC = 12. What is the length of CD?
AB = √(9² + 12²) = 15. The area of the triangle is (1/2)(9)(12) = 54, and also (1/2)(AB)(CD) = (1/2)(15)(CD), so CD = 108/15 = 7.2. (5.4 and 9.6 are AD and BD.)
End of the free lesson
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