Three angles that always total 180°, and three lengths that only sometimes close up.
What is the value of the angle?
In triangle ABC, AB = AC. Point D lies on line BC, where C
is between B and D, and the measure of angle ACD is
118°. What is the measure, in degrees, of angle BAC?
∠ACD and ∠ACB are a linear pair, so ∠ACB = 180 − 118 = 62°. AB = AC makes the triangle isosceles with equal angles at B and C, so ∠ABC = 62° too, and ∠BAC = 180 − 62 − 62 = 56°. The one-step route: 118° is the exterior angle at C, so it equals ∠A + ∠B = ∠A + 62. B halves the exterior angle (118 ÷ 2 = 59), which is what you get by reading AB = AC as “angles A and B are equal”; the equal angles face the equal sides, so they are at B and C. C is ∠ACB, one line short of the answer. D is the angle you were given.
1
The three angles of any triangle total 180°.
Two angles given, the third is a subtraction. Angles as expressions, and it is one equation.
∠A + ∠B + ∠C = 180°
true for every triangle, every time
the angles are x°, (2x − 10)° and (x + 30)°
a typical SAT setup
x + (2x − 10) + (x + 30) = 180
one equation from the one rule
4x + 20 = 180, so x = 40
collect, then − 20 and ÷ 4
angles = 40°, 70°, 70°
and 40 + 70 + 70 = 180 ✓
The sum is the only fact you always have. Everything else in this lesson is a
shortcut built on top of it.
Read what the question wants
Here x = 40 and the largest angle is 70°. Both numbers will be in the choices.
The stem asks for one of them — the value of x, the measure of the largest angle,
the smallest angle — and the choices are built so that stopping early still lands
on something that looks like an answer.
2
An exterior angle equals the two remote interior angles added.
Skip the subtraction: the angle outside equals the two angles it does not touch.
slow: ∠C = 180 − 35 − 85 = 60°
angle sum first
then 180 − 60 = 120°
linear pair along AD — two steps
fast: 35 + 85 = 120°
the two remote interior angles — one step
exterior angle at C = 120°
remote — use these
∠A and ∠B, the two it does not touch
adjacent — not this one
∠C, the one sitting beside it on the line
It is not a new fact, just the other two combined: the interior angles make 180°
and so does the linear pair, so subtracting cancels the adjacent angle. Whenever a
figure shows three points on a line with a triangle standing on them, look for it —
it saves a line every single time.
3
Equal sides face equal angles.
In an isosceles triangle the two equal angles are the ones opposite the two equal sides — across from them, not next to them.
PQ = PR
the tick marks say so
∠Q = ∠R
PQ is opposite ∠R, PR is opposite ∠Q
180 − 76 = 104, each base angle = 52°
what is left, split in two
It runs both ways. Two equal angles force the two sides opposite them to be equal,
which is how the SAT hands you a length without ever mentioning it.
Which angle is the odd one out?
If a stem says an isosceles triangle has a 40° angle and does not say where, there
are two triangles: 40° as the apex gives 70-70-40, and 40° as a base angle gives
40-40-100. Read the figure or the naming to see which angle was named — and if a
question really is ambiguous, the choices will only contain one of the two.
4
Chain the rules: the opening problem in three lines.
AB = AC, points B, C and D lie on a line, and ∠ACD = 118°. Find ∠BAC.
∠ACB = 180 − 118 = 62°
linear pair on line BD
∠ABC = 62°
AB = AC, so the angles at B and C match
x = 180 − 62 − 62 = 56
the angle sum finishes it
∠BAC = 56°
Or use the shortcut and save a line: 118° is the exterior angle at C, so it equals
the two remote interior angles, ∠A + ∠B. With ∠B = 62° that is
118 = x + 62, giving x = 56 straight away. Same answer, one deduction fewer.
5
Three lengths only close up if the two shortest beat the longest.
Can a triangle have sides 6, 9 and 16 inches?
6 + 9 = 15
the two shorter sides
15 < 16
they fall short of the longest side
no such triangle exists
whatever angle you hinge them at
Equal is no good either. If the two short sides summed to exactly 16 they would lie
flat along the long one — no height, no area, no triangle. The inequality is strict.
Strictly, every pair must beat the third side, but only the two shortest can ever
fail. Sort the three numbers, add the small two, compare with the big one, done.
6
“Which could be the third side?” is a range question.
Two sides measure 5 and 12. The third side is trapped between their difference and their sum.
x < 5 + 12 = 17
less than the sum
x > 12 − 5 = 7
more than the difference
7 < x < 17
both ends strict
7 < x < 17
Scan the four choices and pick the one inside the range. If the question asks for
whole numbers, count 8 through 16 — that is 9 values, not 10 and not 17. The
endpoints are exactly the values the distractors are made of.
Difference on the low end, sum on the high end
Swapping them is the standard error: a third side of 17 would lie flat, and a
third side of 7 would too. Anything smaller than 7 or larger than 17 does not
reach at all.
✓
Check: do the totals close?
angle sum56 + 62 + 62 = 180the opening triangle ✓
linear pair62 + 118 = 180along line BD ✓
shortcut56 + 62 = 118the exterior angle agrees ✓
Remember the flow
Sum is 180 · Exterior = the two far ones · Equal sides, equal angles
For lengths: difference < third side < sum.
Now you try
Practice problem
How many values are possible?
Two sides of a triangle have lengths 8 and
15 centimeters. If the length of the third side is an
integer number of centimeters, how many possible values are there for that length?
The third side x satisfies 15 − 8 < x < 15 + 8, so 7 < x < 23. The integers are 8 through 22, and 22 − 8 + 1 = 15 values. A is the off-by-one, 22 − 8, from subtracting the endpoints without adding 1. C counts 7 and 23 as allowed (7 through 23 is 17 values), but both make a flat triangle — the inequality is strict. D drops the lower bound and counts every integer from 1 to 22.
Hint: write the range first, then count. Counting whole numbers from a to b inclusive
is b − a + 1, and the “+ 1” is where this question is won or lost. A calculator is
allowed but will not help; the difference and the sum are the whole calculation.
