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Triangle angles & sides: Geometry Regents Practice

Practice Triangle angles & sides for the Geometry Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 4 · 2 pts

In a triangle, two interior angles measure $43^\circ$ and $68^\circ$. What is the measure of the third angle?

Worked solution
$43 + 68 + x = 180$ Interior angles of a triangle sum to 180°.
$x = 180 - 111 = 69^\circ$
Answer: 69^\circ

The three interior angles of any triangle add to 180°.

Question 1 · 2 pts

In $\triangle ABC$, $m\angle A = 52^\circ$ and $m\angle B = 71^\circ$. What is $m\angle C$?

Worked solution
$m\angle A + m\angle B + m\angle C = 180^\circ$ Triangle Angle Sum Theorem
$52^\circ + 71^\circ + m\angle C = 180^\circ$
$m\angle C = 180^\circ - 123^\circ = 57^\circ$
Answer: $m\angle C = 57^\circ$

The three interior angles of any triangle sum to $180^\circ$, so the third angle equals $180^\circ$ minus the sum of the other two.

Question 2 · 2 pts

An exterior angle of a triangle measures $115^\circ$. One of its remote (nonadjacent) interior angles measures $48^\circ$. What is the measure of the other remote interior angle?

Worked solution
$\text{exterior angle} = \text{remote}_1 + \text{remote}_2$ Exterior Angle Theorem
$115^\circ = 48^\circ + x$
$x = 115^\circ - 48^\circ = 67^\circ$
Answer: $67^\circ$

An exterior angle of a triangle equals the sum of the two nonadjacent (remote) interior angles.

Question 2 · 2 pts

The measures of the angles of a triangle are represented by $(x + 10)°$, $(2x)°$, and $(3x - 40)°$. What is the measure of the largest angle of the triangle?

Worked solution
$(x + 10) + 2x + (3x - 40) = 180$ The angles of a triangle sum to 180 degrees.
$6x - 30 = 180$
$6x = 210, \quad x = 35$
$x + 10 = 45, \quad 2x = 70, \quad 3x - 40 = 65$ Check: 45 + 70 + 65 = 180.
$\text{The largest angle is } 70°.$
Answer: $70°$

The angle measures of a triangle sum to 180 degrees, giving 6x - 30 = 180 and x = 35. The three angles are 45, 70, and 65 degrees, so the largest is 70 degrees. Choosing 35 is the common error of answering with x instead of the requested angle.

Question 1 · 2 pts

The measures of the three angles of a triangle are in the ratio $2:3:4$. What is the measure of the smallest angle of the triangle?

Worked solution
$2x + 3x + 4x = 180$ The angles of a triangle sum to 180 degrees.
$9x = 180$
$x = 20$
$2x = 2(20) = 40$ The smallest angle corresponds to the smallest ratio number.
Answer: $40°$

The angle sum of a triangle is $180°$. Writing the angles as $2x$, $3x$, and $4x$ gives $9x = 180$, so $x = 20$. The smallest angle is $2x = 40°$. A common error is stopping at $x = 20$ and choosing $20°$.

Question 13 · 2 pts

In triangle $ABC$, side $BC$ is extended through $C$ to point $D$. If $m\angle ACD = 112^\circ$ and $m\angle A = 46^\circ$, what is $m\angle B$?

Worked solution
$m\angle ACD = m\angle A + m\angle B$ An exterior angle of a triangle equals the sum of the two remote interior angles.
$112 = 46 + m\angle B$
$m\angle B = 112 - 46 = 66^\circ$
Answer: $66^\circ$

The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two nonadjacent interior angles. Since the exterior angle at $C$ is $112^\circ$ and angle $A$ is $46^\circ$, angle $B$ must be $112 - 46 = 66^\circ$.

Question 15 · 2 pts

Two sides of a triangle measure $7$ and $12$. Which of the following could be the length of the third side?

Worked solution
$12 - 7 < x < 12 + 7$ Triangle inequality: the third side must be greater than the difference and less than the sum of the other two sides.
$5 < x < 19$
$x = 10 \text{ satisfies } 5 < 10 < 19$
Answer: $10$

By the triangle inequality theorem, the third side must be strictly between $12 - 7 = 5$ and $12 + 7 = 19$. Only $10$ lies strictly in that interval; $5$ and $19$ fail because the inequality is strict, and $4$ is too small.

Question 2 · 2 pts

In triangle $ABC$, $m\angle A = 40°$ and the exterior angle at vertex $C$ measures $115°$. What is $m\angle B$?

Worked solution
$\text{An exterior angle equals the sum of the two remote interior angles.}$
$115° = m\angle A + m\angle B = 40° + m\angle B$
$m\angle B = 115° - 40° = 75°$
Answer: $75°$

The exterior angle theorem states that an exterior angle of a triangle equals the sum of the two nonadjacent interior angles. Since the exterior angle at C is 115 degrees and angle A is 40 degrees, angle B must be 115 minus 40, which is 75 degrees. A common error is first computing the interior angle at C (65 degrees) and stopping there.

Question 25 · 4 pts · constructed response

In triangle $ABC$, $m\angle A = 2x + 10$, $m\angle B = 3x$, and $m\angle C = x + 50$. Determine and state $m\angle B$.

Model answer: $m\angle B = 60^\circ$
Worked solution
$(2x+10) + 3x + (x+50) = 180$ The angles of a triangle sum to 180 degrees.
$6x + 60 = 180$
$6x = 120, \quad x = 20$
$m\angle B = 3x = 3(20) = 60^\circ$
Answer: $m\angle B = 60^\circ$

The three interior angles of any triangle sum to $180^\circ$. Setting the sum of the three expressions equal to $180$ gives $x = 20$, so $m\angle B = 3(20) = 60^\circ$. As a check, the angles are $50^\circ$, $60^\circ$, and $70^\circ$, which sum to $180^\circ$.

Question 3 · 2 pts

Two sides of a triangle measure $5$ and $9$. Which could be the length of the third side?

Worked solution
$9 - 5 < s < 9 + 5$ Triangle inequality: the third side is between the difference and the sum of the other two.
$4 < s < 14$
$\text{Only } 10 \text{ satisfies } 4 < s < 14.$ The values 4 and 14 fail because the inequality is strict.
Answer: $10$

By the triangle inequality, the third side must be strictly greater than 9 minus 5 and strictly less than 9 plus 5, so it must lie between 4 and 14. Of the choices, only 10 is in that open interval; 4 and 14 would make the points collinear.

Question 13 · 2 pts

In triangle $ABC$, the measure of angle $A$ is $47°$. The exterior angle at vertex $C$ measures $115°$. What is the measure of angle $B$?

Worked solution
$m\angle A + m\angle B = 115°$ An exterior angle of a triangle equals the sum of the two remote interior angles.
$47° + m\angle B = 115°$
$m\angle B = 115° - 47° = 68°$
Answer: $68°$

The exterior angle theorem states that an exterior angle of a triangle equals the sum of the two nonadjacent interior angles. Since the exterior angle at $C$ is $115°$ and angle $A$ is $47°$, angle $B$ must be $115° - 47° = 68°$. Choosing $65°$ comes from subtracting the exterior angle from $180°$, which gives angle $C$ instead.

Question 3 · 2 pts

In an isosceles triangle, the vertex (apex) angle measures $40^\circ$. What is the measure of each base angle?

Worked solution
$2b + 40^\circ = 180^\circ$ the two base angles b are congruent
$2b = 140^\circ$
$b = 70^\circ$
Answer: $70^\circ$

In an isosceles triangle the two base angles are congruent, so subtract the vertex angle from $180^\circ$ and split the remainder in half.

Question 4 · 2 pts

Two sides of a triangle have lengths $7$ and $10$. Which of the following could be the length of the third side?

Worked solution
$10 - 7 < x < 10 + 7$ Triangle Inequality Theorem
$3 < x < 17$
$x = 12 \text{ satisfies } 3 < 12 < 17$ must be strictly between the bounds
Answer: $12$

By the Triangle Inequality, the third side must be greater than the difference and less than the sum of the other two sides.

Question 5 · 2 pts

In $\triangle ABC$, $m\angle A = 45^\circ$ and $m\angle B = 65^\circ$. Which side of the triangle is the longest?

Worked solution
$m\angle C = 180^\circ - 45^\circ - 65^\circ = 70^\circ$
$\angle C \text{ is the largest angle } (70^\circ)$
$\text{longest side is opposite } \angle C = \overline{AB}$ longest side faces largest angle
Answer: $\overline{AB}$

In any triangle the longest side lies opposite the largest angle.

Question 6 · 2 pts

The measures of the three angles of a triangle are $x$, $x+20$, and $x+40$ degrees. What is the measure of the largest angle?

Worked solution
$x + (x+20) + (x+40) = 180$ angles of a triangle sum to 180
$3x + 60 = 180$
$3x = 120 \Rightarrow x = 40$
$\text{largest} = x + 40 = 80^\circ$
Answer: $80^\circ$

Set the sum of the angle expressions equal to $180^\circ$, solve for $x$, then evaluate the largest expression.

Question 20 · 2 pts

In isosceles triangle $ABC$, $AB = AC$. The exterior angle at vertex $B$ measures $130°$. What is the measure of angle $A$?

Worked solution
$m\angle ABC = 180° - 130° = 50°$ An interior angle and its exterior angle are supplementary.
$m\angle ACB = m\angle ABC = 50°$ Since AB = AC, the base angles at B and C are congruent.
$m\angle A = 180° - 50° - 50° = 80°$
Answer: $80°$

The exterior angle at B is supplementary to interior angle ABC, so angle ABC measures 50 degrees. Because AB = AC, the base angles at B and C are both 50 degrees, leaving 80 degrees for angle A. Choosing 100 comes from mistakenly treating the vertex angle as the supplement of a base angle pair, and 65 from halving 130.

Question 10 · 2 pts

Two sides of a triangle have lengths $5$ and $9$. Which of the following could be the length of the third side?

Worked solution
$9 - 5 < x < 9 + 5$ Triangle inequality: the third side must be between the difference and the sum of the other two sides.
$4 < x < 14$
$x = 10 \text{ satisfies } 4 < 10 < 14$ The values 3, 4, and 14 all fail the strict inequality.
Answer: $10$

By the triangle inequality, the third side must be strictly between $9 - 5 = 4$ and $9 + 5 = 14$. Only $10$ lies in the interval $4 < x < 14$. The values $4$ and $14$ are tempting because they are the boundary values, but a triangle cannot have a side equal to the difference or the sum of the other two.

Question 24 · 2 pts

In triangle $ABC$, medians $\overline{AM}$, $\overline{BN}$, and $\overline{CP}$ intersect at point $G$. If $AM = 24$, what is the length of $\overline{AG}$?

Worked solution
$G \text{ is the centroid of } \triangle ABC$ The medians of a triangle intersect at the centroid.
$AG = \frac{2}{3} AM$ The centroid divides each median in a 2:1 ratio, measured from the vertex.
$AG = \frac{2}{3}(24) = 16$
Answer: $16$

The point where the medians of a triangle intersect is the centroid, which divides each median so that the segment from the vertex to the centroid is twice the segment from the centroid to the midpoint. Therefore $AG = \frac{2}{3}(24) = 16$. Choosing $8$ gives the shorter piece $GM$, and $12$ incorrectly assumes the centroid bisects the median.

Question 7 · 2 pts

In $\triangle DEF$, $\overline{DE} \cong \overline{DF}$. If $m\angle E = (2x+10)^\circ$ and $m\angle F = (3x-5)^\circ$, what is $m\angle D$?

Worked solution
$\overline{DE} \cong \overline{DF} \Rightarrow m\angle E = m\angle F$ base angles opposite congruent sides
$2x + 10 = 3x - 5$
$x = 15,\quad m\angle E = m\angle F = 40^\circ$
$m\angle D = 180^\circ - 40^\circ - 40^\circ = 100^\circ$
Answer: $m\angle D = 100^\circ$

Congruent sides force congruent opposite (base) angles; find those angles, then use the angle sum to get the vertex angle.

Question 8 · 2 pts

In $\triangle ABC$, $AB = 8$, $BC = 5$, and $CA = 7$. Which statement correctly orders the angle measures from least to greatest?

Worked solution
$\text{opp } \angle A = BC = 5,\ \text{opp } \angle B = CA = 7,\ \text{opp } \angle C = AB = 8$ each angle faces the side of the same letter's opposite vertices
$5 < 7 < 8$ order the opposite sides
$\text{shorter side} \Rightarrow \text{smaller opposite angle}$
$m\angle A < m\angle B < m\angle C$
Answer: $m\angle A < m\angle B < m\angle C$

By the angle-side inequality, larger angles lie opposite longer sides, so the angles order the same way as the sides opposite them.

Question 32 · 4 pts · constructed response

Given: In triangle $ABC$, $\overline{AB} \cong \overline{AC}$, and $D$ is the midpoint of $\overline{BC}$. Prove: $\overline{AD} \perp \overline{BC}$.

Model answer: Since $D$ is the midpoint, $\overline{BD} \cong \overline{DC}$. With $\overline{AB} \cong \overline{AC}$ and $\overline{AD} \cong \overline{AD}$ (reflexive), $\triangle ABD \cong \triangle ACD$ by SSS. Then $\angle ADB \cong \angle ADC$ (CPCTC). Since these angles form a linear pair, $m\angle ADB + m\angle ADC = 180^\circ$, so each measures $90^\circ$. Therefore $\overline{AD} \perp \overline{BC}$.
Worked solution
$\overline{BD} \cong \overline{DC}$ D is the midpoint of BC, and a midpoint divides a segment into two congruent segments.
$\overline{AB} \cong \overline{AC}$ Given.
$\overline{AD} \cong \overline{AD}$ Reflexive property.
$\triangle ABD \cong \triangle ACD \text{ by SSS}$
$\angle ADB \cong \angle ADC$ Corresponding parts of congruent triangles are congruent (CPCTC).
$m\angle ADB + m\angle ADC = 180^\circ$ Angles ADB and ADC form a linear pair along line BC.
$2\,m\angle ADB = 180^\circ \Rightarrow m\angle ADB = 90^\circ$
$\therefore \overline{AD} \perp \overline{BC}$ Segments meeting at a right angle are perpendicular.
Answer: Since $D$ is the midpoint, $\overline{BD} \cong \overline{DC}$. With $\overline{AB} \cong \overline{AC}$ and $\overline{AD} \cong \overline{AD}$ (reflexive), $\triangle ABD \cong \triangle ACD$ by SSS. Then $\angle ADB \cong \angle ADC$ (CPCTC). Since these angles form a linear pair, $m\angle ADB + m\angle ADC = 180^\circ$, so each measures $90^\circ$. Therefore $\overline{AD} \perp \overline{BC}$.

The median to the base of an isosceles triangle creates two triangles congruent by SSS. The congruent angles at $D$ form a linear pair, so each must be $90^\circ$, which proves $\overline{AD}$ is perpendicular to $\overline{BC}$.

Question 32 · 4 pts · constructed response

In triangle $ABC$, $m\angle A = (2x+10)^\circ$ and $m\angle B = (3x)^\circ$. Side $\overline{BC}$ is extended through $C$ to point $D$, and the exterior angle at $C$, angle $ACD$, measures $(6x-20)^\circ$. Determine and state the value of $x$, and find the measure of each interior angle of triangle $ABC$. Justify your answer.

Model answer: $x = 30$; $m\angle A = 70^\circ$, $m\angle B = 90^\circ$, $m\angle ACB = 20^\circ$. The exterior angle theorem gives $(2x+10)+(3x) = 6x-20$, so $x = 30$; the interior angles then sum to $180^\circ$ as required.
Worked solution
$m\angle A + m\angle B = m\angle ACD$ An exterior angle of a triangle equals the sum of the two remote interior angles.
$(2x+10) + 3x = 6x - 20$
$5x + 10 = 6x - 20, \quad x = 30$
$m\angle A = 2(30)+10 = 70^\circ, \quad m\angle B = 3(30) = 90^\circ$
$m\angle ACB = 180 - (70 + 90) = 20^\circ$ Check: exterior angle $6(30)-20 = 160^\circ$ and $160 + 20 = 180^\circ$, consistent with a linear pair.
Answer: $x = 30$; $m\angle A = 70^\circ$, $m\angle B = 90^\circ$, $m\angle ACB = 20^\circ$. The exterior angle theorem gives $(2x+10)+(3x) = 6x-20$, so $x = 30$; the interior angles then sum to $180^\circ$ as required.

The exterior angle theorem states that an exterior angle of a triangle equals the sum of the two remote interior angles, so $(2x+10) + 3x = 6x-20$, giving $x = 30$. The interior angles are then $70^\circ$, $90^\circ$, and $20^\circ$, which sum to $180^\circ$, and the exterior angle $160^\circ$ forms a linear pair with the $20^\circ$ angle at $C$, confirming the solution.

Question 12 · 2 pts

In right triangle $RST$, the right angle is at $S$, and the altitude from $S$ meets hypotenuse $\overline{RT}$ at point $H$. If $RH = 4$ and $HT = 9$, what is the length of altitude $\overline{SH}$?

Worked solution
$SH^2 = RH \cdot HT$ The altitude to the hypotenuse is the geometric mean of the two segments of the hypotenuse.
$SH^2 = 4 \cdot 9 = 36$
$SH = 6$
Answer: $6$

The altitude drawn to the hypotenuse of a right triangle creates two smaller triangles similar to the original, which makes the altitude the geometric mean of the two hypotenuse segments. So SH squared equals 4 times 9, giving SH = 6. Averaging the segments gives 6.5 and adding them gives 13, both common mistakes.

Question 32 · 4 pts · constructed response

In triangle $ABC$, $m\angle A = (2x+10)^\circ$ and $m\angle B = (3x-5)^\circ$. The exterior angle at vertex $C$ measures $(6x-15)^\circ$. Determine and state the value of $x$, and find $m\angle C$. Justify your work.

Model answer: $x = 20$ and $m\angle C = 75^\circ$, by the exterior angle theorem and the triangle angle sum.
Worked solution
$(2x+10) + (3x-5) = 6x-15$ The exterior angle of a triangle equals the sum of the two nonadjacent interior angles.
$5x + 5 = 6x - 15$
$x = 20$
$m\angle A = 2(20)+10 = 50^\circ, \quad m\angle B = 3(20)-5 = 55^\circ$
$m\angle C = 180 - (50 + 55) = 75^\circ$ Check: exterior angle at $C$ is $6(20)-15 = 105^\circ$, and $180 - 105 = 75^\circ$.
Answer: $x = 20$ and $m\angle C = 75^\circ$, by the exterior angle theorem and the triangle angle sum.

By the exterior angle theorem, the exterior angle at $C$ equals the sum of the two remote interior angles $A$ and $B$. Solving $5x+5 = 6x-15$ gives $x = 20$, so angles $A$ and $B$ are $50^\circ$ and $55^\circ$, and the triangle angle sum gives $m\angle C = 75^\circ$, consistent with the $105^\circ$ exterior angle.

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