Calculating an angle in an acute / obtuse triangle - GMAT Quantitative

Card 0 of 168

Which of the following cannot be the measure of a base angle of an isosceles triangle?

Tap to see back →

An isosceles triangle has two congruent angles by the Isosceles Triangle Theorem; these are the base angles. Since at least two angles of any triangle must be acute, a base angle must be acute - that is, it must measure under . The only choice that does not fit this criterion is $100 ^{\circ }$ , making this the correct choice.

Let the three interior angles of a triangle measure , and . Which of the following statements is true about the triangle?

Tap to see back →

If these are the measures of the interior angles of a triangle, then they total . Add the expressions, and solve for .

$x +\left ( x + 30 \right ) +\left ( 2x - 10 \right ) = 180$

One angle measures . The others measure:

$x + 30 = 40 + 30 = $70^{\circ }$$

$2x-10 = 2 \cdot 40 - 10 = 80 - 10 = 70 ^{\circ }$

All three angles measure less than $90 ^{\circ }$ , so the triangle is acute. Also, there are two congruent angles, so by the converse of the Isosceles Triangle Theorem, two sides are congruent, and the triangle is isosceles.

Two angles of an isosceles triangle measure $(x + 20) ^{\circ}$ and $\left ( 100-x\right $)^{ \circ}$$ . What are the possible values of ?

Tap to see back →

In an isosceles triangle, at least two angles measure the same. Therefore, one of three things happens:

Case 1: The two given angles have the same measure.

$2x \div 2 =80 \div 2$

The angle measures are $60 ^{\circ}, 60 ^{\circ}, 60 ^{\circ}$ , making the triangle equianglular and, subsequently, equilateral. An equilateral triangle is considered isosceles, so this is a possible scenario.

Case 2: The third angle has measure $(x + 20) ^{\circ}$ .

Then, since the sum of the angle measures is 180,

$\left (x + 20 \right ) +\left (x +20 \right )+ \left ( 100-x \right ) = 180$

as before

Case 3: The third angle has measure $\left ( 100-x\right $)^{ \circ}$$

$\left (x + 20 \right ) +\left (100-x \right )+ \left ( 100-x \right ) = 180$

as before.

Thus, the only possible value of is 40.

Which of the following is true of a triangle with three angles whose measures have an arithmetic mean of ?

Tap to see back →

The sum of the measures of three angles of any triangle is 180; therefore, their mean is $180 \div 3 = 60$ , making a triangle with angles whose measures have mean 90 impossible.

Which of the following is true of a triangle with two $20 ^{\circ }$ angles?

Tap to see back →

The sum of the measures of three angles of any triangle is 180; therefore, if two angles have measure $20 ^{\circ }$ , the third must have measure $180 - 20 - 20 = $140^{\circ }$ > $90^{\circ }$$ . This makes the triangle obtuse. Also, since the triangle has two congruent angles, it is isosceles by the Converse of the Isosceles Triangle Theorem.

Two angles of an isosceles triangle measure $\left (x + 30 \right $)^{\circ }$$ and $\left (x + 36 \right $)^{\circ }$$ . What are the possible value(s) of ?

Tap to see back →

In an isosceles triangle, at least two angles measure the same. Therefore, one of three things happens:

Case 1: The two given angles have the same measure.

This is a false statement, indicating that this situation is impossible.

Case 2: The third angle has measure $\left (x + 30 \right $)^{\circ }$$ .

Then, since the sum of the angle measures is 180,

$\left (x + 30 \right ) +\left (x + 30 \right )+ \left ( x + 36 \right ) = 180$

This makes the angle measures , a plausible scenario.

Case 3: the third angle has measure $\left (x + 36 \right $)^{\circ }$$

Then, since the sum of the angle measures is 180,

$\left (x + 30 \right ) +\left (x + 36 \right )+ \left ( x + 36 \right ) = 180$

This makes the angle measures , a plausible scenario.

Therefore, either or

$\Delta ABC \sim \Delta XYZ$

$m \angle A = 20 ^{\circ }; m \angle Y = 120 ^{\circ }$

Which of the following is true of $\Delta ABC$ ?

Tap to see back →

By similarity, $m \angle B = m \angle Y = 120 ^{\circ }$ .

Since measures of the interior angles of a triangle total $180 ^{\circ }$ ,

$m \angle A + m \angle B + m \angle C = 180$

$20 + 120 + m \angle C = 180$

$140 + m \angle C = 180$

$140 + m \angle C -140 = 180-140$

$m \angle C $=40^{\circ }$$

Since the three angle measures of $\Delta ABC$ are all different, no two sides measure the same; the triangle is scalene. Also, since $m \angle B = 120 ^{\circ } > $90^{\circ }$$ , the angle is obtuse, and $\Delta ABC$ is an obtuse triangle.

Two angles of a triangle measure $47 ^{\circ }$ and . What is the measure of the third angle?

Tap to see back →

The sum of the degree measures of the angles of a triangle is 180, so we can subtract the two angle measures from 180 to get the third:

The angles of a triangle measure . Evaluate .

Tap to see back →

The sum of the measures of the angles of a triangle total , so we can set up and solve for in the following equation:

$3x\div 3 =234 \div 3$

An exterior angle of $\Delta ABC$ with vertex measures ; an exterior angle of $\Delta ABC$ with vertex measures . Which is the following is true of $\Delta ABC$ ?

Tap to see back →

An interior angle of a triangle measures $180 ^{\circ }$ minus the degree measure of its exterior angle. Therefore:

$m \angle A = 180 ^{\circ } - 150 ^{\circ } = 30 ^{\circ }$

$m \angle B= 180 ^{\circ } - 110 ^{\circ } = 70 ^{\circ }$

The sum of the degree measures of the interior angles of a triangle is $180 ^{\circ }$ , so

$m \angle C = 180 ^{\circ } - (m \angle A + m \angle B) = 180 ^{\circ } - (30 ^{\circ }+ 70 ^{\circ }) = 180 ^{\circ } -100 ^{\circ } = 80 ^{\circ }$ .

Each angle is acute, so the triangle is acute; each angle is of a different measure, so the triangle has three sides of different measure, making it scalene.

Note: Figure NOT drawn to scale.

Refer to the above diagram.

$m\angle DCF = 140 ^{\circ } , m \angle ABF = 134 ^{\circ }$

Evaluate $m \angle EFB$ .

Tap to see back →

The sum of the exterior angles of a triangle, one per vertex, is $360 ^{\circ}$ . $\angle DCF$ , $\angle ABF$ and $\angle EFB$ are exterior angles at different vertices, so

$m \angle EFB + m\angle DCF + m \angle ABF = 360 ^{\circ }$

$m \angle EFB + 140 ^{\circ } + 134 ^{\circ } = 360 ^{\circ }$

$m \angle EFB + 274 ^{\circ } = 360 ^{\circ }$

$m \angle EFB + 274 ^{\circ }- 274 ^{\circ } = 360 ^{\circ } - 274 ^{\circ }$

$m \angle EFB = $86^{\circ }$$

In the following triangle:

The angle degrees

The angle degrees

(Figure not drawn on scale)

Find the value of .

Tap to see back →

Since , the following triangles are isoscele: .

If ADC, BDC, and BDA are all isoscele; then:

The angle degrees

The angle degrees, and

The angle degrees

Therefore:

The angle

The angle degrees, and

The angle

Since the sum of angles of a triangle is equal to 180 degrees then:

. So:

.

Now let us solve the equation for x:

$2x + 150 = 180 \rightarrow x = 15$

(See image below - not drawn on scale)

$\angle 1$ is an exterior angle of $\bigtriangleup ABC$ at $\angle B AC$ .

Is $\bigtriangleup ABC$ an acute triangle, a right triangle, or an obtuse triangle?

Statement 1: $\angle 1$ is acute.

Statement 2: $\angle B$ and $\angle C$ are both acute.

Tap to see back →

Exterior angle $\angle 1$ forms a linear pair with its interior angle $\angle B AC$ . Either both are right, or one is acute and one is obtuse. From Statement 1 alone, since $\angle 1$ is acute, $\angle B AC$ is obtuse, and $\bigtriangleup ABC$ is an obtuse triangle.

Statement 2 alone is insufficient. Every triangle has at least two acute angles, and Statement 2 only establishes that $\angle B$ and $\angle C$ are both acute; the third angle, $\angle B AC$ , can be acute, right, or obtuse, so the question of whether $\bigtriangleup ABC$ is an acute, right, or obtuse triangle is not settled.

Is $\bigtriangleup ABC$ an acute triangle, a right triangle, or an obtuse triangle?

Statement 1: $m \angle A + m \angle B < 90 ^{\circ }$

Statement 2:

Tap to see back →

Assume Statement 1 alone. The sum of the measures of interior angles of a triangle is $180 ^{\circ }$ ;

$m \angle A + m \angle B < 90 ^{\circ }$ , or, equivalently, for some positive number ,

$m \angle A + m \angle B =( 90 - N) ^{\circ }$ ,

so

$m \angle A + m \angle B + m \angle C = 180 ^{\circ }$

$( 90 - N) ^{\circ }+ m \angle C = 180 ^{\circ }$

$m \angle C = 180 ^{\circ } - ( 90 - N) ^{\circ } = ( 90+N) ^{\circ }$

Therefore, $m \angle C > 90 ^{\circ }$ , making $\angle C$ obtuse, and $\bigtriangleup ABC$ an obtuse triangle.

Assume Statement 2 alone. Since the sum of the squares of the lengths of two sides exceeds the square of the length of the third, it follows that $\bigtriangleup ABC$ is an obtuse triangle.

$\angle 1$ , $\angle 2$ , and $\angle 3$ are all exterior angles of $\bigtriangleup ABC$ with vertices , , and , respectively.

Is $\bigtriangleup ABC$ an acute triangle, a right triangle, or an obtuse triangle?

Statement 1: $\angle 1$ , $\angle 2$ , and $\angle 3$ are all obtuse angles.

Statement 2: $\angle A \cong \angle B$ .

Note: For purposes of this problem, $\angle A$ , $\angle B$ , and $\angle C$ will refer to the interior angles of the triangle at these vertices.

Tap to see back →

Assume Statement 1 alone. An exterior angle of a triangle forms a linear pair with the interior angle of the triangle of the same vertex. The two angles, whose measures total , must be two right angles or one acute angle and one obtuse angle. Since $\angle 1$ , $\angle 2$ , and $\angle 3$ are all obtuse angles, it follows that their respective interior angles - the three angles of $\bigtriangleup ABC$ - are all acute. This makes $\bigtriangleup ABC$ an acute triangle.

Statement 2 alone provides insufficient information to answer the question. For example, if $\angle A$ and $\angle B$ each measure and $\angle C$ measures , the sum of the angle measures is , $\angle A$ and $\angle B$ are congruent, and $\angle C$ is an obtuse angle (measuring more than ); this makes $\bigtriangleup ABC$ an obtuse triangle. But if $\angle A$ , $\angle B$ , and $\angle C$ each measure , the sum of the angle measures is again , $\angle A$ and $\angle B$ are again congruent, and all three angles are acute (measuring less than ); this makes $\bigtriangleup ABC$ an acute triangle.

The measures of the interior angles of a triangle are , , and . Also,

.

Evaluate .

Tap to see back →

The measures of the interior angles of a triangle have sum , so

Along with , a system of linear equations is formed that can be solved by adding:

$\underline{x+y = 108}$

$2y \div 2 = 86 \div 2$

The interior angles of a triangle have measures , , and . Also,

.

Which of the following is closest to ?

Tap to see back →

The measures of the interior angles of a triangle have sum , so

, or

Along with , a system of linear equations is formed that can be solved by adding:

$\underline{x - y = 32}$

$4x \div 4 = 212 \div 4$

Of the given choices, 50 comes closest to the correct measure.

A triangle has interior angles whose measures are , , and $42 ^{\circ }$ . A second triangle has interior angles, two of whose measures are and . What is the measure of the third interior angle of the second triangle?

Tap to see back →

The measures of the interior angles of a triangle have sum , so

, or, equivalently,

$$\frac{1}{2}$\left (x+ y \right )= $\frac{1}{2}$ \cdot 138$

$$\frac{1}{2}$ x+ $\frac{1}{2}$ y =69$

and are the measures of two interior angles of the second triangle, so if we let be the measure of the third angle, then

$$\frac{1}{2}$ x+ $\frac{1}{2}$ y +Z= 180$

By substitution,

and

.

The correct response is .

The measures of the interior angles of Triangle 1 are , , and . The measures of two of the interior angles of Triangle 2 are and . Which of the following is the measure of the third interior angle of Triangle 2?

Tap to see back →

The measures of the interior angles of a triangle have sum , so

, or, equivalently,

$2 (x+ y )= 2 \cdot 73$

and are the measures of two interior angles of the second triangle, so if we let be the measure of the third angle, then

By substitution,

The correct response is .

Triangle 1 has three interior angles with measures , , and . Triangle 1 has three interior angles with measures $\frac{3}{2}$ $x^{\circ }$ , $\frac{3}{2}$ $x^{\circ }$ , and .

Express in terms of .

Tap to see back →

The sum of the measures of the interior angles of a triangle is , so it can be determined from Triangle 1 that

$$\frac{1}{2}$ \cdot 2x = $\frac{1}{2}$ \cdot (180-y)$

$x =90-$\frac{1}{2}$ y$

From Triangle 2, we can deduce that

$$\frac{3}{2}$ $x^{\circ }$ + $\frac{3}{2}$ $x^{\circ }$ + z = 180$

By substitution:

$z = 180 -3\left ( 90-$\frac{1}{2}$ y \right )$

$z = 180 -270 +\frac{3}{2}$ y$

$z = $\frac{3}{2}$ y - 90$