Acute / Obtuse Triangles - GMAT Quantitative

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Question

The largest angle of an obtuse isosceles triangle is $120^{\circ}$ . If two of the sides have an equal length of , what is the height of the triangle?

Answer

If the largest angle of the obtuse isosceles triangle is $120^{\circ}$ , then this is the unique angle in between the two sides with an equal length of . We can imagine that the height of this isosceles triangle is simply the third side of a triangle formed by half of its base and the length of either equal side. That is, if we bisected the $120^{\circ}$ angle with a line perpendicular to the base of the obtuse isosceles triangle, this line would be the height of the triangle. If we bisected the $120^{\circ}$ angle, we would have two congruent triangles with angles of $60^{\circ}$ between the height and each side of equal length. This means the cosine of that angle will be equal to the length of the height over the length of either equal side, which gives us:

$\cos (60^{\circ})=\frac{h}{18}$

$h=18\cos(60^{\circ})=18\left(\frac{1}{2}\right)=9$

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Question

Given: $\bigtriangleup ABC$ with , construct two altitudes of $\bigtriangleup ABC$ : one from to a point on $\overleftrightarrow{BC}$ , and another from to a point on $\overleftrightarrow{AC}$ . Which of the following is true of the relationship of the lengths of $\overline{AM}$ and $\overline{BN}$ ?

Answer

The area of a triangle is one half the product of the length of any base and its corresponding height; this is $\frac{1}{2} \cdot AC \cdot BN$ , but it is also $\frac{1}{2} \cdot BC \cdot AM$ . Set these equal, and note the following:

$\frac{1}{2} \cdot AC \cdot BN= \frac{1}{2} \cdot BC \cdot AM$

$\frac{1}{2} \cdot 16 \cdot BN= \frac{1}{2} \cdot 12 \cdot AM$

$8 \cdot BN= 6 \cdot AM$

$\frac{8 \cdot BN}{8}= \frac{6 \cdot AM}{8}$

$BN = \frac{3}{4} \cdot AM$

That is, the length of $\overline{BN}$ is three fourths that of that of $\overline{AM}$ .

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Question

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

Answer

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 $90^{\circ }$ . The only choice that does not fit this criterion is $100 ^{\circ }$ , making this the correct choice.

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Question

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

Answer

If these are the measures of the interior angles of a triangle, then they total $180^{\circ }$ . Add the expressions, and solve for .

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

One angle measures $40^{\circ }$ . 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.

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Question

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

Answer

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.

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Question

Which of the following is true of a triangle with three angles whose measures have an arithmetic mean of $90^{\circ }$ ?

Answer

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.

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Question

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

Answer

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.

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Question

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 ?

Answer

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 $58^{\circ },58^{\circ },64^{\circ }$ , 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 $56^{\circ }, 62^{\circ }, 62^{\circ }$ , a plausible scenario.

Therefore, either or

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Question

$\Delta ABC \sim \Delta XYZ$

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

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

Answer

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.

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Question

Two angles of a triangle measure $47 ^{\circ }$ and $23^{\circ }$ . What is the measure of the third angle?

Answer

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:

$(180-47-23)^{\circ } = 110^{\circ }$

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Question

The angles of a triangle measure $x^{\circ }, x^{\circ },\left ( x -54 \right ) ^{\circ }$ . Evaluate .

Answer

The sum of the measures of the angles of a triangle total $180^{\circ }$ , so we can set up and solve for in the following equation:

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

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Question

An exterior angle of $\Delta ABC$ with vertex measures $150^{\circ }$ ; an exterior angle of $\Delta ABC$ with vertex measures $110^{\circ }$ . Which is the following is true of $\Delta ABC$ ?

Answer

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.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram.

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

Evaluate $m \angle EFB$ .

Answer

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

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

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

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

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

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

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Question

In the following triangle:

The angle degrees

The angle degrees

(Figure not drawn on scale)

Find the value of .

Answer

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)

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Question

$\angle 1$ 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.

Answer

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.

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Question

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: $(AC)^{2} + (BC ) ^{2}< (AB) ^{2}$

Answer

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.

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Question

$\angle 1$ , $\angle 2$ , 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.

Answer

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 $180^{\circ }$ , 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 $10^{\circ }$ and $\angle C$ measures $160^{\circ }$ , the sum of the angle measures is $180^{\circ }$ , $\angle A$ and $\angle B$ are congruent, and $\angle C$ is an obtuse angle (measuring more than $90^{\circ }$ ); this makes $\bigtriangleup ABC$ an obtuse triangle. But if $\angle A$ , $\angle B$ , and $\angle C$ each measure $60^{\circ }$ , the sum of the angle measures is again $180^{\circ }$ , $\angle A$ and $\angle B$ are again congruent, and all three angles are acute (measuring less than $90^{\circ }$ ); this makes $\bigtriangleup ABC$ an acute triangle.

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Question

The measures of the interior angles of a triangle are $x^{\circ }$ , $y^{\circ }$ , and $72^{\circ }$ . Also,

.

Evaluate .

Answer

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

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

$\underline{x+y = 108}$

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

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Question

The interior angles of a triangle have measures $x^{\circ }$ , $y^{\circ }$ , and $2x^{\circ }$ . Also,

.

Which of the following is closest to ?

Answer

The measures of the interior angles of a triangle have sum $180^{\circ }$ , 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.

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Question

A triangle has interior angles whose measures are $x^{\circ }$ , $y^{\circ }$ , and $42 ^{\circ }$ . A second triangle has interior angles, two of whose measures are $\frac{1}{2}x^{\circ }$ and $\frac{1}{2}y^{\circ }$ . What is the measure of the third interior angle of the second triangle?

Answer

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

, or, equivalently,

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

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

$\frac{1}{2}x^{\circ }$ and $\frac{1}{2}y^{\circ }$ 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 $111^{\circ }$ .

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