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SAT II Math II : Finding Angles

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Example Questions

Example Question #3 : Linear Functions

Solve for $\displaystyle x$ and $\displaystyle y$.

(Figure not drawn to scale).

Possible Answers:

$\small x=15^o;\ y=7.5^o$ $\displaystyle \small x=15^o;\ y=7.5^o$

$\small x=15^o;\ y=37.5^o$ $\displaystyle \small x=15^o;\ y=37.5^o$

$\small x=7.5^o;\ y=18.75^o$ $\displaystyle \small x=7.5^o;\ y=18.75^o$

$\small x=15^o;\ y=52.5^o$ $\displaystyle \small x=15^o;\ y=52.5^o$

Correct answer:

$\small x=15^o;\ y=37.5^o$ $\displaystyle \small x=15^o;\ y=37.5^o$

Explanation:

The angles containing the variable $\displaystyle x$ all reside along one line, therefore, their sum must be 180^o $\displaystyle 180^o$.

$\small 5x+4x+3x=180^o$ $\displaystyle \small 5x+4x+3x=180^o$

$\small 12x=180^o$ $\displaystyle \small 12x=180^o$

$\small x=15^o$ $\displaystyle \small x=15^o$

Because $\displaystyle 2y$ and $\displaystyle 5x$ are opposite angles, they must be equal.

$\small 2y=5x$ $\displaystyle \small 2y=5x$

$\small x=15^o$ $\displaystyle \small x=15^o$

$\small 2y=5(15^o)=75^o$ $\displaystyle \small 2y=5(15^o)=75^o$

$\small y=\frac{75^o}{2}=37.5^o$ $\displaystyle \small y=\frac{75^o}{2}=37.5^o$

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Example Question #1 : Finding Angles

What angle do the minute and hour hands of a clock form at 6:15?

Possible Answers:

$92 \frac{1}{2}^{\circ }$ $\displaystyle 92 \frac{1}{2}^{\circ }$

$95^{\circ }$ $\displaystyle 95^{\circ }$

$90^{\circ }$ $\displaystyle 90^{\circ }$

$97 \frac{1}{2}^{\circ }$ $\displaystyle 97 \frac{1}{2}^{\circ }$

$87 \frac{1}{2}^{\circ }$ $\displaystyle 87 \frac{1}{2}^{\circ }$

Correct answer:

$97 \frac{1}{2}^{\circ }$ $\displaystyle 97 \frac{1}{2}^{\circ }$

Explanation:

There are twelve numbers on a clock; from one to the next, a hand rotates $\left (\frac{360}{12} \right )^{\circ } = 30^{\circ }$ $\displaystyle \left (\frac{360}{12} \right )^{\circ } = 30^{\circ }$. At 6:15, the minute hand is exactly on the "3" - that is, on the $\left (30 \times 3 \right )^{\circ }= 90^{\circ }$ $\displaystyle \left (30 \times 3 \right )^{\circ }= 90^{\circ }$ position. The hour hand is one-fourth of the way from the "6" to the "7" - that is, on the $\left (6 \frac{1}{4} \times 30 \right )^{\circ } = 187 \frac{1}{2} ^{\circ }$ $\displaystyle \left (6 \frac{1}{4} \times 30 \right )^{\circ } = 187 \frac{1}{2} ^{\circ }$ position. Therefore, the difference is the angle they make:

$187 \frac{1}{2} ^{\circ } - 90^{\circ } = 97 \frac{1}{2}^{\circ }$ $\displaystyle 187 \frac{1}{2} ^{\circ } - 90^{\circ } = 97 \frac{1}{2}^{\circ }$.

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Example Question #2 : Finding Angles

In triangle $\Delta ABC$ $\displaystyle \Delta ABC$, $m\angle A = 50^{\circ }$ $\displaystyle m\angle A = 50^{\circ }$ and $m \angle B = 80^{\circ }$ $\displaystyle m \angle B = 80^{\circ }$. Which of the following describes the triangle?

Possible Answers:

None of the other responses is correct.

$\Delta ABC$ $\displaystyle \Delta ABC$ is acute and scalene.

$\Delta ABC$ $\displaystyle \Delta ABC$ is acute and isosceles.

$\Delta ABC$ $\displaystyle \Delta ABC$ is obtuse and scalene.

$\Delta ABC$ $\displaystyle \Delta ABC$ is obtuse and isosceles.

Correct answer:

$\Delta ABC$ $\displaystyle \Delta ABC$ is acute and isosceles.

Explanation:

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

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

$m \angle C + 50^{\circ } + 80^{\circ } = 180^{\circ }$ $\displaystyle m \angle C + 50^{\circ } + 80^{\circ } = 180^{\circ }$

$m \angle C + 130^{\circ } = 180^{\circ }$ $\displaystyle m \angle C + 130^{\circ } = 180^{\circ }$

$m \angle C = 50^{\circ }$ $\displaystyle m \angle C = 50^{\circ }$

All three angles have measure less than $90^{\circ }$ $\displaystyle 90^{\circ }$, making the triangle acute. Also, by the Isosceles Triangle Theorem, since $m \angle A= m \angle C = 50^{\circ }$ $\displaystyle m \angle A= m \angle C = 50^{\circ }$, BC = AB $\displaystyle BC = AB$; the triangle has two congruent sides and is isosceles.

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Example Question #3 : Finding Angles

In $\Delta DEF$ $\displaystyle \Delta DEF$, $\angle D$ $\displaystyle \angle D$ and $\angle E$ $\displaystyle \angle E$ are complementary, and $m\angle D > m \angle E$ $\displaystyle m\angle D > m \angle E$. Which of the following is true of $\Delta DEF$ $\displaystyle \Delta DEF$ ?

Possible Answers:

None of the other responses is correct.

$\Delta DEF$ $\displaystyle \Delta DEF$ is acute and scalene.

$\Delta DEF$ $\displaystyle \Delta DEF$ is acute and isosceles.

$\Delta DEF$ $\displaystyle \Delta DEF$ is right and scalene.

$\Delta DEF$ $\displaystyle \Delta DEF$ is right and isosceles.

Correct answer:

$\Delta DEF$ $\displaystyle \Delta DEF$ is right and scalene.

Explanation:

$\angle D$ $\displaystyle \angle D$ and $\angle E$ $\displaystyle \angle E$ are complementary, so, by definition, $m\angle D + m\angle E = 90^{\circ }$ $\displaystyle m\angle D + m\angle E = 90^{\circ }$.

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

$m \angle F + m \angle D + m \angle E = 180^{\circ }$ $\displaystyle m \angle F + m \angle D + m \angle E = 180^{\circ }$

$m \angle F + 90^{\circ } = 180^{\circ }$ $\displaystyle m \angle F + 90^{\circ } = 180^{\circ }$

$m \angle F = 90^{\circ }$ $\displaystyle m \angle F = 90^{\circ }$

$\angle F$ $\displaystyle \angle F$ is a right angle, so $\Delta DEF$ $\displaystyle \Delta DEF$ is a right triangle.

$\angle D$ $\displaystyle \angle D$ and $\angle E$ $\displaystyle \angle E$ must be acute, so neither is congruent to $\angle F$ $\displaystyle \angle F$; also, $\angle D$ $\displaystyle \angle D$ and $\angle E$ $\displaystyle \angle E$ are not congruent to each other. Therefore, all three angles have different measure. Consequently, all three sides have different measure, and $\Delta DEF$ $\displaystyle \Delta DEF$ is scalene.

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Example Question #4 : Finding Angles

Decagon

The above figure is a regular decagon. Evaluate $m \angle ACB$ $\displaystyle m \angle ACB$.

Possible Answers:

$18^{\circ }$ $\displaystyle 18^{\circ }$

$15^{\circ }$ $\displaystyle 15^{\circ }$

$12^{\circ }$ $\displaystyle 12^{\circ }$

$20^{\circ }$ $\displaystyle 20^{\circ }$

$24^{\circ }$ $\displaystyle 24^{\circ }$

Correct answer:

$18^{\circ }$ $\displaystyle 18^{\circ }$

Explanation:

As an interior angle of a regular decagon, $\angle B$ $\displaystyle \angle B$ measures

$m \angle B = \left [\frac{180(10-2)}{10} \right ] ^{\circ }= 144^{\circ }$ $\displaystyle m \angle B = \left [\frac{180(10-2)}{10} \right ] ^{\circ }= 144^{\circ }$.

Since $\overline{AB}$ $\displaystyle \overline{AB}$ and $\overline{BC}$ $\displaystyle \overline{BC}$ are two sides of a regular polygon, they are congruent. Therefore, by the Isosceles Triangle Theorem,

$m \angle ACB = m \angle CAB$ $\displaystyle m \angle ACB = m \angle CAB$

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

$m \angle ACB + m \angle CAB + m \angle ABC =180^{\circ }$ $\displaystyle m \angle ACB + m \angle CAB + m \angle ABC =180^{\circ }$

$m \angle ACB + m \angle ACB + 144 ^{\circ } =180^{\circ }$ $\displaystyle m \angle ACB + m \angle ACB + 144 ^{\circ } =180^{\circ }$

$2 m \angle ACB + 144 ^{\circ } =180^{\circ }$ $\displaystyle 2 m \angle ACB + 144 ^{\circ } =180^{\circ }$

$2 m \angle ACB =36^{\circ }$ $\displaystyle 2 m \angle ACB =36^{\circ }$

$m \angle ACB =18^{\circ }$ $\displaystyle m \angle ACB =18^{\circ }$

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Example Question #5 : Finding Angles

Hexagon

The above hexagon is regular. What is $\displaystyle x$?

Possible Answers:

x = 69 $\displaystyle x = 69$

x = 79 $\displaystyle x = 79$

x = 64 $\displaystyle x = 64$

x = 74 $\displaystyle x = 74$

None of the other responses is correct.

Correct answer:

x = 79 $\displaystyle x = 79$

Explanation:

Two of the angles of the quadrilateral formed are angles of a regular hexagon, so each measures

$\frac{180^{\circ } (6-2)}{6}= 120^{\circ }$ $\displaystyle \frac{180^{\circ } (6-2)}{6}= 120^{\circ }$.

The four angles of the quadrilateral are $120 ^{\circ }, 120^{\circ }, 41^{\circ }, x^{\circ }$ $\displaystyle 120 ^{\circ }, 120^{\circ }, 41^{\circ }, x^{\circ }$. Their sum is $360^{\circ }$ $\displaystyle 360^{\circ }$, so we can set up, and solve for $\displaystyle x$ in, the equation:

$\displaystyle x + 120 + 120 + 41 =360$

$\displaystyle x + 281 =360$

x = 79 $\displaystyle x = 79$

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Example Question #61 : 2 Dimensional Geometry

What angle do the minute and hour hands of a clock form at 4:15?

Possible Answers:

$30 ^{\circ }$ $\displaystyle 30 ^{\circ }$

$40 ^{\circ }$ $\displaystyle 40 ^{\circ }$

$37\frac{1}{2} ^{\circ }$ $\displaystyle 37\frac{1}{2} ^{\circ }$

$32\frac{1}{2} ^{\circ }$ $\displaystyle 32\frac{1}{2} ^{\circ }$

$27\frac{1}{2} ^{\circ }$ $\displaystyle 27\frac{1}{2} ^{\circ }$

Correct answer:

$37\frac{1}{2} ^{\circ }$ $\displaystyle 37\frac{1}{2} ^{\circ }$

Explanation:

There are twelve numbers on a clock; from one to the next, a hand rotates $\left (\frac{360}{12} \right )^{\circ } = 30^{\circ }$ $\displaystyle \left (\frac{360}{12} \right )^{\circ } = 30^{\circ }$. At 4:15, the minute hand is exactly on the "3" - that is, on the $\left (30 \times 3 \right )^{\circ }= 90^{\circ }$ $\displaystyle \left (30 \times 3 \right )^{\circ }= 90^{\circ }$ position. The hour hand is one-fourth of the way from the "4" to the "5" - that is, on the $\left (4 \frac{1}{4} \times 30 \right )^{\circ } = 127 \frac{1}{2} ^{\circ }$ $\displaystyle \left (4 \frac{1}{4} \times 30 \right )^{\circ } = 127 \frac{1}{2} ^{\circ }$ position. Therefore, the difference is the angle they make:

$127 \frac{1}{2} ^{\circ } - 90^{\circ } = 37 \frac{1}{2}^{\circ }$ $\displaystyle 127 \frac{1}{2} ^{\circ } - 90^{\circ } = 37 \frac{1}{2}^{\circ }$.

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Example Question #62 : 2 Dimensional Geometry

If the vertical angles of intersecting lines are: 2x-9 $\displaystyle 2x-9$ and 3x+6 $\displaystyle 3x+6$, what must be the value of $\displaystyle x$?

Possible Answers:

$\displaystyle -15$

$\frac{183}{5}$ $\displaystyle \frac{183}{5}$

$\displaystyle -3$

$\frac{93}{5}$ $\displaystyle \frac{93}{5}$

$\displaystyle 3$

Correct answer:

$\displaystyle -15$

Explanation:

Vertical angles of intersecting lines are always equal.

Set the two expressions equal to each other and solve for $\displaystyle x$.

$\displaystyle 2x-9 = 3x+6$

Subtract $\displaystyle 2x$ from both sides.

$\displaystyle 2x-9 -2x= 3x+6-2x$

$\displaystyle -9 = x+6$

Subtract 6 from both sides.

$\displaystyle -9 -6= x+6-6$

The answer is: $\displaystyle -15$

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Example Question #63 : 2 Dimensional Geometry

If the angles in degrees are x+3 $\displaystyle x+3$ and x+7 $\displaystyle x+7$ which are complementary to each other, what is three times the value of the smallest angle?

Possible Answers:

129 $\displaystyle 129$

141 $\displaystyle 141$

264 $\displaystyle 264$

255 $\displaystyle 255$

120 $\displaystyle 120$

Correct answer:

129 $\displaystyle 129$

Explanation:

Complementary angles add up to 90 degrees.

Set up an equation such that the sum of both angles equal to 90.

$\displaystyle x+3+x+7 = 90$

$\displaystyle 2x+10 = 90$

Subtract 10 from both sides.

$\displaystyle 2x+10 -10= 90-10$

2x=80 $\displaystyle 2x=80$

Divide by 2 on both sides.

$\frac{2x}{2}=\frac{80}{2}$ $\displaystyle \frac{2x}{2}=\frac{80}{2}$

x=40 $\displaystyle x=40$

The angles are:

$\displaystyle x+3 = 40+3 = 43$

$\displaystyle x+7 = 40+7 = 47$

Three times the value of the smallest angle is:

$43\times 3 = 129$ $\displaystyle 43\times 3 = 129$

The answer is: 129 $\displaystyle 129$

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Example Question #64 : 2 Dimensional Geometry

If the angles -10x-5 $\displaystyle -10x-5$ and $\displaystyle 5x$ are supplementary, what must be the value of $\displaystyle x$?

Possible Answers:

$\frac{37}{3}$ $\displaystyle \frac{37}{3}$

$\displaystyle 37$

$-\frac{37}{3}$ $\displaystyle -\frac{37}{3}$

$\displaystyle -37$

$\displaystyle 19$

Correct answer:

$\displaystyle -37$

Explanation:

Supplementary angles sum up to 180 degrees.

$\displaystyle -10x-5+5x = 180$

$\displaystyle -5x-5 = 180$

Add five on both sides.

$\displaystyle -5x-5 +5= 180+5$

$\displaystyle -5x = 185$

Divide by negative five on both sides to determine $\displaystyle x$.

$\frac{-5x}{-5} = \frac{185}{-5}$ $\displaystyle \frac{-5x}{-5} = \frac{185}{-5}$

x=-37 $\displaystyle x=-37$

The answer is: $\displaystyle -37$

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SAT II Math II : Finding Angles

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #3 : Linear Functions

Example Question #1 : Finding Angles

Example Question #2 : Finding Angles

Example Question #3 : Finding Angles

Example Question #4 : Finding Angles

Example Question #5 : Finding Angles

Example Question #61 : 2 Dimensional Geometry

Example Question #62 : 2 Dimensional Geometry

Example Question #63 : 2 Dimensional Geometry

Example Question #64 : 2 Dimensional Geometry

All SAT II Math II Resources

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