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SAT II Math II : Finding Angles

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Example Question #1 : Angles

Solve for and .

(Figure not drawn to scale).

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

$\small 12x=180^o$

$\small x=15^o$

Because and are opposite angles, they must be equal.

$\small 2y=5x$

$\small x=15^o$

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

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

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Example Question #2 : Finding Angles

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

Possible Answers:

$90^{\circ }$

$95^{\circ }$

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #3 : Finding Angles

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

Possible Answers:

$\Delta ABC$ is acute and isosceles.

$\Delta ABC$ is obtuse and scalene.

$\Delta ABC$ is acute and scalene.

$\Delta ABC$ is obtuse and isosceles.

None of the other responses is correct.

Correct answer:

$\Delta ABC$ is acute and isosceles.

Explanation:

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

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

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

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

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

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

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Example Question #1 : Finding Angles

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

Possible Answers:

$\Delta DEF$ is right and isosceles.

$\Delta DEF$ is acute and scalene.

$\Delta DEF$ is right and scalene.

None of the other responses is correct.

$\Delta DEF$ is acute and isosceles.

Correct answer:

$\Delta DEF$ is right and scalene.

Explanation:

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

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

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

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

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

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

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

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Example Question #5 : Finding Angles

Decagon

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

Possible Answers:

$15^{\circ }$

$12^{\circ }$

$24^{\circ }$

$20^{\circ }$

$18^{\circ }$

Correct answer:

$18^{\circ }$

Explanation:

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

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

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

$m \angle ACB = m \angle CAB$

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

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

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

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

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

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

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Example Question #6 : Finding Angles

Hexagon

The above hexagon is regular. What is ?

Possible Answers:

x = 64

x = 69

None of the other responses is correct.

x = 74

x = 79

Correct answer:

x = 79

Explanation:

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

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

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

x + 120 + 120 + 41 =360

x + 281 =360

x = 79

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Example Question #7 : Finding Angles

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

Possible Answers:

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

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

$30 ^{\circ }$

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

$40 ^{\circ }$

Correct answer:

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

Explanation:

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

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

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Example Question #1 : Finding Angles

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

Possible Answers:

$\frac{183}{5}$

$\frac{93}{5}$

Correct answer:

Explanation:

Vertical angles of intersecting lines are always equal.

Set the two expressions equal to each other and solve for .

2x-9 = 3x+6

Subtract from both sides.

2x-9 -2x= 3x+6-2x

-9 = x+6

Subtract 6 from both sides.

-9 -6= x+6-6

The answer is:

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Example Question #1 : Finding Angles

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

Possible Answers:

129

255

264

120

141

Correct answer:

129

Explanation:

Complementary angles add up to 90 degrees.

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

x+3+x+7 = 90

2x+10 = 90

Subtract 10 from both sides.

2x+10 -10= 90-10

2x=80

Divide by 2 on both sides.

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

x=40

The angles are:

x+3 = 40+3 = 43

x+7 = 40+7 = 47

Three times the value of the smallest angle is:

$43\times 3 = 129$

The answer is: 129

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Example Question #1 : Finding Angles

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

Possible Answers:

$-\frac{37}{3}$

$\frac{37}{3}$

Correct answer:

Explanation:

Supplementary angles sum up to 180 degrees.

-10x-5+5x = 180

-5x-5 = 180

Add five on both sides.

-5x-5 +5= 180+5

-5x = 185

Divide by negative five on both sides to determine .

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

x=-37

The answer is:

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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 #1 : Angles

Example Question #2 : Finding Angles

Example Question #3 : Finding Angles

Example Question #1 : Finding Angles

Example Question #5 : Finding Angles

Example Question #6 : Finding Angles

Example Question #7 : Finding Angles

Example Question #1 : Finding Angles

Example Question #1 : Finding Angles

Example Question #1 : Finding Angles

All SAT II Math II Resources

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