SAT Math - Finding Angles

Hexagon

The above hexagon is regular. What is ?

None of the other responses is correct.

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:

If the angles and are supplementary, what must be the value of ?

$-\frac{37}{3}$

$\frac{37}{3}$

Explanation

Supplementary angles sum up to 180 degrees.

Add five on both sides.

Divide by negative five on both sides to determine .

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

The answer is:

If a set of angles are supplementary, what must be the other angle if a given angle is ?

$145^\circ$

$215^\circ$

$155^\circ$

$135^\circ$

$55^\circ$

Explanation

Supplementary angles must add up to 180 degrees.

To find the missing angle, subtract the known angle from 180 degrees.

The answer is: $145^\circ$

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

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

$90^{\circ }$

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

$95^{\circ }$

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

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:

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

If two angles of a triangle are $60^{\circ}$ and $75.5^{\circ}$ , find the measurement of the third angle.

$44.5^{\circ}$

$45^{\circ}$

$180^{\circ}$

Explanation

Step 1: Recall the sum of the angles of a triangle...

The sum of the internal angles of a triangle is $180^{\circ}$ .

Step 2: To find the missing angle, subtract the given angles from $180^{\circ}$ ...

$180^{\circ}-60^{\circ}-75.5^{\circ}=44.5^{\circ}$

Suppose a set of intersecting lines. If an angle is 120 degrees, what must be the sum of the adjacent angle and the vertical angle to the given angle?

$180^\circ$

$200^\circ$

$220^\circ$

$360^\circ$

$240^\circ$

Explanation

In an intersecting pair of lines, recall that vertical angles will always equal.

$\textup{Vertical angle } = 120$

The adjacent angle with the given angle will form a straight line, and both of the angles must sum to 180 degrees.

Subtract 120 from 180 to get the adjacent angle.

Sum the two angles.

The answer is:

Hexagon

Note: Figure NOT drawn to scale.

The above hexagon is regular. What is ?

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 $120 ^{\circ }, 120^{\circ } , x^{\circ }, (x - 12)^{\circ }$ . Their sum is $360^{\circ }$ , so we can set up, and solve for in, the equation:

Solve for and .

(Figure not drawn to scale).

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

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

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

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

Explanation

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

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

$\small 12x=180^o$

$\small x=15^o$

Because and are opposite angles, they must be equal.

$\small 2y=5x$

$\small x=15^o$

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

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

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

$\Delta DEF$ is right and scalene.

$\Delta DEF$ is acute and scalene.

$\Delta DEF$ is acute and isosceles.

$\Delta DEF$ is right and isosceles.

None of the other responses is correct.

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.

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

$\Delta ABC$ is acute and isosceles.

$\Delta ABC$ is acute and scalene.

$\Delta ABC$ is obtuse and scalene.

$\Delta ABC$ is obtuse and isosceles.

None of the other responses is correct.

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 $m \angle A= m \angle C = 50^{\circ }$ , ; the triangle has two congruent sides and is isosceles.

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