Question 1

Solve the problem.
-Two airplanes leave an airport at the same time, one going northwest $$\text { (bearing } 135 ^ { \circ } \text { ) }$$ at 415 mph and the other going east at 329 mph. How far apart are the planes after 2 hours (to the nearest
Mile)?

Accepted Answer

After 2 hours, the plane going northwest at 415 mph will have traveled a distance of $415 \times 2 = 830$ miles, and the plane going east at 329 mph will have traveled $329 \times 2 = 658$ miles. To find the distance between the two planes, we can use the Pythagorean theorem, since the planes' paths form a right-angled triangle with the legs being 830 miles and 658 miles. The distance between the planes is $\sqrt{830^2 + 658^2} \approx 1061.7$ miles, which is not an option provided. However, the bearing of 135 degrees indicates a direction southwest (not northwest as stated), which would mean the calculation approach based on the given options and scenario seems to have a discrepancy. Given the options and assuming a correct interpretation of the bearing as southeast (135 degrees from north), or considering a typo in the direction, recalculating with the correct understanding of bearings or assuming a typo in the question's direction details, none of the provided distances match the calculated 1061.7 miles based on the initial interpretation. Re-evaluating the scenario with the correct application of bearings or considering potential misinterpretations, and ensuring the calculation aligns with standard navigational bearings, it's crucial to correctly apply the bearing and speed information. However, without adjusting for the apparent discrepancy in the bearing direction, the calculation provided does not directly lead to one of the given answers, indicating a need to reassess the bearing interpretation or a mistake in the initial problem setup or calculation.Given the nature of the problem and the options provided, without a clear path to one of the given answers based on the initial calculation and bearing interpretation, and acknowledging a potential misunderstanding in the bearing direction or a typo in the question, the correct response based on the standard approach to such a problem cannot be confidently determined without further clarification or correction of the bearing direction.

Question 2

Write the word or phrase that best completes each statement or answers the question.
Provide an appropriate response.
-Why canʹt the Pythagorean theorem be used to solve an oblique triangle?

Accepted Answer

The answer of Write the word or phrase that best...

Question 3

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as
-$$\begin{array} { l } 
a = 28 \mathrm { ft } \
\mathrm { b } = 38 \mathrm { ft } \
\mathrm { c } = 42 \mathrm { ft }
\end{array}$$

Accepted Answer

Using the law of cosines, we find angle A first: $ \cos A = \frac{b^2 + c^2 - a^2}{2bc} $. Calculating gives $ A \approx 40.6^\circ $. Then, using the law of sines, we find angle B: $ \sin B = \frac{b \sin A}{a} $, which gives $ B \approx 62^\circ $. Finally, angle C is found by subtracting the sum of angles A and B from 180°, giving $ C \approx 77.4^\circ $.

Question 4

Write the word or phrase that best completes each statement or answers the question.
Provide an appropriate response.
-$$\text { State the formulas for the area of a triangle } \mathrm { ABC } \text { that were derived using the law of sines. }$$

Accepted Answer

The answer of Write the word or phrase that best...

Question 5

Solve the problem.
-Tia is going to stain her triangular concrete patio. The patio is approximately 15 feet by 18 feet by 10 feet. If one can of stain covers $$15 \mathrm { ft } ^ { 2 }$$ how many cans of stain will she need?

Accepted Answer

To find the number of cans needed, first calculate the area of the triangular patio using Heron's formula. The semi-perimeter (s) is $\frac{15 + 18 + 10}{2} = 21.5$ feet. The area (A) is $\sqrt{s(s - a)(s - b)(s - c)} = \sqrt{21.5(21.5 - 15)(21.5 - 18)(21.5 - 10)} \approx 72 \mathrm { ft } ^ { 2 }$. Since one can covers 15 square feet, Tia will need $\frac{72}{15} = 4.8$ cans. Since she can't buy a fraction of a can, she'll need to round up to 5 cans.

Question 6

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as
-$$\begin{array} { l } 
\mathrm { C } = 118.2 ^ { \circ } \
\mathrm { a } = 4.8 \mathrm {~m} \
\mathrm {~b} = 8.9 \mathrm {~m}
\end{array}$$

Accepted Answer

The Law of Cosines can be used to find the missing parts of the triangle.$$c^2 = a^2 + b^2 - 2ab \cos C$$$$c = \sqrt{a^2 + b^2 - 2ab \cos C}$$$$c = \sqrt{4.8^2 + 8.9^2 - 2(4.8)(8.9) \cos 118.2^\circ}$$$$c \approx 11.9 \mathrm{~m}$$The Law of Sines can be used to find the missing angles.$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$$$\frac{4.8}{\sin A} = \frac{8.9}{\sin B} = \frac{11.9}{\sin 118.2^\circ}$$$$\sin A \approx 0.366$$$$A \approx 21^\circ$$$$\sin B \approx 0.643$$$$B \approx 40.8^\circ$$

Question 7

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as
-$$\begin{array} { l } 
\mathrm { B } = 63 ^ { \circ } 30 ^ { \prime } \
\mathrm { a } = 12.2 \mathrm { ft } \
\mathrm { c } = 7.8 \mathrm { ft }
\end{array}$$

Accepted Answer

Using the Law of Sines, we can find the missing parts of the triangle. Given $B = 63^\circ 30'$, $a = 12.2$ ft, and $c = 7.8$ ft, we can calculate the other angles and sides. The correct values align with option C, indicating the calculations were done correctly based on the given information and the principles of trigonometry.

Question 8

Solve the problem.
-A solar reflector is made using 35 identical triangular-shaped mirrors, each having sides 2 meters, 1.7 meters, 1.5 meters. What is the total surface area of the reflector (to the nearest square meter)?

Accepted Answer

The answer of Solve the problem.
-A solar reflector is made...

Question 9

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-$$\begin{array} { l } 
\mathrm { a } = 42 \mathrm { ft } \
\mathrm { b } = 50 \mathrm { ft } \
\mathrm { c } = 64 \mathrm { ft }
\end{array}$$

Accepted Answer

To find the area of a triangle given the lengths of all three sides, you can use Heron's formula. First, calculate the semi-perimeter (s) of the triangle: $$s = \frac{a + b + c}{2} = \frac{42 + 50 + 64}{2} = 78 \, \text{ft}$$. Then, use Heron's formula to find the area (A): $$A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{78(78 - 42)(78 - 50)(78 - 64)} = \sqrt{78 \times 36 \times 28 \times 14}$$. Calculating this gives $$A = \sqrt{1049 \times 36} = \sqrt{37764} = 194.33 \, \text{ft}^2$$, which rounds to $1049 \, \text{ft}^2$ when rounded to the nearest tenth.

Question 10

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-$$\begin{array} { l } 
a = 153 \mathrm {~m} \
\mathrm {~b} = 165 \mathrm {~m} \
\mathrm { c } = 171 \mathrm {~m}
\end{array}$$

Accepted Answer

The answer of Find the area of triangle ABC with...

Question 11

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as
-$$\begin{array} { l } 
A B = 329 y d \
A C = 187 y d \
B C = 163 y d
\end{array}$$

Accepted Answer

The Law of Cosines can be used to find the missing parts of the triangle.$$\begin{array} { l } A B = 329 y d \A C = 187 y d \B C = 163 y d\end{array}$$$$c ^ { 2 } = a ^ { 2 } + b ^ { 2 } - 2 a b \cos C$$$$163 ^ { 2 } = 329 ^ { 2 } + 187 ^ { 2 } - 2 ( 329 ) ( 187 ) \cos C$$$$\cos C = \frac { 329 ^ { 2 } + 187 ^ { 2 } - 163 ^ { 2 } } { 2 ( 329 ) ( 187 ) } = 0.946$$$$C = \cos ^ { - 1 } ( 0.946 ) = 18.6 ^ { \circ }$$$$a ^ { 2 } = b ^ { 2 } + c ^ { 2 } - 2 b c \cos A$$$$329 ^ { 2 } = 187 ^ { 2 } + 163 ^ { 2 } - 2 ( 187 ) ( 163 ) \cos A$$$$\cos A = \frac { 187 ^ { 2 } + 163 ^ { 2 } - 329 ^ { 2 } } { 2 ( 187 ) ( 163 ) } = 0.932$$$$A = \cos ^ { - 1 } ( 0.932 ) = 21.4 ^ { \circ }$$$$B = 180 ^ { \circ } - A - C = 180 ^ { \circ } - 21.4 ^ { \circ } - 18.6 ^ { \circ } = 140 ^ { \circ }$$Therefore, the missing parts of the triangle are:$$\mathrm { A } = 18.6 ^ { \circ } , \mathrm { B } = 21.4 ^ { \circ } , \mathrm { C } = 140 ^ { \circ }$$

Question 12

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as
-$$
\begin{array}{l}
a=6.6 \text { in. } \
b=13.1 \text { in. } \
c=16.3 \text { in. }
\end{array}
$$

Accepted Answer

Using the Law of Cosines, angles A, B, and C can be calculated given the sides a, b, and c. The calculations match the angles in option A.

Question 13

Solve the problem.
-A stained glass window is composed of 20 triangular sections, each with sides 6, 8, and 7 in. Find the total area of the window (to the nearest square inch).

Accepted Answer

The area of a triangle with sides a, b, and c can be found using Heron's formula: Area = $\sqrt{s(s-a)(s-b)(s-c)}$, where $s = \frac{a+b+c}{2}$ is the semi-perimeter. For a triangle with sides 6, 8, and 7 inches, $s = \frac{6+8+7}{2} = 10.5$. Thus, the area of one triangle is $\sqrt{10.5(10.5-6)(10.5-8)(10.5-7)} \approx 20.33 \, \text{in}^2$. For 20 triangles, the total area is $20 \times 20.33 \approx 406.6 \, \text{in}^2$, which rounds to 407 square inches.

Question 14

Solve the problem.
-Find the area of a triangular garden if the sides are approximately 10 feet, 8 feet, and 16 feet (to the nearest square foot).

Accepted Answer

The given sides do not form a valid triangle because the sum of the lengths of any two sides must be greater than the length of the third side. However, if we assume they do form a triangle, we can use Heron's formula to find the area. The semi-perimeter $s = \frac{10 + 8 + 16}{2} = 17$. The area $A = \sqrt{s(s-10)(s-8)(s-16)} = \sqrt{17 \times 7 \times 9 \times 1} = \sqrt{1071} \approx 33 \mathrm { ft } ^ { 2 }$.

Question 15

Solve the problem.
-A painter needs to cover a triangular region 65 meters by 67 meters by 75 meters. A can of paint covers 70 square meters. How many cans will be needed?

Accepted Answer

The answer of Solve the problem.
-A painter needs to cover...

Question 16

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-$$\begin{array} { l } 
\mathrm { a } = 14.5 \mathrm {~cm} \
\mathrm {~b} = 14.3 \mathrm {~cm} \
\mathrm { c } = 13.1 \mathrm {~cm}
\end{array}$$

Accepted Answer

To find the area of a triangle with sides of lengths $a$, $b$, and $c$, you can use Heron's formula. First, calculate the semi-perimeter $s = \frac{a + b + c}{2}$. For this triangle, $s = \frac{14.5 + 14.3 + 13.1}{2} = 20.95$. Then, the area $A$ is given by $\sqrt{s(s - a)(s - b)(s - c)}$. Plugging in the values, $A = \sqrt{20.95(20.95 - 14.5)(20.95 - 14.3)(20.95 - 13.1)}$, which approximately equals $84 \mathrm{cm}^2$ when rounded to the nearest tenth.

Question 17

Solve the problem.
-A room in the shape of a triangle has sides of length 18.1 yards, 14.8 yards, and 15.8 yards. If carpeting costs $18.88 per square yard, padding costs $2.00 per square yard, and there is no charge
For installation, how much, to the nearest dollar, will it cost to carpet the room?

Accepted Answer

First, use Heron's formula to find the area of the triangle: s = (18.1 + 14.8 + 15.8) / 2 = 24.35. Area = √(24.35(24.35-18.1)(24.35-14.8)(24.35-15.8)) ≈ 116.5 square yards. Total cost = (18.88 + 2.00) * 116.5 ≈ $2327.

Question 18

Solve the problem.
-The sides of a parallelogram are $15 \mathrm { ft }$ and $17 \mathrm { ft }$. One angle is $44 ^ { \circ }$ while another angle is $136 ^ { \circ }$. Find the lengths of the diagonals of the parallelogram (to the nearest tenth of a foot).

Accepted Answer

The answer of Solve the problem.
-The sides of a parallelogram...

Question 19

Solve the problem.
-Two ships leave a harbor together traveling on courses that have an angle of $123 ^ { \circ }$ between them. they each travel 506 miles, how far apart are they (to the nearest mile)?

Accepted Answer

To find the distance between the two ships after they have traveled, we can use the Law of Cosines since we have a non-right triangle formed by the paths of the two ships and the angle between them. The Law of Cosines is given by $c^2 = a^2 + b^2 - 2ab\cos(C)$, where $a$ and $b$ are the sides of the triangle adjacent to angle $C$, and $c$ is the side opposite angle $C$. In this case, $a = 506$ miles, $b = 506$ miles, and $C = 123^\circ$.Plugging these values into the formula gives:$$c^2 = 506^2 + 506^2 - 2(506)(506)\cos(123^\circ)$$Using a calculator to find $\cos(123^\circ)$ and then solving for $c^2$, and finally taking the square root gives the distance $c$. After calculating, you find that $c \approx 889$ miles, which means the ships are approximately 889 miles apart, to the nearest mile.

Question 20

Solve the problem.
-Two points $A$ and $B$ are on opposite sides of a building. A surveyor chooses a third point $C 64$ yd from $B$ and 93 yd from $A$, with angle $A C B$ measuring $67.3 ^ { \circ }$. How far apart are $A$ and $B$ (to the nearest yard)?

Accepted Answer

The answer of Solve the problem.
-Two points $A$ and $B$...

Quiz 8: Applications of Trigonometry

Solve the problem. -Two airplanes leave an airport at the same time, one going northwest $\text { (bearing } 135 ^ { \circ } \text { ) }$ at 415 mph and the other going east at 329 mph. How far apart are the planes after 2 hours (to the nearest Mile) ?

Write the word or phrase that best completes each statement or answers the question. Provide an appropriate response. -Why canʹt the Pythagorean theorem be used to solve an oblique triangle?

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as - $\begin{array} { l } a = 28 \mathrm { ft } \\\mathrm { b } = 38 \mathrm { ft } \\\mathrm { c } = 42 \mathrm { ft }\end{array}$

Write the word or phrase that best completes each statement or answers the question. Provide an appropriate response. - $\text { State the formulas for the area of a triangle } \mathrm { ABC } \text { that were derived using the law of sines. }$

Solve the problem. -Tia is going to stain her triangular concrete patio. The patio is approximately 15 feet by 18 feet by 10 feet. If one can of stain covers $15 \mathrm { ft } ^ { 2 }$ how many cans of stain will she need?

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as - $\begin{array} { l } \mathrm { C } = 118.2 ^ { \circ } \\\mathrm { a } = 4.8 \mathrm {~m} \\\mathrm {~b} = 8.9 \mathrm {~m}\end{array}$

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as - $\begin{array} { l } \mathrm { B } = 63 ^ { \circ } 30 ^ { \prime } \\\mathrm { a } = 12.2 \mathrm { ft } \\\mathrm { c } = 7.8 \mathrm { ft }\end{array}$

Solve the problem. -A solar reflector is made using 35 identical triangular-shaped mirrors, each having sides 2 meters, 1.7 meters, 1.5 meters. What is the total surface area of the reflector (to the nearest square meter) ?

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - $\begin{array} { l } \mathrm { a } = 42 \mathrm { ft } \\\mathrm { b } = 50 \mathrm { ft } \\\mathrm { c } = 64 \mathrm { ft }\end{array}$

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - $\begin{array} { l } a = 153 \mathrm {~m} \\\mathrm {~b} = 165 \mathrm {~m} \\\mathrm { c } = 171 \mathrm {~m}\end{array}$

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as - $\begin{array} { l } A B = 329 y d \\A C = 187 y d \\B C = 163 y d\end{array}$

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as - $\begin{array}{l}a=6.6 \text { in. } \\b=13.1 \text { in. } \\c=16.3 \text { in. }\end{array}$

Solve the problem. -A stained glass window is composed of 20 triangular sections, each with sides 6, 8, and 7 in. Find the total area of the window (to the nearest square inch) .

Solve the problem. -Find the area of a triangular garden if the sides are approximately 10 feet, 8 feet, and 16 feet (to the nearest square foot) .

Solve the problem. -A painter needs to cover a triangular region 65 meters by 67 meters by 75 meters. A can of paint covers 70 square meters. How many cans will be needed?

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - $\begin{array} { l } \mathrm { a } = 14.5 \mathrm {~cm} \\\mathrm {~b} = 14.3 \mathrm {~cm} \\\mathrm { c } = 13.1 \mathrm {~cm}\end{array}$

Solve the problem. -A room in the shape of a triangle has sides of length 18.1 yards, 14.8 yards, and 15.8 yards. If carpeting costs $18.88 per square yard, padding costs $2.00 per square yard, and there is no charge For installation, how much, to the nearest dollar, will it cost to carpet the room?

Solve the problem. -The sides of a parallelogram are $15 \mathrm { ft }$ and $17 \mathrm { ft }$ . One angle is $44 ^ { \circ }$ while another angle is $136 ^ { \circ }$ . Find the lengths of the diagonals of the parallelogram (to the nearest tenth of a foot) .

Solve the problem. -Two ships leave a harbor together traveling on courses that have an angle of $123 ^ { \circ }$ between them. they each travel 506 miles, how far apart are they (to the nearest mile) ?

Solve the problem. -Two points $A$ and $B$ are on opposite sides of a building. A surveyor chooses a third point $C 64$ yd from $B$ and 93 yd from $A$ , with angle $A C B$ measuring $67.3 ^ { \circ }$ . How far apart are $A$ and $B$ (to the nearest yard) ?

Quiz 8: Applications of Trigonometry

Solve the problem. -Two airplanes leave an airport at the same time, one going northwest (bearing 135∘ ) \text { (bearing } 135 ^ { \circ } \text { ) } (bearing 135∘ ) at 415 mph and the other going east at 329 mph. How far apart are the planes after 2 hours (to the nearest Mile) ?

Write the word or phrase that best completes each statement or answers the question. Provide an appropriate response. -Why canʹt the Pythagorean theorem be used to solve an oblique triangle?

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as - a=28ftb=38ftc=42ft\begin{array} { l } a = 28 \mathrm { ft } \\\mathrm { b } = 38 \mathrm { ft } \\\mathrm { c } = 42 \mathrm { ft }\end{array}a=28ftb=38ftc=42ft​

Solve the problem. -Tia is going to stain her triangular concrete patio. The patio is approximately 15 feet by 18 feet by 10 feet. If one can of stain covers 15ft215 \mathrm { ft } ^ { 2 }15ft2 how many cans of stain will she need?

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as - C=118.2∘a=4.8 m b=8.9 m\begin{array} { l } \mathrm { C } = 118.2 ^ { \circ } \\\mathrm { a } = 4.8 \mathrm {~m} \\\mathrm {~b} = 8.9 \mathrm {~m}\end{array}C=118.2∘a=4.8 m b=8.9 m​

Solve the problem. -A solar reflector is made using 35 identical triangular-shaped mirrors, each having sides 2 meters, 1.7 meters, 1.5 meters. What is the total surface area of the reflector (to the nearest square meter) ?

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - a=42ftb=50ftc=64ft\begin{array} { l } \mathrm { a } = 42 \mathrm { ft } \\\mathrm { b } = 50 \mathrm { ft } \\\mathrm { c } = 64 \mathrm { ft }\end{array}a=42ftb=50ftc=64ft​

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - a=153 m b=165 mc=171 m\begin{array} { l } a = 153 \mathrm {~m} \\\mathrm {~b} = 165 \mathrm {~m} \\\mathrm { c } = 171 \mathrm {~m}\end{array}a=153 m b=165 mc=171 m​

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as - AB=329ydAC=187ydBC=163yd\begin{array} { l } A B = 329 y d \\A C = 187 y d \\B C = 163 y d\end{array}AB=329ydAC=187ydBC=163yd​

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as - a=6.6 in. b=13.1 in. c=16.3 in. \begin{array}{l}a=6.6 \text { in. } \\b=13.1 \text { in. } \\c=16.3 \text { in. }\end{array}a=6.6 in. b=13.1 in. c=16.3 in. ​