Question 1

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-A solar reflector is made using 32 identical triangular-shaped mirrors, each having sides $1.7$ meters, $2.3$ meters, $1.4$ meters. What is the total surface area of the reflector (to the nearest square meter)?

Accepted Answer

To find the area of a triangle with sides of lengths $a$, $b$, and $c$, we can use Heron's formula, which is $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semi-perimeter of the triangle, $s = \frac{a+b+c}{2}$. For the given triangle, $a = 1.7$ meters, $b = 2.3$ meters, and $c = 1.4$ meters, so the semi-perimeter $s = \frac{1.7+2.3+1.4}{2} = 2.7$ meters. Plugging these values into Heron's formula gives $A = \sqrt{2.7(2.7-1.7)(2.7-2.3)(2.7-1.4)} \approx 1.183$ square meters for one triangular mirror. The total surface area for 32 mirrors is $32 \times 1.183 \approx 37.856$, which rounds to $38$ square meters.

Question 2

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate.
-$$\begin{array} { l } 
\mathrm { AB } = 324 \mathrm { yd } \
\mathrm { AC } = 180 \mathrm { yd } \
\mathrm { BC } = 169 \mathrm { yd }
\end{array}$$

Accepted Answer

Using the Law of Cosines, we can calculate the angles of the triangle. Given the sides, the calculations will show that option C has the correct angles.

Question 3

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-Find the area of a triangular garden if the sides are approximately 6 feet, 11 feet, and 12 feet (to the nearest square foot).

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: $s = \frac{a + b + c}{2}$, where $a$, $b$, and $c$ are the lengths of the sides. For this triangle, $s = \frac{6 + 11 + 12}{2} = 14.5$ feet. Then, use Heron's formula to find the area ($A$): $A = \sqrt{s(s - a)(s - b)(s - c)}$. Plugging in the values, $A = \sqrt{14.5(14.5 - 6)(14.5 - 11)(14.5 - 12)} \approx 33 \mathrm { ft } ^ { 2 }$.

Question 4

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-A room in the shape of a triangle has sides of length $17.8$ yards, $14.1$ yards, and $13.4$ yards. If carpeting costs $\$ 20.83$ per square yard, padding costs $\$ 1.60$ 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

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

Question 5

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate.
-$$\begin{array} { l } 
\mathrm { C } = 105 ^ { \circ } \
\mathrm { a } = 6.7 \mathrm {~m} \
\mathrm {~b} = 9.3 \mathrm {~m}
\end{array}$$

Accepted Answer

Using the law of cosines, $ c = \sqrt{a^2 + b^2 - 2ab \cdot \cos(C)} $, we find $ c \approx 12.8 \, \text{m} $. Then, using the law of sines, $\sin A / a = \sin C / c$, we find $ A \approx 30.4^\circ $ and $ B = 180^\circ - C - A \approx 44.6^\circ $.

Question 6

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-A stained glass window is composed of 20 triangular sections, each with sides 4,7 , and 8 in. Find the total area of the window (to the nearest square inch).

Accepted Answer

To find the area of a triangle with sides 4, 7, and 8 inches, we can use Heron's formula. First, calculate the semi-perimeter (s) as $s = \frac{4 + 7 + 8}{2} = 9.5$. Then, the area ($A$) of one triangle is given by $\sqrt{s(s - 4)(s - 7)(s - 8)}$, which simplifies to $\sqrt{9.5 \times 5.5 \times 2.5 \times 1.5} \approx 14 \, \text{in}^2$. Since the window is composed of 20 such triangles, the total area is $20 \times 14 = 280 \, \text{in}^2$.

Question 7

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate.
-$$\begin{array} { l } 
a = 7.2 \mathrm { in } . \
\mathrm { b } = 13.7 \mathrm { in } . \
\mathrm { c } = 15.6 \mathrm { in } .
\end{array}$$

Accepted Answer

To find the angles of the triangle, we can use the Law of Cosines. Given the sides $a$, $b$, and $c$, the angles opposite these sides can be found using the formula $\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}$, and similarly for angles $B$ and $C$. Calculating these for the given side lengths and converting the cosine values to degrees will give the angles as in option D.

Question 8

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-Tia is going to stain her triangular concrete patio. The patio is approximately 14 feet by 18 feet by 10 feet. If one can of stain covers $12 \mathrm { ft } ^ { 2 }$, how many cans of stain will she need?

Accepted Answer

To find the area of a triangle given three sides, we can use Heron's formula. First, calculate the semi-perimeter (s) as $s = \frac{a + b + c}{2}$, where $a$, $b$, and $c$ are the lengths of the sides. Here, $s = \frac{14 + 18 + 10}{2} = 21$. Then, the area ($A$) can be found using $A = \sqrt{s(s - a)(s - b)(s - c)}$, which gives $A = \sqrt{21(21 - 14)(21 - 18)(21 - 10)} = \sqrt{21 \times 7 \times 3 \times 11} = \sqrt{4851}$ square feet. The exact area is approximately 69.6 square feet. Since one can covers $12 \mathrm { ft } ^ { 2 }$, she needs $\frac{69.6}{12} \approx 5.8$ cans. Rounding up, Tia needs 6 cans of stain.

Question 9

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-$$\begin{array} { l } 
a = 18.6 \mathrm {~cm} \
b = 17.2 \mathrm {~cm} \
c = 15.0 \mathrm {~cm}
\end{array}$$

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: $s = \frac{a + b + c}{2}$. Then, use Heron's formula: $\sqrt{s(s - a)(s - b)(s - c)}$. Plugging in the given values: $s = \frac{18.6 + 17.2 + 15.0}{2} = 25.4$. The area is then $\sqrt{25.4(25.4 - 18.6)(25.4 - 17.2)(25.4 - 15.0)} \approx 121 \mathrm{cm}^2$.

Question 10

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-Two airplanes leave an airport at the same time, one going northwest (bearing $135 ^ { \circ }$ ) at $410 \mathrm { mph }$ and the other going east at $333 \mathrm { mph }$. How far apart are the planes after 4 hours (to the nearest mile)?

Accepted Answer

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

Question 11

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-A painter needs to cover a triangular region 63 meters by 67 meters by 71 meters. A can of paint covers 70 square meters. How many cans will be needed?

Accepted Answer

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

Question 12

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-The sides of a parallelogram are $14 \mathrm { ft }$ and $16 \mathrm { ft }$. One angle is $41 ^ { \circ }$ while another angle is $139 ^ { \circ }$. Find the lengths of the diagonals of the parallelogram (to the nearest tenth of a foot).

Accepted Answer

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

Question 13

Provide an appropriate response.
-Why can't the Pythagorean theorem be used to solve an oblique triangle?

Accepted Answer

The answer of Provide an appropriate response.
-Why can't the Pythagorean...

Question 14

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-Two ships leave a harbor together traveling on courses that have an angle of $128 ^ { \circ }$ between them. If they each travel 543 miles, how far apart are they (to the nearest mile)?

Accepted Answer

Using the Law of Cosines, $c^2 = a^2 + b^2 - 2ab\cos(C)$, where $a = 543$, $b = 543$, and $C = 128^\circ$, we find the distance between the ships to be approximately 976 miles.

Question 15

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-$$\begin{array} { l } 
a = 47 \mathrm { ft } \
\mathrm { b } = 59 \mathrm { ft } \
\mathrm { c } = 65 \mathrm { ft }
\end{array}$$

Accepted Answer

Using Heron's formula, first calculate the semi-perimeter $s = \frac{a + b + c}{2} = \frac{47 + 59 + 65}{2} = 85.5$ ft. Then, the area $A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{85.5(85.5 - 47)(85.5 - 59)(85.5 - 65)} \approx 1337 \mathrm { ft } ^ { 2 }$.

Question 16

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-Two points A and B are on opposite sides of a building. A surveyor chooses a third point C 67 yd from B and 104 yd from A, with angle ACB measuring 68.3°. How far apart are A and B (to the
Nearest yard)?

Accepted Answer

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

Question 17

Provide an appropriate response.
-State the formulas for the area of a triangle ABC that were derived using the law of sines.

Accepted Answer

The answer of Provide an appropriate response.
-State the formulas for...

Question 18

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate.
-$$\begin{array} { l } 
a = 20 \mathrm { ft } \
b = 38 \mathrm { ft } \
c = 42 \mathrm { ft }
\end{array}$$

Accepted Answer

Using the law of cosines, we find angle C first, then use the law of sines to find angles A and B. The angles add up to 180 degrees, confirming the triangle.

Question 19

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

Accepted Answer

To solve this problem, the Law of Sines can be used, which states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. Given $B = 63^\circ 30'$, $a = 12.2$ ft, and $c = 7.8$ ft, we can find angle $A$ and side $b$ using the Law of Sines, and then angle $C$ by subtracting the sum of angles $A$ and $B$ from $180^\circ$. The correct values are $b = 11.2$ ft, $A = 77^\circ 49'$, and $C = 38^\circ 41'$, which match choice A.

Question 20

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-$$\begin{array} { l } 
a = 150 \mathrm {~m} \
\mathrm {~b} = 164 \mathrm {~m} \
\mathrm { c } = 172 \mathrm {~m}
\end{array}$$

Accepted Answer

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

Quiz 9: Applications of Trigonometry

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate. - $\begin{array} { l } \mathrm { AB } = 324 \mathrm { yd } \\\mathrm { AC } = 180 \mathrm { yd } \\\mathrm { BC } = 169 \mathrm { yd }\end{array}$

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. -Find the area of a triangular garden if the sides are approximately 6 feet, 11 feet, and 12 feet (to the nearest square foot) .

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate. - $\begin{array} { l } \mathrm { C } = 105 ^ { \circ } \\\mathrm { a } = 6.7 \mathrm {~m} \\\mathrm {~b} = 9.3 \mathrm {~m}\end{array}$

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. -A stained glass window is composed of 20 triangular sections, each with sides 4,7 , and 8 in. Find the total area of the window (to the nearest square inch) .

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate. - $\begin{array} { l } a = 7.2 \mathrm { in } . \\\mathrm { b } = 13.7 \mathrm { in } . \\\mathrm { c } = 15.6 \mathrm { in } .\end{array}$

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - $\begin{array} { l } a = 18.6 \mathrm {~cm} \\b = 17.2 \mathrm {~cm} \\c = 15.0 \mathrm {~cm}\end{array}$

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. -A painter needs to cover a triangular region 63 meters by 67 meters by 71 meters. A can of paint covers 70 square meters. How many cans will be needed?

Provide an appropriate response. -Why can't the Pythagorean theorem be used to solve an oblique triangle?

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. -Two ships leave a harbor together traveling on courses that have an angle of $128 ^ { \circ }$ between them. If they each travel 543 miles, how far apart are they (to the nearest mile) ?

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - $\begin{array} { l } a = 47 \mathrm { ft } \\\mathrm { b } = 59 \mathrm { ft } \\\mathrm { c } = 65 \mathrm { ft }\end{array}$

Provide an appropriate response. -State the formulas for the area of a triangle ABC that were derived using the law of sines.

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate. - $\begin{array} { l } a = 20 \mathrm { ft } \\b = 38 \mathrm { ft } \\c = 42 \mathrm { ft }\end{array}$

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

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - $\begin{array} { l } a = 150 \mathrm {~m} \\\mathrm {~b} = 164 \mathrm {~m} \\\mathrm { c } = 172 \mathrm {~m}\end{array}$

Quiz 9: Applications of Trigonometry

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. -Find the area of a triangular garden if the sides are approximately 6 feet, 11 feet, and 12 feet (to the nearest square foot) .

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. -A stained glass window is composed of 20 triangular sections, each with sides 4,7 , and 8 in. Find the total area of the window (to the nearest square inch) .

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - a=18.6 cmb=17.2 cmc=15.0 cm\begin{array} { l } a = 18.6 \mathrm {~cm} \\b = 17.2 \mathrm {~cm} \\c = 15.0 \mathrm {~cm}\end{array}a=18.6 cmb=17.2 cmc=15.0 cm​

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. -A painter needs to cover a triangular region 63 meters by 67 meters by 71 meters. A can of paint covers 70 square meters. How many cans will be needed?

Provide an appropriate response. -Why can't the Pythagorean theorem be used to solve an oblique triangle?

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. -Two ships leave a harbor together traveling on courses that have an angle of 128∘128 ^ { \circ }128∘ between them. If they each travel 543 miles, how far apart are they (to the nearest mile) ?

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - a=47ftb=59ftc=65ft\begin{array} { l } a = 47 \mathrm { ft } \\\mathrm { b } = 59 \mathrm { ft } \\\mathrm { c } = 65 \mathrm { ft }\end{array}a=47ftb=59ftc=65ft​

Provide an appropriate response. -State the formulas for the area of a triangle ABC that were derived using the law of sines.

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate. - a=20ftb=38ftc=42ft\begin{array} { l } a = 20 \mathrm { ft } \\b = 38 \mathrm { ft } \\c = 42 \mathrm { ft }\end{array}a=20ftb=38ftc=42ft​

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - a=150 m b=164 mc=172 m\begin{array} { l } a = 150 \mathrm {~m} \\\mathrm {~b} = 164 \mathrm {~m} \\\mathrm { c } = 172 \mathrm {~m}\end{array}a=150 m b=164 mc=172 m​

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - $\begin{array} { l } a = 18.6 \mathrm {~cm} \\b = 17.2 \mathrm {~cm} \\c = 15.0 \mathrm {~cm}\end{array}$

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. -Two ships leave a harbor together traveling on courses that have an angle of $128 ^ { \circ }$ between them. If they each travel 543 miles, how far apart are they (to the nearest mile) ?

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - $\begin{array} { l } a = 47 \mathrm { ft } \\\mathrm { b } = 59 \mathrm { ft } \\\mathrm { c } = 65 \mathrm { ft }\end{array}$

Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropriate. - $\begin{array} { l } a = 20 \mathrm { ft } \\b = 38 \mathrm { ft } \\c = 42 \mathrm { ft }\end{array}$

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary. - $\begin{array} { l } a = 150 \mathrm {~m} \\\mathrm {~b} = 164 \mathrm {~m} \\\mathrm { c } = 172 \mathrm {~m}\end{array}$