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Precalculus : Trigonometric Applications

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #18 : Area Of A Triangle

Find the area of this triangle:

Tri area e

Possible Answers:

15.23 un^ 2

37.67 un^ 2

124.56 un^ 2

38.50 un^ 2

Correct answer:

38.50 un^ 2

Explanation:

To solve, use Heron's Formula,

$area = \sqrt{ s (s - a ) (s - b )( s- c )}$

where , , and are the side lengths and

$s = \frac{ 1}{2} ( a + b + c )$ .

In this particular case,

a=13, b=11, c=7

thus,

$s = \frac{ 1}{2} ( 13 + 11 + 7 ) = \frac{ 1}{2} \cdot 31 = 15.5$ .

Plugging these values into the Heron's Formula we arrive at our answer.

$\\ area = \sqrt{ 15.5 ( 15.5 - 11 ) (15.5 - 13 ) ( 15.5 - 7 )} \\ \\= \sqrt{ 15.5 \cdot 4.5 \cdot 2.5 \cdot 8. 5 } \approx 38.50$

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Example Question #19 : Area Of A Triangle

Find the area of this triangle:

Tri area c

Possible Answers:

36.33 un^ 2

15.19 un^ 2

72.99 un^2

24 un^2

Correct answer:

24 un^2

Explanation:

To find the area, use Heron's Formula

$area = \sqrt{ s (s-a ) (s-b )(s-c)}$

where , , and are the side lengths and $s = \frac{ 1}{2} (a + b + c )$ .

In this case,

a=6, b=8, c=10

thus,

$s= \frac{ 1}{2} (6 + 8 +10)= \frac{ 1}{2} \cdot 24 = 12$ .

Now plugging these values into the Heron's Formula, we arrive at our final answer,

$area = \sqrt{12(12-6 ) (12 - 8)(12-10) } = \sqrt{12 \cdot 6 \cdot 4 \cdot 2 } = 24$

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Example Question #20 : Area Of A Triangle

Given a triangle with lengths 4,8, and 10, find the area of the triangle.

Possible Answers:

$A=5\sqrt{17}$

$A=\sqrt{22}$

A=11

$A=\sqrt{231}$

$A=2\sqrt{10}$

Correct answer:

$A=\sqrt{231}$

Explanation:

Use Heron's formula.

First find S using

$S=\frac{a+b+c}{2}=\frac{4+8+10}{2}=11$

Then apply the Heron's area formula:

$A=\sqrt{S(S-a)(S-b)(S-c)}$

$A=\sqrt{11(11-4)(11-8)(11-10)}=\sqrt{231}$

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Example Question #21 : Use Trigonometric Functions To Calculate The Area Of A Triangle

You have a triangluar space in your backyard that you want to cover with mulch and the side lengths of the space measure 5 meters, 6 meters, and 8 meters. If a bag of mulch covers 2 square meters of ground and cost $10, how much money will you need to spend to complete the job? We will assume the sides of the triangle are designated a, b, and c based on the order presented above.

Possible Answers:

$\$105$

$\$15$

$\$75$

$\$40$

$\$80$

Correct answer:

$\$80$

Explanation:

Use Heron's formula to compute the area of the oblique triangle.

$A=\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{a+b+c}{2}$ .

$s=\frac{5+6+8}{2}=9.5$

$A=\sqrt{9.5(9.5-5)(9.5-6)(9.5-8)}=15m^2$

Since the bags of mulch cover 2 meters squared, the job will require 7.5 bags. However, the store does not sell half bags so 8 are required and $80 dollars must be spent.

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Example Question #22 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Find the area of a triangle with sides 25 meters, 42 meters, and 23 meters.

Possible Answers:

234m^2

265m^2

None of these.

220m^2

244m^2

Correct answer:

244m^2

Explanation:

We use Heron's formula for finding the area of oblique triangles or triangles without right angles.

Heron's formula is

$\sqrt{s(s-a)(s-b)(s-c)}$ .

We find s by

$s=\frac{a+b+c}{2}$

Find s.

$s=\frac{a+b+c}{2}=\frac{25+42+23}{2}=45$

Input into Heron's formula:

$\sqrt{45(45-25)(45-42)(45-23)}=244m^2$

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Example Question #21 : Area Of A Triangle

On a Cartesian plane plot a circle centered at the origin of radius . Now draw a segment from (6,0) to (0,6) . Shade the area between the segment and the boundary of the circle, above the segment.

What is the area of the shaded region?

Possible Answers:

$36\pi-36$

$9\pi-18$

$9\pi$

$9\pi-36$

$36\pi-18$

Correct answer:

$9\pi-18$

Explanation:

The shaded area is one quarter of the area of the circle, minus the area of an isosceles right triangle with legs along the radius.

The circle has radius , so the area of the circle is

$A_C=\pi r^2$

$=\pi (6)^2=36\pi$ .

The part of the circle in the first quadrant is $\frac{36\pi}{4}=9\pi$

If we subtract the area under the line connecting (6,0),(0,6) we get the correct answer.

To find this area, recognize the geometry as a triangle. The two points give you the base and height so its area is,

$A_T=\frac{1}{2}bh=\frac{1}{2}\cdot6\cdot 6=18$ .

Therefore, the area of the region of interest is $9\pi-18$ .

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Example Question #21 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Find the area of the shaded segment of the given circle that has a radius of 8 cm:

Varsity log graph

Possible Answers:

$51 \; cm^{2}$

$173.4 \; cm^{2}$

$44.4 \; cm^{2}$

$39.3\; cm^{2}$

$11.6 \; cm^{2}$

Correct answer:

$39.3\; cm^{2}$

Explanation:

The area of a segment can be calculated with the following formula:

$A_{segment}=A_{sector}-A_{triangle}$

Finding the area of the sector is calculated as such:

$A_{sector}=\frac{120^{\circ}}{360^{\circ}}\cdot \pi \cdot8^{2}$

$A_{sector}=67cm^{2}$

To find the area of the triangle, the 120 degree isosceles can be divided into two 30 60 90 triangles:

Varsity log graph

The area for the triangle can be calculated as such:

$A_{triangle}=\frac{1}{2}\cdot8\sqrt{3}\cdot4=16\sqrt{3}$

$A_{triangle}=27.7cm^{2}$

Going back to the original formula for segment area:

$A_{segment}=67cm^{2}-27.7cm2=39.3cm2$

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Example Question #21 : Area Of A Triangle

Find the area of the shaded segment:

Varsity log graph

Possible Answers:

$13.7 \; cm^{2}$

$12.6 \; cm^{2}$

$22.4 \; cm^{2}$

$19.4\; cm^{2}$

$18.3 \; cm^{2}$

Correct answer:

$18.3 \; cm^{2}$

Explanation:

The area of the segment can be calculated with the following equation:

$A_{segment}=A_{sector}-A_{triangle}$

The area of the sector can be found as such:

$A_{sector}=\frac{90^{\circ}}{360^{\circ}}\cdot \pi \cdot(8\; cm)^{2}$

$A_{sector}=16\pi=50.3\; cm^{2}$

The area of the triangle can be determined easily because it is a right triangle, at 90 degrees:

$A_{triangle}=\frac{1}{2}\cdot8\; cm\cdot8\; cm=32\; cm^{2}$

Now that we have both variables needed for the formula, we can determine the area of the segment:

$A_{segment}=50.3\; cm^{2}-32\; cm^{2}=18.3\; cm^{2}$

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Example Question #21 : Area Of A Triangle

Find the area of the shaded segment BC:

Varsity log graph

Possible Answers:

$44.12 \; cm^{2}$

$105.23\; cm^{2}$

$47.1 \; cm^{2}$

$43.94\; cm^{2}$

$37.61\; cm^{2}$

Correct answer:

$44.12 \; cm^{2}$

Explanation:

The area of a segment can be calculated as such:

$A_{segment}=A_{sector}-A_{triangle}$

We can find the area of the sector as such:

$A_{sector}=\frac{160^{\circ}}{360^{\circ}}\cdot \pi \cdot(6\; cm)^{2}=50.27\; cm^{2}$

To find the area of the triangle, we must divide the isosceles into two right triangles with a bisector:

Varsity log graph

To determine base, we calculate:

$sin(80^{\circ})=\frac{1/2base}{6\; cm}$

$0.9848=\frac{1/2base}{6 \; cm}$

$1/2base=5.91\; cm$

$base=11.82\; cm$

To determine height, we calculate:

$sin(10^{\circ})=\frac{height}{6\; cm}$

$0.1736=\frac{height}{6\; cm}$

$height=1.04\; cm$

Now that we have all the needed values, we can calculate the area of the triangle:

$Area=\frac{1}{2}\cdot11.82\; cm\cdot1.04\; cm=6.15\; cm^{2}$

The total segment area for BC can now be calculated:

$A_{segment}=50.27\; cm^{2}-6.15\; cm^{2}=44.12\; cm^{2}$

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Example Question #27 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Find the area of the segment BC:

Varsity log graph

Possible Answers:

$8 \: cm^{2}$

$3.8 \: cm^{2}$

$0.3 \: cm^{2}$

$0.2 \: cm^{2}$

$4.3 \: cm^{2}$

Correct answer:

$0.2 \: cm^{2}$

Explanation:

To solve for the area of the segment, we must first find the area of the sector and the triangle, to fulfill the given formula:

$A_{segment}=A_{sector}-A_{triangle}$

To find the area of the sector:

$Area=\frac{30^{\circ}}{360^{\circ}}\cdot \pi \cdot(4cm)^{2}=4.2cm^{2}$

To find the area of the triangle, we first must divide it into two right triangles:

Varsity log graph

We can find the height of the triangle as such:

$sin(75^{\circ})=\frac{height}{4\: cm}$

$0.9659=\frac{height}{4\: cm}$

$height=3.86\: cm$

We can find the base employing a similar method:

$sin(15^{\circ})=\frac{1/2base}{4\: cm}$

$0.2588=\frac{1/2base}{4\: cm}$

$1/2base=1.035\: cm$

$base=2.07\: cm$

The area of the triangle can then be calculated as such:

$Area=\frac{1}{2}\cdot2.07\: cm\cdot3.86\: cm=4\: cm^2$

To find the Area of the segment, we can now subtract the triangle area from the sector area:

$A_{segment}=4.2\: cm^{2}-4\: cm^2=0.2\: cm^{2}$

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Precalculus : Trigonometric Applications

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #18 : Area Of A Triangle

Example Question #19 : Area Of A Triangle

Example Question #20 : Area Of A Triangle

Example Question #21 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Example Question #22 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Example Question #21 : Area Of A Triangle

Example Question #21 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Example Question #21 : Area Of A Triangle

Example Question #21 : Area Of A Triangle

Example Question #27 : Use Trigonometric Functions To Calculate The Area Of A Triangle

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