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Precalculus : Trigonometric Applications

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Example Questions

Example Question #11 : Solving Right Triangles

Find the area of the given isosceles triangle:

Varsity log graph

Possible Answers:

$24.5\; cm^{2}$

$67.32 \; cm^{2}$

$35.7 \; cm^{2}$

$49 \; cm^{2}$

$17.85\; cm^{2}$

Correct answer:

$17.85\; cm^{2}$

Explanation:

The first step is to divide this isosceles triangle into 2 right triangles, making it easier to solve:

Varsity log graph

The equation for area is $A=\frac{1}{2}base\cdot height$

We already know the base, so we need to solve for height to get the area.

$tan20^{\circ}=\frac{height}{7\; cm}$

$0.364=\frac{height}{7\; cm}$

$height=2.55\; cm$

Then we plug in all values for the equation:

$Area=\frac{1}{2}\cdot14\; cm\cdot2.55\; cm=17.85\; cm^{2}$

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Example Question #12 : Solving Right Triangles

Find the area of the given isosceles triangle:

Varsity log graph

Possible Answers:

$3.34 \: cm^{2}$

$3.11 \: cm^{2}$

$6.71 \: cm^{2}$

$25.55\: cm^{2}$

$6.25 \: cm^{2}$

Correct answer:

$6.25 \: cm^{2}$

Explanation:

The first step toward finding the area is to divide this isosceles triangle into two right triangles:

Varsity log graph

Trigonometric ratios can be used to find both the height and the base, which are needed to calculate area:

$sin(75^{\circ})=\frac{height}{5\: cm}$

$0.9659=\frac{height}{5\: cm}$

$height=4.83\: cm$

$sin(15^{\circ})=\frac{1/2base}{5\: cm}$

$0.2588=\frac{1/2base}{5\: cm}$

$Base=2.59\: cm$

With both of those values calculated, we can now calculate the area of the triangle:

$Area=\frac{1}{2}\cdot2.59\: cm\cdot4.83\: cm=6.25\: cm^2$

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Example Question #13 : Solving Right Triangles

Pcq1

Solve the right triangle given that a=5, b=12, and A=22.620°

Possible Answers:

None of these answers are correct.

B=67.380°

C=90°

c=13

B=90°

C=67.380°

c=169

B=90°

C=67.380°

c=13

B=67.380°

C=90°

c=17

Correct answer:

B=67.380°

C=90°

c=13

Explanation:

Pcq1

C is given as 90°.

A is given as 22.620°

a is given as 5

b is given as 12

$a^{2}+b^{2}=c^{2}$

Therefore...

$5^{2}+12^{2}=c^{2}$

$169=c^{2}$

$\sqrt{169}=\sqrt{c^{2}}$

13=c

All angles of a triangle add up to equal 180°.

$90^{\circ}+22.620^{\circ}+B=180^{\circ}$

$112.620^{\circ}+B=180^{\circ}$

$B=180^{\circ}-112.620^{\circ}$

$B=67.380^{\circ}$

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Example Question #1 : Area Of A Triangle

In triangle ABC , $m\angle C = 30^\circ$ , BC = 40 , and AC = 20 . Find the area of the triangle.

Possible Answers:

100

800

400

1600

200

Correct answer:

200

Explanation:

When given the lengths of two sides and the measure of the angle included by the two sides, the area formula is:

$A = \frac {1}{2}ab\sin C \\$

Plugging in the given values we are able to calculate the area.

$A = \frac {1}{2}\cdot 40\cdot 20 \cdot \sin 30^\circ= 200$

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Example Question #2 : Area Of A Triangle

Find the area of this triangle:

Tri area f

Possible Answers:

68 un^ 2

11.81 un^ 2

136 un^ 2

66.97 un^ 2

Correct answer:

68 un^ 2

Explanation:

To find the area, use the formula associated with side, angle, side triangles which states,

$area = \frac{ 1}{2} a b \sin C$

where and are side lengths and is the included angle.

In our case,

$a=8, b=17, C=90^\circ$ .

Plug the values into the area formula and solve.

$area = \frac{ 1}{2} (8 )( 17 ) \cdot \sin( 90 ^ o ) = 68$

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Example Question #1 : Area Of A Triangle

Find the area of this triangle:

Tri area d

Possible Answers:

88.11 un^ 2

53.96 un^ 2

82. 99 un ^ 2

212.13 un^ 2

Correct answer:

82. 99 un ^ 2

Explanation:

Use the area formula to find area that is associated with the side angle side theorem for triangles.

$area= \frac{ 1}{2 } a b \sin C$

where and are side lengths and is the included angle.

Plugging these values into the formula above, we arrive at our final answer.

$area = \frac{ 1}{2} (11) (23 ) \sin (41^o ) \approx 82.99$

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Example Question #1 : Area Of A Triangle

Find the area of this triangle:

Tri area b

Possible Answers:

25.38 un^ 2

28 un^ 2

23.67 un^ 2

11.83 un^ 2

Correct answer:

11.83 un^ 2

Explanation:

To solve, use the formula for area that is associated with the side angle side theorem for triangles,

$area = \frac{ 1}{2} ab \sin C$

where and are side lengths and is the included angle.

Here we are using 25^o and not 90^o since that is the angle between and .

Therefore,

$a=8,b=7,C=25^\circ$ .

Plugging the above values into the area formula we arrive at our final answer.

$area = \frac{ 1}{2} (8)(7) \sin (25^o ) \approx 11.83$

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Example Question #1 : Area Of A Triangle

Find the area of this triangle:

Tri area a

Possible Answers:

48.74 un^ 2

133.91 un^2

48.84 un^2

66.95 un^ 2

Correct answer:

48.74 un^ 2

Explanation:

Find the area using the formula associated the side angle side theorem of a triangle,

$area = \frac{ 1}{2} ab \sin C$

where and are side lengths and is the included angle.

In this particular case,

$a=15, b=19, C=20^\circ$

therefore the area is found to be,

$area = \frac {1}{2} (15)(19) \sin (20^o ) \approx 48.74$ .

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Example Question #2 : Area Of A Triangle

Find the exact area of a triangle with side lengths of , , and .

Possible Answers:

1.5

$\frac{\sqrt{17}}{2}$

$\frac{\sqrt{135}}{8}$

$\frac{\sqrt {135}}{4}$

2.5

Correct answer:

$\frac{\sqrt {135}}{4}$

Explanation:

Use the Heron's Formula:

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

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

Solve for .

$s=\frac{a+b+c}{2}=\frac{2+3+4}{2}=\frac{9}{2}$

Solve for the area.

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

$A=\sqrt{\frac{9}{2}\left(\frac{9}{2}-2\right)\left(\frac{9}{2}-3\right)\left(\frac{9}{2}-4\right)}$

$A=\sqrt{\frac{9}{2}\left(\frac{5}{2}\right)\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)}=\sqrt{\frac{135}{16}}=\frac{\sqrt{135}}{4}$

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Example Question #3 : Area Of A Triangle

What is the area of a triangle with side lengths , , and ?

Possible Answers:

$area=4\sqrt{6}$

area=16

$area=16\sqrt{6}$

$area=\sqrt{92}$

area=4

Correct answer:

$area=4\sqrt{6}$

Explanation:

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

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

where , , are the sides of a triangle.

Then the area is

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

So if we plug in

$s=\frac{1}{2}(4+5+7)$

$s=\frac{1}{2}(16)$

s=8

So the area is

$area=\sqrt{8(8-4)(8-5)(8-7)}$

$area=\sqrt{96}$

$area=4\sqrt{6}$

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Precalculus : Trigonometric Applications

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #11 : Solving Right Triangles

Example Question #12 : Solving Right Triangles

Example Question #13 : Solving Right Triangles

Example Question #1 : Area Of A Triangle

Example Question #2 : Area Of A Triangle

Example Question #1 : Area Of A Triangle

Example Question #1 : Area Of A Triangle

Example Question #1 : Area Of A Triangle

Example Question #2 : Area Of A Triangle

Example Question #3 : Area Of A Triangle

All Precalculus Resources

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