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Precalculus : Use trigonometric functions to calculate the area of a triangle

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Example Questions

Example Question #1 : Area Of A Triangle

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

Possible Answers:

1600 $\displaystyle 1600$

200 $\displaystyle 200$

400 $\displaystyle 400$

100 $\displaystyle 100$

800 $\displaystyle 800$

Correct answer:

200 $\displaystyle 200$

Explanation:

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

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

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

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

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Example Question #2 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Find the area of this triangle:

Tri area f

Possible Answers:

$\displaystyle 136 un^ 2$

$\displaystyle 11.81 un^ 2$

$\displaystyle 68 un^ 2$

$\displaystyle 66.97 un^ 2$

Correct answer:

$\displaystyle 68 un^ 2$

Explanation:

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

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

where $\displaystyle a$ and $\displaystyle b$ are side lengths and $\displaystyle C$ is the included angle.

In our case,

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

Plug the values into the area formula and solve.

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

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Example Question #1 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Find the area of this triangle:

Tri area d

Possible Answers:

$\displaystyle 53.96 un^ 2$

$\displaystyle 88.11 un^ 2$

$\displaystyle 82. 99 un ^ 2$

$\displaystyle 212.13 un^ 2$

Correct answer:

$\displaystyle 82. 99 un ^ 2$

Explanation:

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

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

where $\displaystyle a$ and $\displaystyle b$ are side lengths and $\displaystyle C$ is the included angle.

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

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

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Example Question #4 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Find the area of this triangle:

Tri area b

Possible Answers:

$\displaystyle 25.38 un^ 2$

$\displaystyle 28 un^ 2$

$\displaystyle 11.83 un^ 2$

$\displaystyle 23.67 un^ 2$

Correct answer:

$\displaystyle 11.83 un^ 2$

Explanation:

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

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

where $\displaystyle a$ and $\displaystyle b$ are side lengths and $\displaystyle C$ is the included angle.

Here we are using 25^o $\displaystyle 25^o$ and not 90^o $\displaystyle 90^o$ since that is the angle between $\displaystyle 8$ and $\displaystyle 7$.

Therefore,

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

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

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

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Example Question #5 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Find the area of this triangle:

Tri area a

Possible Answers:

$\displaystyle 48.84 un^2$

$\displaystyle 133.91 un^2$

$\displaystyle 48.74 un^ 2$

$\displaystyle 66.95 un^ 2$

Correct answer:

$\displaystyle 48.74 un^ 2$

Explanation:

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

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

where $\displaystyle a$ and $\displaystyle b$ are side lengths and $\displaystyle C$ is the included angle.

In this particular case,

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

therefore the area is found to be,

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

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Example Question #1 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Find the exact area of a triangle with side lengths of $\displaystyle 2$, $\displaystyle 3$, and $\displaystyle 4$.

Possible Answers:

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

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

1.5 $\displaystyle 1.5$

2.5 $\displaystyle 2.5$

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

Correct answer:

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

Explanation:

Use the Heron's Formula:

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

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

Solve for $\displaystyle s$.

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

Solve for the area.

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

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

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

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Example Question #1 : Use Trigonometric Functions To Calculate The Area Of A Triangle

What is the area of a triangle with side lengths $\displaystyle 4$ , $\displaystyle 5$, and $\displaystyle 7$ ?

Possible Answers:

area=4 $\displaystyle area=4$

$area=\sqrt{92}$ $\displaystyle area=\sqrt{92}$

$area=4\sqrt{6}$ $\displaystyle area=4\sqrt{6}$

$area=16\sqrt{6}$ $\displaystyle area=16\sqrt{6}$

area=16 $\displaystyle area=16$

Correct answer:

$area=4\sqrt{6}$ $\displaystyle area=4\sqrt{6}$

Explanation:

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

The semiperimeter is

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

where $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ are the sides of a triangle.

Then the area is

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

So if we plug in

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

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

s=8 $\displaystyle s=8$

So the area is

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

$area=\sqrt{96}$ $\displaystyle area=\sqrt{96}$

$area=4\sqrt{6}$ $\displaystyle area=4\sqrt{6}$

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Example Question #21 : Trigonometric Applications

What is the area of a triangle with sides $\displaystyle 5$, $\displaystyle 6$, and $\displaystyle 9$ ?

Possible Answers:

area=10 $\displaystyle area=10$

$\displaystyle area=200$

$area=10\sqrt{2}$ $\displaystyle area=10\sqrt{2}$

$area=20\sqrt{2}$ $\displaystyle area=20\sqrt{2}$

$area=100\sqrt{2}$ $\displaystyle area=100\sqrt{2}$

Correct answer:

$area=10\sqrt{2}$ $\displaystyle area=10\sqrt{2}$

Explanation:

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

The semiperimeter is

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

where $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ are the sides of a triangle.

Then the area is

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

So if we plug in

$s=\frac{1}{2}(6+5+9)$ $\displaystyle s=\frac{1}{2}(6+5+9)$

$s=\frac{1}{2}(20)$ $\displaystyle s=\frac{1}{2}(20)$

s=10 $\displaystyle s=10$

So the area is

$area=\sqrt{10(10-6)(10-5)(10-9)}$ $\displaystyle area=\sqrt{10(10-6)(10-5)(10-9)}$

$area=\sqrt{200}$ $\displaystyle area=\sqrt{200}$

$area=10\sqrt{2}$ $\displaystyle area=10\sqrt{2}$

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Example Question #22 : Trigonometric Applications

What is the area of a triangle with side lengths of $\displaystyle 3$, $\displaystyle 4$, and $\displaystyle 5$ ?

Possible Answers:

$\displaystyle area= 12$

$\displaystyle area= 7.5$

area=6 $\displaystyle area=6$

area= 5 $\displaystyle area= 5$

area= 7 $\displaystyle area= 7$

Correct answer:

area=6 $\displaystyle area=6$

Explanation:

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

The semiperimeter is

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

where $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ are the sides of a triangle.

Then the area is

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

So if we plug in

$s=\frac{1}{2}(3+4+5)$ $\displaystyle s=\frac{1}{2}(3+4+5)$

$s=\frac{1}{2}(12)$ $\displaystyle s=\frac{1}{2}(12)$

s=6 $\displaystyle s=6$

So the area is

$area=\sqrt{6(6-3)(6-4)(6-5)}$ $\displaystyle area=\sqrt{6(6-3)(6-4)(6-5)}$

$area=\sqrt{36}$ $\displaystyle area=\sqrt{36}$

area=6 $\displaystyle area=6$

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Example Question #10 : Use Trigonometric Functions To Calculate The Area Of A Triangle

What is the area of a triangle with side lengths $\displaystyle 5$, $\displaystyle 12$, and $\displaystyle 13$ ?

Possible Answers:

area=20 $\displaystyle area=20$

area=33 $\displaystyle area=33$

$\displaystyle area=32.5$

$\displaystyle area=900$

area=30 $\displaystyle area=30$

Correct answer:

area=30 $\displaystyle area=30$

Explanation:

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

The semiperimeter is

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

where $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ are the sides of a triangle.

Then the area is

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

So if we plug in

$s=\frac{1}{2}(5+12+13)$ $\displaystyle s=\frac{1}{2}(5+12+13)$

$s=\frac{1}{2}(30)$ $\displaystyle s=\frac{1}{2}(30)$

s=15 $\displaystyle s=15$

So the area is

$area=\sqrt{15(15-5)(15-12)(15-13)}$ $\displaystyle area=\sqrt{15(15-5)(15-12)(15-13)}$

$area=\sqrt{900}$ $\displaystyle area=\sqrt{900}$

area=30 $\displaystyle area=30$

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Precalculus : Use trigonometric functions to calculate the area of a triangle

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #1 : Area Of A Triangle

Example Question #2 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Example Question #1 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Example Question #4 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Example Question #5 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Example Question #1 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Example Question #1 : Use Trigonometric Functions To Calculate The Area Of A Triangle

Example Question #21 : Trigonometric Applications

Example Question #22 : Trigonometric Applications

Example Question #10 : Use Trigonometric Functions To Calculate The Area Of A Triangle

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