Area of a Triangle - Trigonometry

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Question

Find the area of $\bigtriangleup ABC$ given that cm, $\angle A=27^\circ$ , and $\angle C =13^\circ$ .

Answer

We don't have this triangle's height to use the formula $Area=\frac{1}{2}base\cdot height$ , so we'll need to use trigonometry to find the area of this triangle. We can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

$\angle B = 180^\circ-(27^\circ+13^\circ)=140^\circ$

$R=\frac{c^2\sin A\sin B}{2\sin C}$

$R=\frac{22^2\sin 27^\circ\sin 140^\circ}{2\sin 13^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ given that cm, $\angle A=27^\circ$ , and $\angle B=13^\circ$ .

Answer

We don't have this triangle's height to use the formula $Area=\frac{1}{2}base\cdot height$ , so we'll need to use trigonometry to find the area of this triangle. We can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

$\angle C = 180^\circ-(27^\circ+13^\circ)=140^\circ$

$R=\frac{c^2\sin A\sin B}{2\sin C}$

$R=\frac{22^2\sin 27^\circ\sin 13^\circ}{2\sin 140^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ , given m, m, and $\angle A = 17^\circ$ .

Answer

Since we are not given the height of this triangle, we'll need to use trigonometry to find its area. If you draw and label the triangle, you'll notice that we are given Side-Side-Angle, which is the ambiguous case, and therefore there may be two different values that satisfy $\angle B$ . Let's start by finding those.

Use the Law of Sines to find

$\sin\angle B=\frac{b\sin\angle A}{a}=\frac{215\sin17^\circ}{100}=.6286<1$

Because this value is less than 1, there are two values that satisfy $\angle B$ . We can now find these using inverse trig and knowing that $\angle B'=180^\circ-\angle B$ . We have:

$\angle B=\sin^-^1(.6286)=38.9^\circ$ and $\angle B'=180^\circ-38.9^\circ=141.1^\circ$

Because we have two values that satisfy $\angle B$ , we'll also have two angles that satisfy $\angle C$ , and two unique triangles with two unique areas. Each of these is found below.

$\angle C=180^\circ-(\angle A+\angle B)=180^\circ-(17^\circ+38.9^\circ)=124.1^\circ$

$\angle C'=180^\circ-(\angle A+\angle B')=180^\circ-(17^\circ+141.1^\circ)=21.9^\circ$

Next, we can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

$R=\frac{a^2\sin B\sin C}{2\sin A}$

$R=\frac{100^2\sin 38.9^\circ\sin 124.1^\circ}{2\sin 17^\circ}$

Now, find for our second triangle.

$R'=\frac{a^2\sin B\sin C}{2\sin A}$

$R'=\frac{100^2\sin 141.1^\circ\sin 21.9^\circ}{2\sin 17^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ , given $\angle A =40^\circ$ , feet, and feet.

Answer

Since we are not given the height of this triangle, we'll need to use trigonometry to find its area. If you draw and label the triangle, you'll notice that we are given Side-Side-Angle, which is the ambiguous case, and therefore there may be two different values that satisfy $\angle B$ . Let's start by finding those.

Use the Law of Sines to find

$\sin\angle B=\frac{b\sin\angle A}{a}=\frac{95\sin40^\circ}{125}=.4885$

$\angle B=sin^-^1(.4885)=29.2^\circ$

While this is the ambiguous case, there is only one solution if the side opposite the given angle (a=125 ft) is equal to or greater than the other given side (b=95 ft). Because , there is only one value of $\angle B$ and only one solution to this triangle's area. Therefore, we can continue onward to find this triangle's area.

$\angle C=180^\circ-(\angle A+\angle B)=180^\circ-(40^\circ+29.2^\circ)=110.8^\circ$

Next, we can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Since we know a and b both, we could use either of the first two formulas to find the area. You'll get the same answer either way; feel free to test the other one to see this shown. Applying the formula, we get:

$R=\frac{a^2\sin B\sin C}{2\sin A}$

$R=\frac{125^2\sin 29.2^\circ\sin 110.8^\circ}{2\sin 40^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ given that inches, inches, and $\angle A = 44^\circ$ .

Answer

Using the formula for area of a triangle equal to $\frac{1}{2}base\cdot height$ , drawing $\bigtriangleup ABC$ and labelling its sides, angles, and height h, then using triangle trigonometry and substitution, we can derive the formulae $R=\frac{1}{2}ab\sin C=\frac{1}{2}ac\sin B=\frac{1}{2}bc\sin A$ , where R is equal to area. This can be used to find the area of a triangle when we know two of its sides and the included angle. Plugging in, we get:

$R=\frac{1}{2}bc\sin A=\frac{1}{2}\cdot 31\cdot 12\sin 44^\circ=129.2$

Therefore the area of this triangle is 129.2 square inches.

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Question

Find the area of $\bigtriangleup ABC$ given that cm, cm, and $\angle B=88.2^\circ$ .

Answer

Using the formula for area of a triangle equal to $\frac{1}{2}base\cdot height$ , drawing $\bigtriangleup ABC$ and labelling its sides, angles, and height h, then using triangle trigonometry and substitution, we can derive the formulae $R=\frac{1}{2}ab\sin C=\frac{1}{2}ac\sin B=\frac{1}{2}bc\sin A$ , which can be used to find the area of a triangle when we know two of its sides and the included angle. Plugging in, we get:

$R=\frac{1}{2}ac\sin B=\frac{1}{2}\cdot 11.24\cdot 16.98\cdot \sin88.2^\circ=95.38$

Therefore, the area of the triangle is 95.38 square centimeters.

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Question

Find the area of $\bigtriangleup ABC$ given that meters, meters, and meters.

Answer

Because we do not know the height of this triangle, we need to lean into trigonometry. Luckily, Heron's Formula will help us solve this problem. You can use it when you know three sides of a triangle and want to find the area. Please note that in order to find a triangle's area, you must make sure that the triangle exists by satisfying the Triangle Inequality, which states that the sum of any two sides of the triangle must always be larger than the remaining side. This is true for the lengths 4, 8, and 9 given in this problem. Heron's formula can be described as:

$R=\sqrt{s(s-a)(s-b)(s-c)}$ where R is the area of the triangle and s is the semiperimeter equivalent to $s=\frac{1}{2}(a+b+c)$ . Using these formulas and plugging in, we get:

$s=\frac{1}{2}(a+b+c)$ $=\frac{1}{2}(4+8+9)=\frac{21}{2}=10.5$

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

$=\sqrt{10.5(10.5-4)(10.5-8)(10.5-9)}$

$=\sqrt{10.5(6.5)(2.5)(1.5)}$

$=\sqrt{255.9375}$

Therefore the area of $\bigtriangleup ABC$ is 15.998 meters.

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Question

Find the area of $\bigtriangleup ABC$ given that feet, feet, and feet.

Answer

Because we do not know the height of this triangle, we need to lean into trigonometry. Luckily, Heron's Formula will help us solve this problem. You can use it when you know three sides of a triangle and want to find the area. Please note that in order to find a triangle's area, you must make sure that the triangle exists by satisfying the Triangle Inequality, which states that the sum of any two sides of the triangle must always be larger than the remaining side. This is true for the lengths 1.1, 3.7, and 4.3 given in this problem. Heron's formula can be described as:

$R=\sqrt{s(s-a)(s-b)(s-c)}$ where R is the area of the triangle and s is the semiperimeter equivalent to $s=\frac{1}{2}(a+b+c)$ . Using these formulas and plugging in, we get:

$s=\frac{1}{2}(a+b+c)$ $=\frac{1}{2}(1.1+3.7+4.3)=4.55$

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

$=\sqrt{4.55(4.55-1.1)(4.55-3.7)(4.55-4.3)}$

$=\sqrt{4.55(3.45)(.85)(.25)}$

$=\sqrt{3.3357}$

Therefore the area of $\bigtriangleup ABC$ is 1.83 feet.

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Question

Find the area of $\bigtriangleup ABC$ given that cm, $\angle A=27^\circ$ , and $\angle C =13^\circ$ .

Answer

We don't have this triangle's height to use the formula $Area=\frac{1}{2}base\cdot height$ , so we'll need to use trigonometry to find the area of this triangle. We can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

$\angle B = 180^\circ-(27^\circ+13^\circ)=140^\circ$

$R=\frac{c^2\sin A\sin B}{2\sin C}$

$R=\frac{22^2\sin 27^\circ\sin 140^\circ}{2\sin 13^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ given that cm, $\angle A=27^\circ$ , and $\angle B=13^\circ$ .

Answer

We don't have this triangle's height to use the formula $Area=\frac{1}{2}base\cdot height$ , so we'll need to use trigonometry to find the area of this triangle. We can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

$\angle C = 180^\circ-(27^\circ+13^\circ)=140^\circ$

$R=\frac{c^2\sin A\sin B}{2\sin C}$

$R=\frac{22^2\sin 27^\circ\sin 13^\circ}{2\sin 140^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ , given m, m, and $\angle A = 17^\circ$ .

Answer

Since we are not given the height of this triangle, we'll need to use trigonometry to find its area. If you draw and label the triangle, you'll notice that we are given Side-Side-Angle, which is the ambiguous case, and therefore there may be two different values that satisfy $\angle B$ . Let's start by finding those.

Use the Law of Sines to find

$\sin\angle B=\frac{b\sin\angle A}{a}=\frac{215\sin17^\circ}{100}=.6286<1$

Because this value is less than 1, there are two values that satisfy $\angle B$ . We can now find these using inverse trig and knowing that $\angle B'=180^\circ-\angle B$ . We have:

$\angle B=\sin^-^1(.6286)=38.9^\circ$ and $\angle B'=180^\circ-38.9^\circ=141.1^\circ$

Because we have two values that satisfy $\angle B$ , we'll also have two angles that satisfy $\angle C$ , and two unique triangles with two unique areas. Each of these is found below.

$\angle C=180^\circ-(\angle A+\angle B)=180^\circ-(17^\circ+38.9^\circ)=124.1^\circ$

$\angle C'=180^\circ-(\angle A+\angle B')=180^\circ-(17^\circ+141.1^\circ)=21.9^\circ$

Next, we can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

$R=\frac{a^2\sin B\sin C}{2\sin A}$

$R=\frac{100^2\sin 38.9^\circ\sin 124.1^\circ}{2\sin 17^\circ}$

Now, find for our second triangle.

$R'=\frac{a^2\sin B\sin C}{2\sin A}$

$R'=\frac{100^2\sin 141.1^\circ\sin 21.9^\circ}{2\sin 17^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ , given $\angle A =40^\circ$ , feet, and feet.

Answer

Since we are not given the height of this triangle, we'll need to use trigonometry to find its area. If you draw and label the triangle, you'll notice that we are given Side-Side-Angle, which is the ambiguous case, and therefore there may be two different values that satisfy $\angle B$ . Let's start by finding those.

Use the Law of Sines to find

$\sin\angle B=\frac{b\sin\angle A}{a}=\frac{95\sin40^\circ}{125}=.4885$

$\angle B=sin^-^1(.4885)=29.2^\circ$

While this is the ambiguous case, there is only one solution if the side opposite the given angle (a=125 ft) is equal to or greater than the other given side (b=95 ft). Because , there is only one value of $\angle B$ and only one solution to this triangle's area. Therefore, we can continue onward to find this triangle's area.

$\angle C=180^\circ-(\angle A+\angle B)=180^\circ-(40^\circ+29.2^\circ)=110.8^\circ$

Next, we can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Since we know a and b both, we could use either of the first two formulas to find the area. You'll get the same answer either way; feel free to test the other one to see this shown. Applying the formula, we get:

$R=\frac{a^2\sin B\sin C}{2\sin A}$

$R=\frac{125^2\sin 29.2^\circ\sin 110.8^\circ}{2\sin 40^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ given that inches, inches, and $\angle A = 44^\circ$ .

Answer

Using the formula for area of a triangle equal to $\frac{1}{2}base\cdot height$ , drawing $\bigtriangleup ABC$ and labelling its sides, angles, and height h, then using triangle trigonometry and substitution, we can derive the formulae $R=\frac{1}{2}ab\sin C=\frac{1}{2}ac\sin B=\frac{1}{2}bc\sin A$ , where R is equal to area. This can be used to find the area of a triangle when we know two of its sides and the included angle. Plugging in, we get:

$R=\frac{1}{2}bc\sin A=\frac{1}{2}\cdot 31\cdot 12\sin 44^\circ=129.2$

Therefore the area of this triangle is 129.2 square inches.

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Question

Find the area of $\bigtriangleup ABC$ given that cm, cm, and $\angle B=88.2^\circ$ .

Answer

Using the formula for area of a triangle equal to $\frac{1}{2}base\cdot height$ , drawing $\bigtriangleup ABC$ and labelling its sides, angles, and height h, then using triangle trigonometry and substitution, we can derive the formulae $R=\frac{1}{2}ab\sin C=\frac{1}{2}ac\sin B=\frac{1}{2}bc\sin A$ , which can be used to find the area of a triangle when we know two of its sides and the included angle. Plugging in, we get:

$R=\frac{1}{2}ac\sin B=\frac{1}{2}\cdot 11.24\cdot 16.98\cdot \sin88.2^\circ=95.38$

Therefore, the area of the triangle is 95.38 square centimeters.

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Question

Find the area of $\bigtriangleup ABC$ given that meters, meters, and meters.

Answer

Because we do not know the height of this triangle, we need to lean into trigonometry. Luckily, Heron's Formula will help us solve this problem. You can use it when you know three sides of a triangle and want to find the area. Please note that in order to find a triangle's area, you must make sure that the triangle exists by satisfying the Triangle Inequality, which states that the sum of any two sides of the triangle must always be larger than the remaining side. This is true for the lengths 4, 8, and 9 given in this problem. Heron's formula can be described as:

$R=\sqrt{s(s-a)(s-b)(s-c)}$ where R is the area of the triangle and s is the semiperimeter equivalent to $s=\frac{1}{2}(a+b+c)$ . Using these formulas and plugging in, we get:

$s=\frac{1}{2}(a+b+c)$ $=\frac{1}{2}(4+8+9)=\frac{21}{2}=10.5$

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

$=\sqrt{10.5(10.5-4)(10.5-8)(10.5-9)}$

$=\sqrt{10.5(6.5)(2.5)(1.5)}$

$=\sqrt{255.9375}$

Therefore the area of $\bigtriangleup ABC$ is 15.998 meters.

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Question

Find the area of $\bigtriangleup ABC$ given that feet, feet, and feet.

Answer

Because we do not know the height of this triangle, we need to lean into trigonometry. Luckily, Heron's Formula will help us solve this problem. You can use it when you know three sides of a triangle and want to find the area. Please note that in order to find a triangle's area, you must make sure that the triangle exists by satisfying the Triangle Inequality, which states that the sum of any two sides of the triangle must always be larger than the remaining side. This is true for the lengths 1.1, 3.7, and 4.3 given in this problem. Heron's formula can be described as:

$R=\sqrt{s(s-a)(s-b)(s-c)}$ where R is the area of the triangle and s is the semiperimeter equivalent to $s=\frac{1}{2}(a+b+c)$ . Using these formulas and plugging in, we get:

$s=\frac{1}{2}(a+b+c)$ $=\frac{1}{2}(1.1+3.7+4.3)=4.55$

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

$=\sqrt{4.55(4.55-1.1)(4.55-3.7)(4.55-4.3)}$

$=\sqrt{4.55(3.45)(.85)(.25)}$

$=\sqrt{3.3357}$

Therefore the area of $\bigtriangleup ABC$ is 1.83 feet.

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Question

Find the area of $\bigtriangleup ABC$ given that cm, $\angle A=27^\circ$ , and $\angle C =13^\circ$ .

Answer

We don't have this triangle's height to use the formula $Area=\frac{1}{2}base\cdot height$ , so we'll need to use trigonometry to find the area of this triangle. We can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

$\angle B = 180^\circ-(27^\circ+13^\circ)=140^\circ$

$R=\frac{c^2\sin A\sin B}{2\sin C}$

$R=\frac{22^2\sin 27^\circ\sin 140^\circ}{2\sin 13^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ given that cm, $\angle A=27^\circ$ , and $\angle B=13^\circ$ .

Answer

We don't have this triangle's height to use the formula $Area=\frac{1}{2}base\cdot height$ , so we'll need to use trigonometry to find the area of this triangle. We can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

$\angle C = 180^\circ-(27^\circ+13^\circ)=140^\circ$

$R=\frac{c^2\sin A\sin B}{2\sin C}$

$R=\frac{22^2\sin 27^\circ\sin 13^\circ}{2\sin 140^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ , given m, m, and $\angle A = 17^\circ$ .

Answer

Since we are not given the height of this triangle, we'll need to use trigonometry to find its area. If you draw and label the triangle, you'll notice that we are given Side-Side-Angle, which is the ambiguous case, and therefore there may be two different values that satisfy $\angle B$ . Let's start by finding those.

Use the Law of Sines to find

$\sin\angle B=\frac{b\sin\angle A}{a}=\frac{215\sin17^\circ}{100}=.6286<1$

Because this value is less than 1, there are two values that satisfy $\angle B$ . We can now find these using inverse trig and knowing that $\angle B'=180^\circ-\angle B$ . We have:

$\angle B=\sin^-^1(.6286)=38.9^\circ$ and $\angle B'=180^\circ-38.9^\circ=141.1^\circ$

Because we have two values that satisfy $\angle B$ , we'll also have two angles that satisfy $\angle C$ , and two unique triangles with two unique areas. Each of these is found below.

$\angle C=180^\circ-(\angle A+\angle B)=180^\circ-(17^\circ+38.9^\circ)=124.1^\circ$

$\angle C'=180^\circ-(\angle A+\angle B')=180^\circ-(17^\circ+141.1^\circ)=21.9^\circ$

Next, we can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Applying the formula, we get:

$R=\frac{a^2\sin B\sin C}{2\sin A}$

$R=\frac{100^2\sin 38.9^\circ\sin 124.1^\circ}{2\sin 17^\circ}$

Now, find for our second triangle.

$R'=\frac{a^2\sin B\sin C}{2\sin A}$

$R'=\frac{100^2\sin 141.1^\circ\sin 21.9^\circ}{2\sin 17^\circ}$

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Question

Find the area of $\bigtriangleup ABC$ , given $\angle A =40^\circ$ , feet, and feet.

Answer

Since we are not given the height of this triangle, we'll need to use trigonometry to find its area. If you draw and label the triangle, you'll notice that we are given Side-Side-Angle, which is the ambiguous case, and therefore there may be two different values that satisfy $\angle B$ . Let's start by finding those.

Use the Law of Sines to find

$\sin\angle B=\frac{b\sin\angle A}{a}=\frac{95\sin40^\circ}{125}=.4885$

$\angle B=sin^-^1(.4885)=29.2^\circ$

While this is the ambiguous case, there is only one solution if the side opposite the given angle (a=125 ft) is equal to or greater than the other given side (b=95 ft). Because , there is only one value of $\angle B$ and only one solution to this triangle's area. Therefore, we can continue onward to find this triangle's area.

$\angle C=180^\circ-(\angle A+\angle B)=180^\circ-(40^\circ+29.2^\circ)=110.8^\circ$

Next, we can use the formula $R=\frac{a^2\sin B\sin C}{2\sin A}=\frac{b^2\sin A\sin C}{2\sin B}=\frac{c^2\sin A\sin B}{2\sin C}$ , where R is the area of a triangle. This formula comes from the fact that $Area=\frac{1}{2}base\cdot height$ and also incorporates triangle trigonometry and the Law of Sines. Since we know a and b both, we could use either of the first two formulas to find the area. You'll get the same answer either way; feel free to test the other one to see this shown. Applying the formula, we get:

$R=\frac{a^2\sin B\sin C}{2\sin A}$

$R=\frac{125^2\sin 29.2^\circ\sin 110.8^\circ}{2\sin 40^\circ}$

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