- Demonstrate the ability to solve oblique triangles (SAS) using the law of cosines
- Demonstrate the ability to solve oblique triangles (SSS) using the law of cosines
- Demonstrate the ability to find the area of a triangle using Heron's formula
#1:
Instructions: Solve each triangle. Round your answer to the nearest tenth.
$$a)\hspace{.1em}a=17 \hspace{.1em}\text{ft}, B=112°, c=26 \hspace{.1em}\text{ft}$$ 
$$b)\hspace{.1em}b=23 \hspace{.1em}\text{cm}, C=100°, a=24 \hspace{.1em}\text{cm}$$ 
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#2:
Instructions: Solve each triangle. Round your answer to the nearest tenth.
$$a)\hspace{.1em}a=18 \hspace{.1em}\text{mi}, c=17 \hspace{.1em}\text{mi}, b=30 \hspace{.1em}\text{mi}$$ 
$$b)\hspace{.1em}a=25 \hspace{.1em}\text{ft}, c=27 \hspace{.1em}\text{ft}, b=28 \hspace{.1em}\text{ft}$$ 
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#3:
Instructions: Solve each triangle. Round your answer to the nearest tenth.
$$a)\hspace{.1em}a=19 \hspace{.1em}\text{yd}, B=130°, c=14 \hspace{.1em}\text{yd}$$ 
$$b)\hspace{.1em}a=11 \hspace{.1em}\text{cm}, c=17 \hspace{.1em}\text{cm}, b=10 \hspace{.1em}\text{cm}$$ 
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#4:
Instructions: Find the area using Heron's formula. Round your answer to the nearest tenth.
$$a)\hspace{.1em}a=9 \hspace{.1em}\text{m}, c=15 \hspace{.1em}\text{m}, b=16 \hspace{.1em}\text{m}$$ 
$$b)\hspace{.1em}a=14 \hspace{.1em}\text{cm}, c=13 \hspace{.1em}\text{cm}, b=16 \hspace{.1em}\text{cm}$$ 
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#5:
Instructions: Find the area using Heron's formula. Round your answer to the nearest tenth.
$$a)\hspace{.1em}b=6.8 \hspace{.1em}\text{in}, C=120°, a=5 \hspace{.1em}\text{in}$$ 
$$b)\hspace{.1em}a=15 \hspace{.1em}\text{mi}, B=78°, c=12 \hspace{.1em}\text{mi}$$ 
$$c)\hspace{.1em}\text{Derive Heron's Formula}$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.1em}C=42°, A=26°, b=36 \hspace{.1em}\text{ft}$$
$$b)\hspace{.1em}A=41°, B=39°, c=36 \hspace{.1em}\text{cm}$$
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#2:
Solutions:
$$a)\hspace{.1em}C=30°, A=32°, B=118°$$
$$b)\hspace{.1em}C=61°, A=54°, B=65°$$
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#3:
Solutions:
$$a)\hspace{.1em}C=21°, A=29°, b=30 \hspace{.1em}\text{yd}$$
$$b)\hspace{.1em}A=38°, B=34°, C=108°$$
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#4:
Solutions:
$$a)\hspace{.1em}66.3 \hspace{.1em}\text{m}^2$$
$$b)\hspace{.1em}86.8 \hspace{.1em}\text{cm}^2$$
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#5:
Solutions:
$$a)\hspace{.1em}14.7 \hspace{.1em}\text{in}^2$$ Note: Due to rounding, you may get an answer of 14.6 square inches
$$b)\hspace{.1em}88 \hspace{.1em}\text{mi}^2$$
$$c)\hspace{.1em}\text{Area}=\frac{1}{2}a \hspace{.1em}b \cdot \text{sin}C$$ $$\text{Area}^2=\left(\frac{1}{2}a \hspace{.1em}b \cdot \text{sin}C\right)^2$$ $$=\frac{1}{4}a^2 \hspace{.1em}b^2 \cdot \text{sin}^2 C$$ Use the Pythagorean Identity: $$\text{sin}^2 C=1 - \text{cos}^2 C$$ $$=\frac{1}{4}a^2 \hspace{.1em}b^2 \cdot (1 - \text{cos}^2 C)$$ Use the difference of squares formula: $$x^2 - y^2=(x - y)(x + y)$$ $$=\frac{1}{4}a^2 \hspace{.1em}b^2 \cdot (1 - \text{cos}\hspace{.1em}C)(1 + \text{cos}\hspace{.1em}C)$$ For the next part, we need to solve our law of cosines formula for the cos C: $$c^2=a^2 + b^2 - 2ab \cdot \text{cos}\hspace{.1em}C$$ $$c^2 - a^2 - b^2=-2ab \cdot \text{cos}\hspace{.1em}C$$ $$\text{cos}\hspace{.1em}C=\frac{c^2 - a^2 - b^2}{-2ab}$$ $$\text{cos}\hspace{.1em}C=\frac{-c^2 + a^2 + b^2}{2ab}$$ $$\text{cos}\hspace{.1em}C=\frac{a^2 + b^2 - c^2}{2ab}$$ Now, we will find 1 - cos C and 1 + cos C: $$1 + \text{cos}\hspace{.1em}C=1 + \frac{a^2 + b^2 - c^2}{2ab}$$ $$=\frac{2ab}{2ab}+ \frac{a^2 + b^2 - c^2}{2ab}$$ $$=\frac{a^2 + 2ab + b^2 - c^2}{2ab}$$ Use the formula: $$(x + y)^2=x^2 + 2xy + y^2$$ $$=\frac{(a^2 + 2ab + b^2) - c^2}{2ab}$$ $$=\frac{(a + b)^2 - c^2}{2ab}$$ Use the difference of squares formula: $$x^2 - y^2=(x - y)(x + y)$$ $$=\frac{(a + b)^2 - c^2}{2ab}$$ $$=\frac{(a + b - c)(a + b + c)}{2ab}$$ $$1 - \text{cos}\hspace{.1em}C=1 - \frac{a^2 + b^2 - c^2}{2ab}$$ $$=\frac{2ab}{2ab}- \frac{a^2 + b^2 - c^2}{2ab}$$ $$=\frac{2ab - (a^2 + b^2 - c^2)}{2ab}$$ $$=\frac{2ab - a^2 - b^2 + c^2}{2ab}$$ $$=\frac{c^2 - a^2 + 2ab - b^2}{2ab}$$ $$=\frac{c^2 - (a^2 - 2ab + b^2)}{2ab}$$ $$=\frac{c^2 - (a - b)^2}{2ab}$$ $$=\frac{(c + (a - b))(c - (a - b))}{2ab}$$ $$=\frac{(c + a - b)(c - a + b)}{2ab}$$ Let's return to: $$\text{Area}^2=\frac{1}{4}a^2 \hspace{.1em}b^2 \cdot (1 - \text{cos}\hspace{.1em}C)(1 + \text{cos}\hspace{.1em}C)$$ Plug in for 1 - cos C and 1 + cos C: $$=\frac{1}{4}a^2 \hspace{.1em}b^2 \cdot \frac{(c + a - b)(c - a + b)}{2ab}\cdot \frac{(a + b - c)(a + b + c)}{2ab}$$ Cancel common factors: $$\require{cancel}=\frac{1}{4}\cdot \frac{\cancel{a^2}\cdot \cancel{b^2}}{\cancel{a^2}\cdot \cancel{b^2}}\cdot \frac{(c + a - b)(c - a + b)}{2}\cdot \frac{(a + b - c)(a + b + c)}{2}$$ $$=\frac{(a + b + c)}{2}\cdot \frac{(a + b - c)}{2}\cdot \frac{(c + a - b)}{2}\cdot \frac{(c - a + b)}{2}$$ Recall the formula for semiperimeter: $$s=\frac{a + b + c}{2}$$ Note: We will need to write things in terms of (a + b + c)/(2). We will only show one and the remaining two are done in a similar way.
Add 2c and subtract 2c in the numerator. $$\frac{a + b - c}{2}=\frac{a + b - c + 2c - 2c}{2}$$ $$=\frac{a + b + c - 2c}{2}$$ $$=\frac{a + b + c}{2}- \frac{\cancel{2}c}{\cancel{2}}$$ $$=s - c$$ $$\text{Area}^2=s(s - c)(s - b)(s - a)$$ Heron's Formula:$$\text{Area}=\sqrt{s(s - a)(s - b)(s - c)}$$
