# Given a triangle with sides measuring 15in, 10in, and 12in; with angle C across from 10in, angle A across from 12in, and angle T across from 15in. How do you find the angle measures?

May 5, 2018

color(maroon)("Measure of Angles " color(indigo)(hat A = 52.89^@, hat C = 41.62^@, hat T = 85.49^@

#### Explanation:

$\text{Given : " a = 12 " in", c = 10 " in", t = 15 " in}$

color(green)("Knowing three sides of a triangle, we can use Heron's formula to calculate the area."

$s = \frac{a + c + t}{2} = \frac{12 + 10 + 15}{2} = 18.5$

${A}_{t} = \sqrt{18.5 \cdot \left(18.5 - 12\right) \cdot \left(18.5 - 10\right) \cdot \left(18.5 - 3\right)} = 59.8 {\text{ in}}^{2}$

color(crimson)("Using two sides and included angle to find area",

${A}_{t} = \left(\frac{1}{2}\right) \cdot a \cdot c \cdot \sin T$

$\sin T = \frac{2 \cdot 59.8}{12 \cdot 10} = 0.9969$

$\hat{T} = {\sin}^{-} 1 0.9969 \approx 85.49 \circ$

color(maroon)("Using Law of Sines"

$\frac{a}{\sin} A = \frac{c}{\sin} C = \frac{t}{\sin} T$

$\frac{12}{\sin} A = \frac{10}{\sin} C = \frac{15}{\sin} \left({85.49}^{\circ}\right)$

$\sin A = \frac{12 \cdot \sin \left(85.49\right)}{15} = 0.7975$

$\hat{A} = {\sin}^{-} 1 0.7975 = {52.89}^{\circ}$

$\hat{C} = 180 - \hat{A} - \hat{T} = 180 - 85.49 - 52.89 = {41.62}^{\circ}$

May 5, 2018

color(indigo)(hat A = 53.13@, hat C = 41.65@, hat T = 85.22@

#### Explanation:

$a = 12 , c = 10 , t = 15$

$\cos A = \frac{{c}^{2} + {t}^{2} - {a}^{2}}{2 c t}$

$\cos A = \frac{{10}^{2} + {15}^{2} - {12}^{2}}{2 \cdot 10 \cdot 15} = 0.6$

$\hat{A} = {\cos}^{-} 1 0.6 = 53.13 \circ$

$\cos C = \frac{{a}^{2} + {t}^{2} - {c}^{2}}{2 a t}$

$\cos C = \frac{{12}^{2} + {15}^{2} - {10}^{2}}{2 \cdot 12 \cdot 15} = 0.7472$

$\hat{C} = {\cos}^{-} 1 0.7472 = 41.65 \circ$

$\hat{T} = 180 - \hat{A} - \hat{C} = 180 - 53.13 - 41.65 = 85.22 \circ$