How do you solve the triangle given f=10, g=11, h=4?

Feb 24, 2018

It’s a right triangle.

Measure of the Three angles of the triangle are

color(blue)(hat F = (pi/2)^c or 1.57^c or 90^@

color(red)(hat G = 1.1987^c or 68.68^@

color(green)(hat H = 0.3721^c or 21.32^@

Explanation: From the table above, having known three sides, we can find the area of the triangle using semi perimeter formula,

${A}_{t} \sqrt{s \left(s - f\right) \left(s - g\right) \left(s - h\right)}$ where s is the semi perimeter of the triangle and

$s = \frac{a + b + c}{2} = \frac{10 + 11 + 4}{2} = 12.5$

${A}_{t} = \sqrt{12.5 \cdot \left(12.5 - 10\right) \cdot \left(12.5 - 11\right) \cdot \left(12.5 - 4\right)} \approx \textcolor{b r o w n}{20}$

But we know, ${A}_{t} = \left(\frac{1}{2}\right) f g \sin \left(\hat{H}\right)$

$\therefore \sin H = \frac{2 \cdot {A}_{t}}{f \cdot g} = \frac{2 \cdot 20}{10 \cdot 11} = 0.3636$

$\hat{H} = {\sin}^{-} 1 \left(0.3636\right) \approx {0.3721}^{c}$

$\sin \hat{G} = \frac{g \cdot \sin \hat{H}}{g} = \frac{11 \cdot 0.3636}{4} = 1$

$\hat{G} = {\sin}^{- 1} 1 = {90}^{\circ} \mathmr{and} \frac{\pi}{2} ^ c \mathmr{and} {1.57}^{c}$ $\hat{F} = \pi - 0.3721 - \left(\frac{\pi}{2}\right) = {1.1987}^{c}$

It’s a right triangle.