A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/4, the angle between sides B and C is (7pi)/12, and the length of B is 5, what is the area of the triangle?

Oct 29, 2017

Area$= 17.075 u n i t {s}^{2}$

Explanation:

${\cancel{\pi}}^{\textcolor{m a \ge n t a}{1}} / {\cancel{4}}^{\textcolor{m a \ge n t a}{1}} \times {\cancel{180}}^{\textcolor{m a \ge n t a}{45}} / {\cancel{\pi}}^{\textcolor{m a \ge n t a}{1}} = {45}^{\circ} = \angle c$

$\frac{7 {\cancel{\pi}}^{\textcolor{m a \ge n t a}{1}}}{\cancel{12}} ^ \textcolor{m a \ge n t a}{1} \times {\cancel{180}}^{\textcolor{m a \ge n t a}{15}} / {\cancel{\pi}}^{\textcolor{m a \ge n t a}{1}} = {105}^{\circ} = \angle a$

${180}^{\circ} - \left({105}^{\circ} + {45}^{\circ}\right) = {30}^{\circ} = \angle b$

$\frac{A}{\sin \angle a} = \frac{B}{\sin \angle b}$

$A = \frac{\sin {105}^{\circ} \times 5}{\sin} {30}^{\circ}$

$A = \frac{4.829629131}{0.5}$

$A = 9.659258262$

color(magenta)(A=9.659units to the nearest 3 decimal places

$\frac{C}{\sin \angle c} = \frac{B}{\sin \angle b}$

$C = \frac{5 \cdot \sin {45}^{\circ}}{\sin} {30}^{\circ}$

$C = \frac{3.535533906}{0.5}$

$C = 7.071067812$

color(magenta)(C=7.071units to the nearest 3 decimal places

Area$= \frac{1}{2} B C \sin \angle a$

$0.5 \cdot 5 \cdot 7.071067812 \cdot \sin {105}^{\circ}$

Area$= 17.07531755 u n i t {s}^{2}$

Areacolor(magenta)(=17.075units^2 to the nearest 3 decimal places

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Check:-

Hero's formula:-

$S = \frac{a + b + c}{2}$

$S = \frac{9.659 + 5 + 7.071}{2}$

Area$= \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

$S = 10.865$units

Area$\sqrt{10.865 \left(10.865 - 9.659\right) \left(10.865 - 5\right) \left(10.865 - 7.071\right)}$

$A r e a = \sqrt{10.865 \left(1.206\right) \left(5.865\right) \left(3.794\right)}$

Areacolor(magenta)(=17.075units^2 to the nearest 3 decimal places