A triangle has sides A, B, and C. Sides A and B have lengths of 8 and 12, respectively. The angle between A and C is (3pi)/4 and the angle between B and C is  pi/3. What is the area of the triangle?

Apr 24, 2018

This can't be a real triangle the Law of Sines is not satisfied.

Explanation:

No no no! Triangles have sides $a , b , c$ and vertices $A , B , C$. Here $a = 8 , b = 12 ,$ $B = \frac{3 \setminus \pi}{4} ,$ $A = \setminus \frac{\pi}{3}$.

Yet another trig problem based only on angles related to 30 and 45 degrees.

This can't be a real triangle because:

$\frac{a}{\sin} A \ne \frac{b}{\sin} B$

$\frac{a}{\sin} A = \frac{8}{\sin} \left(\frac{\pi}{3}\right) = \frac{8}{\setminus \frac{\sqrt{3}}{2}} = \frac{16}{\sqrt{3}}$

$\frac{b}{\sin} A = \frac{12}{\sin} \left(3 \frac{\pi}{4}\right) = \frac{12}{\frac{\sqrt{2}}{2}} = \frac{24}{\sqrt{2}}$

They're not equal.

That's too bad, because the area would have been calculated as:

$\textrm{a r e a} = \frac{1}{2} a b \sin C$

$= \frac{1}{2} \left(8\right) \left(12\right) \sin \left(\pi - \frac{3 \pi}{4} - \frac{\pi}{3}\right)$

$= 48 \sin \left(\frac{3 \pi}{4} - \frac{\pi}{3}\right)$

$= 48 \left(\sin \left(\frac{3 \pi}{4}\right) \cos \left(\frac{\pi}{3}\right) - \cos \left(\frac{3 \pi}{4}\right) \sin \left(\frac{\pi}{3}\right)\right)$

$= 48 \left(\left(\setminus \frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right) - \left(- \setminus \frac{\sqrt{2}}{2}\right) \left(\setminus \frac{\sqrt{3}}{2}\right)\right)$

$= 12 \left(\sqrt{2} + \sqrt{6}\right)$

But it's not.

Jun 10, 2018

color(crimson)("Such a triangle cannot exist"

Explanation:

$a = 8 , \hat{A} = \frac{\pi}{3} , b = 12 , \hat{B} = \frac{3 \pi}{4}$

color(maroon)("THEOREM. A greater angle of a triangle is opposite a greater side. "
color(brown)("Let ABC be a triangle in which angle ABC is "

color(brown)("greater than angle BCA; then side AC is also greater than side AB"

color(crimson)("But though " a < b, hat A > hat B

color(crimson)("Such a triangle cannot exist"