# A triangle has sides A, B, and C. Sides A and B have lengths of 7 and 2, respectively. The angle between A and C is (11pi)/24 and the angle between B and C is  (3pi)/8. What is the area of the triangle?

Jun 21, 2018

$A = \frac{7}{2}$

#### Explanation:

The third angle is given by
$\pi - \frac{3}{8} \cdot \pi - \frac{11}{24} \cdot \pi = \frac{\pi}{6}$
Using that
$A = \frac{1}{2} \cdot a \cdot b \cdot \sin \left(\gamma\right)$ we get

$A = \frac{1}{2} \cdot 7 \cdot 2 \cdot \sin \left(\frac{\pi}{6}\right)$ or

$A = \frac{7}{2}$

Aug 10, 2018

Hence with given measurements we cannot form a triangle.

#### Explanation:

$a = 7 , b = 2 , \hat{A} = \frac{3 \pi}{8} , \hat{B} = \frac{11 \pi}{24}$

Lawof sines. : a / sin A = b / sin B#

$\frac{a}{\sin} A = \frac{7}{\sin} \left(\frac{3 \pi}{8}\right) \approx 7.5767$

$\frac{b}{\sin} B = \frac{2}{\sin} \left(\frac{3 \pi}{8}\right) \approx 2.1648$

From the above we can see,

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

Aug 10, 2018

We cannot form a triangle with given measurements.

#### Explanation:

$a = 7 , b = 2 , \hat{A} = \frac{3 \pi}{8} = \frac{9 \pi}{24} , \hat{B} = \frac{12 \pi}{24}$

Greater side will have greater angle opposite to it.

Since $a > b \left(7 > 2\right) , \hat{A}$ must be $\hat{B}$

But $\hat{A} \frac{9 \pi}{24} < \hat{B} \frac{11 \pi}{24}$

Since the values do not satisfy the theorem, we cannot form a triangle with given measurements.