Question

Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. (If an answer does not exist, enter DNE. Round your answers to one decimal place. Below, enter your answers so that ∠B1 is larger than ∠B2.) a = 33, c = 48, ∠A = 33°

Answer 1

Answer:

For a triangle with a = 33, c = 48 and the angle $\alpha=33^o$, the missing side is b = 60.4, and the missing angles are $\beta=94.6^o$and $\gamma = 52.4^o$.

Step-by-step explanation:

The Law of Sines says that in any given triangle, the ratio of any side length to the sine of its opposite angle is the same for all three sides of the triangle. This is true for any triangle, not just right triangles.

                                            $\frac{\sin\alpha}{a} = \frac{\cos\beta}{b} = \frac{cos\gamma}{c}$

We know two sides a = 33, c = 48 and the angle $\alpha=33^o$, so we use the law of sines to find the angle $\gamma$ as follows:

$\frac{ 33 }{ \sin(33^o) } =\frac{ 48 }{ \sin(\gamma) }\\\\\sin(\gamma) \cdot 33 = 48 \cdot \sin(33^o)\\\sin(\gamma) \cdot 33 = 48 \cdot 0.5446\\\sin(\gamma) \cdot 33 = 26.1427\\\sin( \gamma ) = 0.7922\\\gamma = 52.4^o$

One of the basic properties of triangles is that the sum of the measure of angles, in every triangle, is 180°.

So,

$180=\alpha+\beta+\gamma\\180=33+\beta+52.4\\\beta+85.4=180\\\beta=94.6^o$

To find the side b we use the law of sines:

$\frac{ 33 }{ \sin(33^o) } =\frac{ b }{ \sin(94.6^o) }\\\\b=\frac{33\sin \left(94.6^{\circ \:}\right)}{\sin \left(33^{\circ \:}\right)}\\\\b=60.4$