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# Sin 85 Degrees

Learning Objectives Use the Law of Sines to solve oblique triangles and applied problems In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.

Any triangle that is not a right triangle is an oblique triangle. It could be an acute triangle all three angles of the triangle are less than right angles or it could be an obtuse triangle one of the three angles is greater than a right angle.

Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides.

We will investigate three possible oblique triangle problem situations: ASA angle-side-angle The measurements of two angles and the included side are known. AAS angle-angle-side The measurements of two angles and a side opposite one of those angles is known. SSA side-side-angle The measurements of two sides and an angle opposite one of those sides is known.

Knowing how to approach each of these situations enables oblique triangles to be solved without having to drop a perpendicular to form two right triangles. Instead, the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side can be used.

Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side.

While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways.

In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.

It appears that there may be a second triangle that will fit the given criteria. Remember that the sine function is positive in both the first and second quadrants. If there is more than one possible solution, show both.

Round your answers to the nearest tenth. It is important to verify the result, as there may be two viable solutions, only one solution the usual case , or no solutions.