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Two oblique triangles with standard labels. Both have a dotted altitude line h extended from angle beta to the horizontal base side b. In the first, which is an acute triangle, the altitude is within the triangle. In the second, which is an obtuse triangle, the altitude h is outside of the triangle.

Thus,

Area = 1 2 ( base ) ( height ) = 1 2 b ( c sin α )

Similarly,

Area = 1 2 a ( b sin γ ) = 1 2 a ( c sin β )

Area of an oblique triangle

The formula for the area of an oblique triangle is given by

Area = 1 2 b c sin α = 1 2 a c sin β = 1 2 a b sin γ

This is equivalent to one-half of the product of two sides and the sine of their included angle.

Finding the area of an oblique triangle

Find the area of a triangle with sides a = 90 , b = 52 , and angle γ = 102° . Round the area to the nearest integer.

Using the formula, we have

Area = 1 2 a b sin γ Area = 1 2 ( 90 ) ( 52 ) sin ( 102° ) Area 2289 square units
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Find the area of the triangle given β = 42° , a = 7.2 ft , c = 3.4 ft . Round the area to the nearest tenth.

about 8.2 square feet

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Solving applied problems using the law of sines

The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.

Finding an altitude

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in [link] . Round the altitude to the nearest tenth of a mile.

A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted altitude line perpendicular to the ground side connecting the airplane vertex with the ground.

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side a , and then use right triangle relationships to find the height of the aircraft, h .

Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.

   sin ( 130° ) 20 = sin ( 35° ) a a sin ( 130° ) = 20 sin ( 35° )                 a = 20 sin ( 35° ) sin ( 130° )                 a 14.98

The distance from one station to the aircraft is about 14.98 miles.

Now that we know a , we can use right triangle relationships to solve for h .

sin ( 15° ) = opposite hypotenuse sin ( 15° ) = h a sin ( 15° ) = h 14.98             h = 14.98 sin ( 15° )            h 3.88

The aircraft is at an altitude of approximately 3.9 miles.

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The diagram shown in [link] represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point B , is 62°, and the distance between the viewing points of the two end zones is 145 yards.

An oblique triangle formed from three vertices A, B, and C. Verticies A and B are points on the ground, and vertex C is the blimp in the air between them. The distance between A and B is 145 yards. The angle at vertex A is 70 degrees, and the angle at vertex B is 62 degrees.

161.9 yd.

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Access these online resources for additional instruction and practice with trigonometric applications.

Key equations

Law of Sines sin α a = sin β b = sin γ c a sin α = b sin β = c sin γ
Area for oblique triangles Area = 1 2 b c sin α         = 1 2 a c sin β         = 1 2 a b sin γ

Key concepts

  • The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
  • According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
  • There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See [link] .
  • The ambiguous case arises when an oblique triangle can have different outcomes.
  • There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See [link] and [link] .
  • The Law of Sines can be used to solve triangles with given criteria. See [link] .
  • The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See [link] .
  • There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See [link] .

Questions & Answers

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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