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In this section, you will:
  • Use the Law of Sines to solve oblique triangles.
  • Find the area of an oblique triangle using the sine function.
  • Solve applied problems using the Law of Sines.

Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation    measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in [link] that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles .

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 line perpendicular to the ground side connecting the airplane vertex with the ground - an altitude line.

Using the law of sines to solve oblique triangles

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    . 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:

  1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See [link] .
    An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma's values are known, as is the side opposite beta, between alpha and gamma.
  2. AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See [link] .
    An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma are known, as is the side opposite alpha, between beta and gamma.
  3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See [link] .
    An oblique triangle consisting of angles alpha, beta, and gamma. Alpha is the only angle known. Two sides are known. The first is opposite alpha, between beta and gamma, and the second is opposite gamma, between alpha and beta.

Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. 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. Let’s see how this statement is derived by considering the triangle shown in [link] .

An oblique triangle consisting of sides a, b, and c, and angles alpha, beta, and gamma. Side c is opposide angle gamma and is the horizontal base of the triangle. Side b is opposite angle beta, and side a is opposite angle alpha. There is a dotted perpendicular line - an altitude - from the gamma angle to the horizontal base c.

Using the right triangle relationships, we know that sin α = h b and sin β = h a . Solving both equations for h gives two different expressions for h .

h = b sin α  and  h = a sin β

We then set the expressions equal to each other.

            b sin α = a sin β   ( 1 a b ) ( b sin α ) = ( a sin β ) ( 1 a b ) Multiply both sides by 1 a b .                sin α a = sin β b

Similarly, we can compare the other ratios.

sin α a = sin γ c  and  sin β b = sin γ c

Collectively, these relationships are called the Law of Sines .

sin α a = sin β b = sin λ c

Note the standard way of labeling triangles: angle α (alpha) is opposite side a ; angle β (beta) is opposite side b ; and angle γ (gamma) is opposite side c . See [link] .

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.

A triangle with standard labels.

Law of sines

Given a triangle with angles and opposite sides labeled as in [link] , the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines    is based on proportions and is presented symbolically two ways.

sin α a = sin β b = sin γ c
a sin α = b sin β = c sin γ

To solve an oblique triangle, use any pair of applicable ratios.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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