Two artillery bases are located at the points A and C represented in the figure. Each of the bases locates an enemy airplane at point B. The base at A measures the angles of elevation to the airplane to be 95∘ and the distance between the two bases is 739 meters. If the distance from the base at C to the plane is 2,517 meters, what is the measure, in degrees, of ∠CBA?

Answer:

the answer is 17 here is the step below

Step-by-step explanation:

Use the Law of Sines to find the distance. For a triangle with angles A, B, and C, and opposite sides a, b, and c, the Law of Sines states that sinAa=sinBb=sinCc.

In this case, the angle at point B, let it be α, needs to be found. The Law of Sines can be used to relate α, the two known sides, and the angle at point A.

The Law of Sines states that sinα739=sin95∘2517.

By simplifying, sinα=739sin95∘2517. Since α is acute, α=sin−1(739sin95∘2517)≈17.0∘.

The measure of the angle is 17 degrees if the A measures the angle of elevation to the airplane to be 95∘ and the distance between the two bases is 739 meters.

In terms of geometry, the triangle is a three-sided polygon with three edges and three vertices. The triangle's interior angles add up to 180°.

We have a triangle shown in the picture

To find the measure of the angle ∠CBA

Applying sin law:

sin95/2517 = sin∠CBA/739

0.292 = sin∠CBA

∠CBA = sin⁻¹(0.292)

∠CBA = 17 degrees

Thus, the measure of the angle is 17 degrees if the A measures the angles of elevation to the airplane to be 95∘ and the distance between the two bases is 739 meters.

Learn more about the triangle here:

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