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In ΔVWX, w = 680 cm, ∠X=80° and ∠V=32°. Find the area of ΔVWX, to the nearest 10th of a square centimeter.

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To find the area of triangle VWX, we can use the formula for the area of a triangle:

Area = (1/2) * base * height

In this case, we have two sides and the included angle, so we can use the formula:

Area = (1/2) * VW * WX * sin(V)

First, let's calculate the lengths of sides VW and WX using the law of sines:

VW / sin(X) = WX / sin(V)

VW / sin(80°) = WX / sin(32°)

VW = (WX * sin(80°)) / sin(32°)

Now, substitute the value of VW in the area formula:

Area = (1/2) * [(WX * sin(80°)) / sin(32°)] * WX * sin(V)

Area = (1/2) * WX^2 * (sin(80°) * sin(V)) / sin(32°)

Area = (1/2) * WX^2 * (sin(80°) * sin(32°)) / sin(32°)

Area ≈ (1/2) * WX^2 * 0.7365 (rounded to four decimal places)

Now, substitute the value of WX and calculate the area:

Area ≈ (1/2) * (680 cm)^2 * 0.7365

Area ≈ 139,850.4 cm^2

Therefore, the area of triangle VWX is approximately 139,850.4 square centimeters to the nearest 10th of a square centimeter.
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User Mark A Johnson
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