in triangle DEF,<D=50 degree, DE=14 meters, and DF=10 meters. find the area of triangle DEFto the nearest tenth of a square meters
If you have a triangle with sides a and b and Ø is the angle between them, then the area is
(1/2)ab sinØ
you have that condition, so
Area = (1/2)(14)(10)sin50° = appr 53.623
or to the nearest tenth ----> 53.6 m^2
To find the area of triangle DEF, we can use the formula for the area of a triangle: A = (1/2) * base * height.
In this case, we know that triangle DEF has a base of DE and a height of DF. We are given that DE = 14 meters and DF = 10 meters.
First, we need to find the length of the height of the triangle, which is the perpendicular distance from vertex D to the base EF. To do this, we can use the sine function.
Using the sine function, we have: sin(D) = opposite / hypotenuse
Since angle D is 50 degrees and DE is the opposite side, we can say: sin(50) = DF / DE
Solving for DF, we have: DF = DE * sin(50)
Now, we can substitute the values into the formula for the area:
A = (1/2) * DE * DF
Substituting the given values, we have: A = (1/2) * 14 * (14 * sin(50))
Calculating this expression, we find the area of triangle DEF to the nearest tenth of a square meter.