The area of the space under the slide is 2.10 square meters
The vertical distance
The diagram in the question is added as an attachment
The vertical distance is represented by h.
So, we have:
sin(70) = h/2
Solve for h
h = 2 * sin(70)
Evaluate
h = 1.88 m
Hence, the vertical distance is 1.88 m
The length of the slide
This is calculated using
sin(40) = h/Slide
This gives
sin(40) = 1.88/Slide
Solve for Slide
Slide = 1.88/sin(40)
Evaluate
Slide = 2.92
Hence, the length of the slide is 2.92 m
The area of the space under the slide
This is calculated using:
A = 0.5 * h * slide * sin(∅)
Where
∅ = 90 - 40 = 50
So, we have:
A = 0.5 * 1.88 * 2.92 * sin(50)
Evaluate
A = 2.10
Hence, the area of the space under the slide is 2.10 square meters
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