Answer:
5 units
Step-by-step explanation:
Let point O be the point of intersection of the kite diagonals.
|OF| = 2, |OH| = 5
|FH| = |OF| + |OH| = 2 + 5 = 7
FH and EG are the diagonals of the kite. Hence the area of thee kite is:
Area of kite EFGH = (FH * EG) / 2
Substituting:
35 = (7 * |EG|) / 2
|EG| * 7 = 70
|EG| = 10 units
The longer diagonal of a kite bisects the shorter one, therefore |GO| = |EO| = 10 / 2 = 5 units
x = |GO| = |EO| = 5 units
The line of sight is the hypotenuse of a right triangle with short leg 400 m, 90 degree angle where the short leg meets the ground, an 83 degree angle at the top, and a 7 degree angle across from the right angle on the ground. Because this 7 degree angle is an alternate interior angle with the angle of depression, they are the same degree measure. Looking for the hypotenuse, we use the sin ratio: sin (7) = 400/x.
The formula is base x height
Answer:
56
Step-by-step explanation:
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