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Answer: approximately 49 feets
Step-by-step explanation:
The diagram of the tree is shown in the attached photo. The tree fell with its tip forming an angle of 36 degrees with the ground. It forms a right angle triangle,ABC. Angle C is gotten by subtracting the sum of angle A and angle B from 180(sum of angles in a triangle is 180 degrees).
To determine the height of the tree, we will apply trigonometric ratio
Tan # = opposite/ adjacent
Where # = 36 degrees
Opposite = x feets
Adjacent = 25 feets
Tan 36 = x/25
x = 25tan36
x = 25 × 0.7265
x = 18.1625
Height of the tree from the ground to the point where it broke = x = 18.1625 meters.
The entire height of the tree would be the the length of the fallen side of the tree, y + 18.1625m
To get y, we will use Pythagoras theorem
y^2 = 25^2 + 18.1625^2
y^2 = 625 + 329.88
y^2 = 954.88
y = √954.88 = 30.9 meters
Height of the tree before falling was
18.1625+30.9 = 49.0625
The height of the tree was approximately 49 feets
9514 1404 393
Answer:
16.8°
Step-by-step explanation:
The Law of Cosines can be rearranged to give the angle measure.
angle Q = arccos((p² +r² -q²)/(2pr))
angle Q = arccos((690² +450² -290²)/(2×690×450))
angle Q = arccos(594500/621000) ≈ 16.7985°
The measure of angle Q is about 16.8°.
