A tree is broken over by the wind. It forms a right angled triangle with the ground. If the broken part makes an angle of 60° with the ground and the top of the tree is now 20m from its base, how tall WAS the tree?
In triangle abc angle c = 60°
tan60°= ab/bc
√3=ab/20
AB=20√3
To solve this problem, we can use trigonometry and the concept of right-angled triangles. Let's label the triangle formed by the broken tree as ABC, where A is the top of the tree, B is the base of the tree, and C is the point where the tree is broken and makes an angle of 60° with the ground.
Now, we need to find the height of the tree, which is the length of segment AB. We already know the length of segment AC, which is 20m.
To find the length of AB, we can use the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle. In this case, AB is the opposite side, and BC is the adjacent side.
So, we have the equation: tan(60°) = AB / BC
The tangent of 60° is √3. Therefore, we can rewrite the equation as: √3 = AB / BC
Since we know that BC is 20m, we can substitute this value into the equation: √3 = AB / 20
To isolate AB, we can multiply both sides of the equation by 20: AB = 20 * √3
Calculating this expression gives us the height of the tree: AB ≈ 34.64m
Therefore, the tree was approximately 34.64 meters tall before it was broken by the wind.
the figure above represents a pole that has broken 9 feet from its base. the top portion has fallen as if hinged. the part of the pole remaining upright is perpendicular to the ground. if the top of the pole hit the ground 12 feet from its base, how tall was the pole before it broke?
height + hypotenuse
height = 10 tan 60
hypotenus = 10/cos 60 = 20
so
17.32 + 20
= 37.32 meters