A 100-foot rope from the top of a tree house to the ground forms a 45∘ angle of elevation from the ground. How high is the top of the tree house? Round your answer to the nearest tenth of a foot
We can compare this problem to a puthagreom therom problem. We will be using the formula a^2+b^2=c^2
So we know that the trees hight a, and the the grounds distance b, form a right angle with the ropes legnth, c.
Since there is a 45 degree angle, and there is a right angle, 90 degrees, we add these together and subtract them from 180 since a triangle has a total of 180 degrees through its 3 angles. This gets us 45 degrees.
From that we know that a and b will be equal, since their angles are equal.
We put the values we know into the equation to get a^2+b^2=100^2=10,000
Since a and b are equal, we can simplify to get 2*(a^2)=10,000
Divide both sides by 2 to get a^2=5000
Find the square root of both sides to get a equals the square root of 5000.
The top of the treehouse is at the square root of 5000 feet, rounded to the nearest 10th of a foot, is 70.7 feet.
Hope this helps!
Oh and this isnt geometry/trig this is pre-algebra
Another way to find that height is to use the TRIG sine function.
It's necessary to posit that the tree is perpendicular to the ground, i.e., forms a 90o angle.
The sine of an angle in a right triangle is the opposite side over the hypotenuse.
So, sin 45o = x/100 where x is the height of the treehouse and 100 is the length of the rope
You have to look up the value for the sin 45o which we find to be 0.7071.
So, plugging in that value for the sine: 0.7071 = x/100
I just happen to prefer the unknown on the left, so I turn it around: x/100 = 0.7071
Multiply both sides by 100: x = 100 times 0.7071
x = 70.7 feet
That's a pretty doggone high treehouse.