algebra-precalculus
How would I solve the following trig questions.
A tree broken over by the winds forms a right angle with the ground.If the broken part makes angle of 50 degrees with the ground and if the top of the tree is now 20 feet from the base how tall was the tree?
Two straight roads intersect to form an angle 0f 75 degrees. Find the shortest distance from one road to a gas station on the other road 1000 m from the junction.
Here is a diagram illustrating the first problem. It should help you understand what calculations you need to do.
I assume that by height to base ratio you mean the ratio of total height (cylinder plus cone) to the diameter of the base.
Let $\theta$ be the angle between the base of the cone and the sloping walls. Then the height of the cone is $r\tan\theta$. In our case $\tan\theta=\frac{1}{\sqrt{3}}$. Kind of a stubby cone. We keep on writing $\tan\theta$ instead of $\frac{1}{\sqrt{3}}$, it makes typing easier.
Let $h$ be the height of the cylinder. Then $h+r\tan\theta=2(2r)$, and therefore $h=r(4-\tan\theta)$.
The combined volume is $\pi r^2h+\frac{1}{3}\pi r^3\tan\theta$. Substituting for $h$, and simplifying a little, we find that the volume is $$\pi\left(4-\frac{2}{3}\tan\theta \right)r^3.$$ Now we can set this equal to $46000$ and solve for $r$. I get something close to $15.94$, but do check!
Best Answer
$1.$ Draw a picture. Let $b$ be the current (broken) height of the tree, that is, the distance from the ground to the break, which is really only a sharp bend. Then $$\frac{b}{20}=\tan(50^\circ).$$ Let $\ell$ be the length of the "leaning" part. Then $$\frac{20}{\ell}=\cos(50^\circ).$$ Finally, the original height of the tree is $b+\ell$.
The calculation: We have $b=20\tan(50^\circ)\approx 23.835$, and $\ell=\frac{20}{\cos(50^\circ)}\approx 31.114$, giving a sum of $\approx 54.95$.
$2.$ Draw a picture. Let $O$ be the point of intersection of the streets, $G$ the gas station, and $P$ the point on the other road nearest to the gas station. Then $\angle OPG$ is a right angle. Thus $$\frac{PG}{OG}=\sin(75^\circ).$$
The calculator gives the distance as approximately $965.9$. Presumably you have instructions about what kind of rounding to do.