Question

right triangles practicename: _______________word problemsdirections: draw a picture for each problem. set up an equation to solve for the missing information. solve.1. tyler stands facing a tree that is 9 meters tall. he places a straight stick on the ground at his feet, angles it up until it points to the top of the tree, and measures the angle as 55 degrees. how many meters away is he standing from the tree?2. ethan stands 28 meters from a tree. he places a straight stick on the ground at his feet, angles it up until it points to the top of the tree, and measures the angle as 37 degrees. how many meters tall is the tree?3. a praire dog runs 40 meters down a straight slope from the top to the bottom of a hill that is 20 meters tall. when she reaches the bottom of the hill and sees a fox, the praire dog digs a tunnel into the hill, parallel with the flat ground, until she is directly under the top of the hill. how many meters long is the tunnel?4. emily leaves her house to buy milk at a nearby story. she walks 182 meters up one street, turns 90 degrees left and walks another 140 meters to the store. to get home she walks straight through a field from the store to the house. how many meters long is the walk home?5. a man decided to patch up the roof of his house. he placed the ladder against the wall of his house. his house measured 9 feet high. when the man looked down, his angle of depression was 41 degrees. how far did the man set his ladder from the house?

Explanation:

Problem 1

Step1: Identify trigonometric ratio

We have opposite side $\text{opp}=9$ m, angle $\theta=55^\circ$, need adjacent side $\text{adj}=x$. Use $\tan\theta=\frac{\text{opp}}{\text{adj}}$.
$\tan(55^\circ)=\frac{9}{x}$

Step2: Rearrange to solve for $x$

$x=\frac{9}{\tan(55^\circ)}$
Calculate $\tan(55^\circ)\approx1.4281$, so $x\approx\frac{9}{1.4281}\approx6.30$

Problem 2

Step1: Identify trigonometric ratio

We have adjacent side $\text{adj}=28$ m, angle $\theta=37^\circ$, need opposite side $\text{opp}=x$. Use $\tan\theta=\frac{\text{opp}}{\text{adj}}$.
$\tan(37^\circ)=\frac{x}{28}$

Step2: Rearrange to solve for $x$

$x=28\times\tan(37^\circ)$
Calculate $\tan(37^\circ)\approx0.7536$, so $x\approx28\times0.7536\approx21.10$

Problem 3

Step1: Apply Pythagorean theorem

We have hypotenuse $c=40$ m, one leg $a=20$ m, need leg $b=x$. Use $a^2+b^2=c^2$.
$20^2+x^2=40^2$

Step2: Rearrange to solve for $x$

$x^2=40^2-20^2=1600-400=1200$
$x=\sqrt{1200}=20\sqrt{3}\approx34.64$

Problem 4

Step1: Apply Pythagorean theorem

We have legs $a=182$ m, $b=140$ m, need hypotenuse $c=x$. Use $a^2+b^2=c^2$.
$182^2+140^2=x^2$

Step2: Calculate and solve for $x$

$x^2=33124+19600=52724$
$x=\sqrt{52724}=229.62$ (or exactly $2\sqrt{13181}$)

Problem 5

Step1: Identify trigonometric ratio

Angle of depression = angle of elevation $\theta=41^\circ$, opposite side $\text{opp}=9$ ft, need adjacent side $\text{adj}=x$. Use $\tan\theta=\frac{\text{opp}}{\text{adj}}$.
$\tan(41^\circ)=\frac{9}{x}$

Step2: Rearrange to solve for $x$

$x=\frac{9}{\tan(41^\circ)}$
Calculate $\tan(41^\circ)\approx0.8693$, so $x\approx\frac{9}{0.8693}\approx10.35$

Answer:

1. $\approx6.30$ meters
2. $\approx21.10$ meters
3. $\approx34.64$ meters (or $20\sqrt{3}$ meters)
4. $\approx229.62$ meters
5. $\approx10.35$ feet