Angles of elevation and depression are used to solve real-world right triangle problems involving heights, distances, and lines of sight. An angle of elevation looks upward from a horizontal line, while an angle of depression looks downward from a horizontal line. These problems use sine, cosine, and tangent to find missing heights, ground distances, and direct line-of-sight distances.
Practice Problems
\(\textbf{1)}\) You are standing 10 feet from the base of a tree. You look up at the top of the tree with an angle of elevation of 60 degrees. How tall is the tree?
The answer is \(\approx 17.3 \) feet
\(\,\,\,\,\,\,\tan{60}=\frac{x}{10}\)
\(\,\,\,\,\,\,\sqrt{3}=\frac{x}{10}\)
\(\,\,\,\,\,\,10\sqrt{3}=x\)
\(\,\,\,\,\,\,x\approx17.3\)
\(\,\,\,\,\,\,\text{The answer is }17.3\text{ feet.}\)
\(\textbf{2)}\) You are in a hot air balloon. You look at Steve with an angle of depression of 30 degrees. Your elevation is 1200 feet. How far apart are you and Steve?
The answer is \( 2400 \) feet
\(\,\,\,\,\,\,\sin{30}=\frac{1200}{x}\)
\(\,\,\,\,\,\,\frac{1}{2}=\frac{1200}{x}\)
\(\,\,\,\,\,\,x=2400\)
\(\,\,\,\,\,\,\text{The answer is } 2400 \text{ feet.}\)
\(\textbf{3)}\) You are flying a kite. You let out 40 feet of string at an angle of elevation of 40 degrees. How high up is the kite?
The answer is \( 25.7 \) feet
\(\,\,\,\,\,\,\sin{40}=\frac{x}{40}\)
\(\,\,\,\,\,\,x=40\sin{40}\)
\(\,\,\,\,\,\,x\approx 25.7\)
\(\,\,\,\,\,\,\text{The answer is } 25.7 \text{ feet.}\)
\(\textbf{4)}\) You are on top of a building. You look down on the neighboring building at an angle of depression of 30 degrees. Your building is 100 feet tall. The buildings are 30 feet apart. How tall is the other building?
\(\text{The neighboring building is } 82.7 \text{ feet tall}\)
\(\,\,\,\,\,\,\tan{30}=\frac{x}{30}\)
\(\,\,\,\,\,\,0.5774=\frac{x}{30}\)
\(\,\,\,\,\,\,x=30 \cdot 0.5774\)
\(\,\,\,\,\,\,x=17.32\)
\(\,\,\,\,\,\,h+x=100\)
\(\,\,\,\,\,\,h+17.32=100\)
\(\,\,\,\,\,\,h=82.68\)
\(\,\,\,\,\,\,\text{The neighboring building is } 82.7 \text{ feet tall.}\)
\(\textbf{5)}\) From the top of a lighthouse, the angle of depression to a boat is 45 degrees. If the lighthouse is 100 feet tall, how far is the boat from the base of the lighthouse?
The answer is \(100\) feet.
\(\,\,\,\,\,\,\tan{45}=\frac{100}{x}\)
\(\,\,\,\,\,\,1=\frac{100}{x}\)
\(\,\,\,\,\,\,x=100\)
\(\,\,\,\,\,\,\text{The boat is }100\text{ feet from the base of the lighthouse.}\)
\(\textbf{6)}\) You are flying a kite and hold the string 80 feet long. The angle of elevation to the kite is 50 degrees. How high is the kite above the ground?
The answer is \(61.3\) feet.
\(\,\,\,\,\,\,\sin{50}=\frac{x}{80}\)
\(\,\,\,\,\,\,x=80\sin{50}\)
\(\,\,\,\,\,\,x\approx61.3\)
\(\,\,\,\,\,\,\text{The kite is about }61.3\text{ feet above the ground.}\)
\(\textbf{7)}\) A person looks up at the top of a cliff with an angle of elevation of 20 degrees. If they are standing 200 feet away from the base of the cliff, how tall is the cliff?
The answer is \(72.8\) feet.
\(\,\,\,\,\,\,\tan{20}=\frac{x}{200}\)
\(\,\,\,\,\,\,x=200\tan{20}\)
\(\,\,\,\,\,\,x\approx72.8\)
\(\,\,\,\,\,\,\text{The cliff is about }72.8\text{ feet tall.}\)
\(\textbf{8)}\) From the top of a building, the angle of depression to a car on the street is 40 degrees. If the building is 80 feet tall, how far is the car from the base of the building?
The answer is \(95.3\) feet.
\(\,\,\,\,\,\,\tan{40}=\frac{80}{x}\)
\(\,\,\,\,\,\,x=\frac{80}{\tan{40}}\)
\(\,\,\,\,\,\,x\approx95.3\)
\(\,\,\,\,\,\,\text{The car is about }95.3\text{ feet from the base of the building.}\)
\(\textbf{9)}\) A ladder is leaning against a wall. The ladder makes a 70 degree angle with the ground and the base of the ladder is 6 feet from the wall. How long is the ladder?
The answer is \(17.5\) feet.
\(\,\,\,\,\,\,\cos{70}=\frac{6}{x}\)
\(\,\,\,\,\,\,x=\frac{6}{\cos{70}}\)
\(\,\,\,\,\,\,x\approx17.5\)
\(\,\,\,\,\,\,\text{The ladder is about }17.5\text{ feet long.}\)
\(\textbf{10)}\) A ramp rises 4 feet over a horizontal distance of 18 feet. What is the angle of elevation of the ramp?
The answer is \(12.5^\circ\).
\(\,\,\,\,\,\,\tan{\theta}=\frac{4}{18}\)
\(\,\,\,\,\,\,\theta=\tan^{-1}\left(\frac{4}{18}\right)\)
\(\,\,\,\,\,\,\theta\approx12.5^\circ\)
\(\,\,\,\,\,\,\text{The angle of elevation is about }12.5^\circ.\)
\(\textbf{11)}\) A person stands 75 feet from the base of a tower and looks up at the top with an angle of elevation of 38 degrees. How tall is the tower?
The answer is \(58.6\) feet.
\(\,\,\,\,\,\,\tan{38}=\frac{x}{75}\)
\(\,\,\,\,\,\,x=75\tan{38}\)
\(\,\,\,\,\,\,x\approx58.6\)
\(\,\,\,\,\,\,\text{The tower is about }58.6\text{ feet tall.}\)
\(\textbf{12)}\) From the top of a 150-foot building, the angle of depression to a person on the ground is 25 degrees. How far is the person from the base of the building?
The answer is \(321.7\) feet.
\(\,\,\,\,\,\,\tan{25}=\frac{150}{x}\)
\(\,\,\,\,\,\,x=\frac{150}{\tan{25}}\)
\(\,\,\,\,\,\,x\approx321.7\)
\(\,\,\,\,\,\,\text{The person is about }321.7\text{ feet from the base of the building.}\)
\(\textbf{13)}\) A drone is flying 90 feet above the ground. The angle of depression from the drone to a landing pad is 35 degrees. What is the direct distance from the drone to the landing pad?
The answer is \(156.9\) feet.
\(\,\,\,\,\,\,\sin{35}=\frac{90}{x}\)
\(\,\,\,\,\,\,x=\frac{90}{\sin{35}}\)
\(\,\,\,\,\,\,x\approx156.9\)
\(\,\,\,\,\,\,\text{The direct distance is about }156.9\text{ feet.}\)
\(\textbf{14)}\) A flagpole casts a shadow 28 feet long. The angle of elevation from the end of the shadow to the top of the flagpole is 52 degrees. How tall is the flagpole?
The answer is \(35.8\) feet.
\(\,\,\,\,\,\,\tan{52}=\frac{x}{28}\)
\(\,\,\,\,\,\,x=28\tan{52}\)
\(\,\,\,\,\,\,x\approx35.8\)
\(\,\,\,\,\,\,\text{The flagpole is about }35.8\text{ feet tall.}\)
\(\textbf{15)}\) A zipline is attached to a platform 45 feet above the ground. The cable makes a 12 degree angle of depression to the ground. How long is the zipline cable?
The answer is \(216.4\) feet.
\(\,\,\,\,\,\,\sin{12}=\frac{45}{x}\)
\(\,\,\,\,\,\,x=\frac{45}{\sin{12}}\)
\(\,\,\,\,\,\,x\approx216.4\)
\(\,\,\,\,\,\,\text{The zipline cable is about }216.4\text{ feet long.}\)
Challenge Problems
\(\textbf{16)}\) Two people are standing on opposite sides of a flagpole. One person is 30 feet from the base and sees the top at an angle of elevation of 55 degrees. The other person is 50 feet from the base. What angle of elevation does the second person see?
The answer is \(40.6^\circ\).
\(\,\,\,\,\,\,\tan{55}=\frac{h}{30}\)
\(\,\,\,\,\,\,h=30\tan{55}\)
\(\,\,\,\,\,\,h\approx42.84\)
\(\,\,\,\,\,\,\tan{\theta}=\frac{42.84}{50}\)
\(\,\,\,\,\,\,\theta=\tan^{-1}\left(\frac{42.84}{50}\right)\)
\(\,\,\,\,\,\,\theta\approx40.6^\circ\)
\(\textbf{17)}\) Two people are standing on the same side of a building. The closer person is 40 feet from the building and sees the top at an angle of elevation of 65 degrees. The farther person is 25 feet behind the closer person. What angle of elevation does the farther person see?
The answer is \(52.8^\circ\).
\(\,\,\,\,\,\,\tan{65}=\frac{h}{40}\)
\(\,\,\,\,\,\,h=40\tan{65}\)
\(\,\,\,\,\,\,h\approx85.78\)
\(\,\,\,\,\,\,\text{The farther person is }40+25=65\text{ feet from the building.}\)
\(\,\,\,\,\,\,\tan{\theta}=\frac{85.78}{65}\)
\(\,\,\,\,\,\,\theta=\tan^{-1}\left(\frac{85.78}{65}\right)\)
\(\,\,\,\,\,\,\theta\approx52.8^\circ\)
\(\textbf{18)}\) From the top of a cliff, the angle of depression to a boat is 18 degrees. After the boat moves 300 feet closer to the cliff, the angle of depression is 33 degrees. How tall is the cliff?
The answer is \(206.8\) feet.
\(\,\,\,\,\,\,\tan{18}=\frac{h}{x+300}\)
\(\,\,\,\,\,\,\tan{33}=\frac{h}{x}\)
\(\,\,\,\,\,\,h=x\tan{33}\)
\(\,\,\,\,\,\,x\tan{33}=(x+300)\tan{18}\)
\(\,\,\,\,\,\,x(\tan{33}-\tan{18})=300\tan{18}\)
\(\,\,\,\,\,\,x\approx318.5\)
\(\,\,\,\,\,\,h=318.5\tan{33}\approx206.8\)
\(\,\,\,\,\,\,\text{The cliff is about }206.8\text{ feet tall.}\)
\(\textbf{19)}\) A helicopter is directly above a road. Two cars are on the same side of the point below the helicopter. The angles of depression to the cars are 60 degrees and 35 degrees. The cars are 400 feet apart. How high is the helicopter?
The answer is \(565.2\) feet.
\(\,\,\,\,\,\,\tan{60}=\frac{h}{x}\)
\(\,\,\,\,\,\,\tan{35}=\frac{h}{x+400}\)
\(\,\,\,\,\,\,h=x\tan{60}\)
\(\,\,\,\,\,\,x\tan{60}=(x+400)\tan{35}\)
\(\,\,\,\,\,\,x(\tan{60}-\tan{35})=400\tan{35}\)
\(\,\,\,\,\,\,x\approx326.3\)
\(\,\,\,\,\,\,h=326.3\tan{60}\approx565.2\)
\(\,\,\,\,\,\,\text{The helicopter is about }565.2\text{ feet high.}\)
\(\textbf{20)}\) A person is standing 60 feet from a tower and sees the top at an angle of elevation of 48 degrees. Their eye level is 5.5 feet above the ground. How tall is the tower?
The answer is \(72.1\) feet.
\(\,\,\,\,\,\,\tan{48}=\frac{x}{60}\)
\(\,\,\,\,\,\,x=60\tan{48}\)
\(\,\,\,\,\,\,x\approx66.6\)
\(\,\,\,\,\,\,\text{Add the person’s eye level.}\)
\(\,\,\,\,\,\,66.6+5.5=72.1\)
\(\,\,\,\,\,\,\text{The tower is about }72.1\text{ feet tall.}\)
\(\textbf{21)}\) Solve for \(h\).
The answer is \(44.4\) feet.
\(\,\,\,\,\,\,\text{Let }x\text{ be the distance from the taller person to the lamp post.}\)
\(\,\,\,\,\,\,\text{Then the shorter person is }x+30\text{ feet from the lamp post.}\)
\(\,\,\,\,\,\,\tan{62^\circ}=\frac{h-6}{x}\)
\(\,\,\,\,\,\,h=6+x\tan{62^\circ}\)
\(\,\,\,\,\,\,\tan{38^\circ}=\frac{h-5}{x+30}\)
\(\,\,\,\,\,\,h=5+(x+30)\tan{38^\circ}\)
\(\,\,\,\,\,\,6+x\tan{62^\circ}=5+(x+30)\tan{38^\circ}\)
\(\,\,\,\,\,\,x(\tan{62^\circ}-\tan{38^\circ})=30\tan{38^\circ}-1\)
\(\,\,\,\,\,\,x=\frac{30\tan{38^\circ}-1}{\tan{62^\circ}-\tan{38^\circ}}\)
\(\,\,\,\,\,\,x\approx20.4\)
\(\,\,\,\,\,\,h=6+20.4\tan{62^\circ}\)
\(\,\,\,\,\,\,h\approx44.4\)
\(\,\,\,\,\,\,\text{The lamp post is about }44.4\text{ feet tall.}\)
\(\textbf{22)}\) Solve for \(h\).
The answer is \(21.0\) feet.
\(\,\,\,\,\,\,\text{Let }x\text{ be the distance from the left person to the flagpole.}\)
\(\,\,\,\,\,\,\text{Then the right person is }40-x\text{ feet from the flagpole.}\)
\(\,\,\,\,\,\,\tan{32^\circ}=\frac{h-5}{x}\)
\(\,\,\,\,\,\,h=5+x\tan{32^\circ}\)
\(\,\,\,\,\,\,\tan{48^\circ}=\frac{h-5}{40-x}\)
\(\,\,\,\,\,\,h=5+(40-x)\tan{48^\circ}\)
\(\,\,\,\,\,\,5+x\tan{32^\circ}=5+(40-x)\tan{48^\circ}\)
\(\,\,\,\,\,\,x\tan{32^\circ}=(40-x)\tan{48^\circ}\)
\(\,\,\,\,\,\,x(\tan{32^\circ}+\tan{48^\circ})=40\tan{48^\circ}\)
\(\,\,\,\,\,\,x=\frac{40\tan{48^\circ}}{\tan{32^\circ}+\tan{48^\circ}}\)
\(\,\,\,\,\,\,x\approx25.6\)
\(\,\,\,\,\,\,h=5+25.6\tan{32^\circ}\)
\(\,\,\,\,\,\,h\approx21.0\)
\(\,\,\,\,\,\,\text{The flagpole is about }21.0\text{ feet tall.}\)
\(\textbf{23)}\) Solve for \(h\).
The answer is \(28.6\) feet.
\(\,\,\,\,\,\,\text{Let }x\text{ be the distance from the shorter person to the building.}\)
\(\,\,\,\,\,\,\text{Then the taller person is }x+18\text{ feet from the building.}\)
\(\,\,\,\,\,\,\tan{43^\circ}=\frac{h-4}{x}\)
\(\,\,\,\,\,\,h=4+x\tan{43^\circ}\)
\(\,\,\,\,\,\,\tan{27^\circ}=\frac{h-6}{x+18}\)
\(\,\,\,\,\,\,h=6+(x+18)\tan{27^\circ}\)
\(\,\,\,\,\,\,4+x\tan{43^\circ}=6+(x+18)\tan{27^\circ}\)
\(\,\,\,\,\,\,x(\tan{43^\circ}-\tan{27^\circ})=2+18\tan{27^\circ}\)
\(\,\,\,\,\,\,x=\frac{2+18\tan{27^\circ}}{\tan{43^\circ}-\tan{27^\circ}}\)
\(\,\,\,\,\,\,x\approx26.4\)
\(\,\,\,\,\,\,h=4+26.4\tan{43^\circ}\)
\(\,\,\,\,\,\,h\approx28.6\)
\(\,\,\,\,\,\,\text{The building is about }28.6\text{ feet tall.}\)
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