Educator

The sun is 40^° above the horizon. Find the length of a shadow cast by a flagpole that is 35 feet tall.

• Start by drawing a picture using the information you are given
• 
• Use SOHCAHTOA to find the length of the shadow
• tan40^° = [35/x] ⇒ x = [35/(tan40^°)]

x ≈ 41.7 feet

A ladder 20 feet long leans against the side of a building. Find the height from the top of the ladder to the ground if the angle of elevation of the ladder is 53^°.

• Start by drawing a picture using the information you are given
• 
• Use SOHCAHTOA to find the height of the ladder to the ground
• sin53^° = [h/20] ⇒ h = 20sin53^°

h ≈ 15.97 feet

The length of a shadow of a house is 185 feet when the angle of elevation of the sun is 48^°. What is the approximate height of the house?

• Start by drawing a picture using the information you are given
• 
• Use SOHCAHTOA to find the height of the house
• tan48^° = [h/185] ⇒ h = 185tan48^°

h ≈ 205.46 feet

A young girl is flying a kite on 300 feet of string. The kite string makes a 35^° angle with the ground. How high is the kite?

• Start by drawing a picture using the information you are given
• 
• Use SOHCAHTOA to find the height of the kite
• sin35^° = [h/300] ⇒ h = 300sin35^°

h ≈ 172.1 feet

An engineer erects a 85 - foot vertical tower for an antenna. Find the angle of elevation to the top of the tower at a point on level ground 60 feet from its base.

• Start by drawing a picture using the information you are given
• 
• Use SOHCAHTOA to find the angle of elevation
• tanθ = [85/60] ⇒ θ = arctan([85/60])

θ ≈ 54.8^°

The sonnar of a navy cruiser detects a submarine that is 3500 feet from the cruiser. The submarine is 2237 feet deep. Find the angle between the water line and the submarine.

• Start by drawing a picture using the information you are given
• 
• Use SOHCAHTOA to find the length of the shadow
• sinθ = [2237/3500] ⇒ θ = arcsin([2237/3500])

θ ≈ 39.7^°

A bridge is built across a lake from A to B. The bearing from B to A is S33^°W. From point C, 150 meters from B, the bearings to B and A are S67^°E and S18^°E respectively. Find the distance from A to B.

• Start by drawing a picture using the information you are given
• 
• C = 67^o − 18^o = 49^o
• B = 180^o − 67^o − 33^o = 80^o
• A = 180^o − 49^o − 80^o = 51^o
• 
• Use Law of Sines
• [(sin51^°)/150] = [(sin49^°)/c] ⇒ c  sin51^° = 150sin49^° ⇒ c = [(150sin49^°)/(sin51^°)]

c ≈ 146 meters

Due to very strong winds, a Magnolia Tree grew so that it is leaning 8^° from the vertical. At a point 45 meters from the tree, the angle of elevation is 32^°. Find the height of the tree.

• Start by drawing a picture using the information you are given
• 
• θ = 180^° − 32^° − 96^° = 52^°
• Use Law of Sines
• [(sin32^°)/h] = [(sin52^°)/45] ⇒ h sin52^° = 45sin32^° ⇒ c = [(45sin32^°)/(sin52^°)]

h ≈ 30.3 meters

A triangular piece of land has sides of length 730 feet, 620 feet, and 560 feet. Find the measure of the largest side.

• Start by drawing a picture using the information you are given
• 
• Use the Law of Cosines
• cosC = [(a² + b² − c²)/2ab] ⇒ cosC = [(620² + 560² − 730²)/2(620)(560)] ⇒ C = arccos([165100/694400])

C ≈ 76.2^°

A land surveyor measures the fences along the edges of a triangular field to be 480 feet, 540 feet, and 570 feet. What is the area of the field?

• Start by drawing a picture using the information you are given
• 
• Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} ⇒ where s = [1/2](a + b + c)
• s = [1/2](480 + 540 + 670) ⇒ s = 845
• A = √{845(845 − 480)(845 − 540)(845 − 670)} ⇒ A = √{845(365)(305)(175)}

A ≈ 128305.04 square feet

1. Intro
2. Formulas to Remember
• SOHCAHTOA
• Law of Sines
• Law of Cosines
• Heron's Formula
3. Example 1: Telephone Pole Height
4. Example 2: Bridge Length
5. Example 3: Area of Triangular Field
6. Extra Example 1: Kite Height
7. Extra Example 2: Roads to a Town