Tenth Grade Mathematics: Single-Triangle Right-Angled Problems
The simplest and most direct application of trigonometry to physical measurement occurs when the physical configuration can be represented by a single right-angled triangle. In these problems, an observer measures one distance and one angle of elevation or depression, enabling the direct calculation of the remaining sides through a single trigonometric ratio. Whether determining the length of a shadow, calculating the altitude of a kite, or assessing structural stability, single-triangle models provide the fundamental blueprint for practical field mathematics.
The Mathematical Framework of Single-Triangle Problems
Every single-triangle problem involves three primary geometric components:
The Three Quantities in Right Triangle ABC:
1. One Known Length: (Opposite leg p, Adjacent leg b, or Hypotenuse h)
2. One Acute Angle: theta (Angle of elevation or depression)
3. One Unknown Length: (The target distance or height to be solved)
Selection of the Appropriate Trigonometric Ratio:
- If Opposite (p) and Adjacent (b) are involved: Use TANGENT
tan(theta) = p / b
- If Opposite (p) and Hypotenuse (h) are involved: Use SINE
sin(theta) = p / h
- If Adjacent (b) and Hypotenuse (h) are involved: Use COSINE
cos(theta) = b / h
Decision Flowchart:
Are you finding or given the Hypotenuse?
|
+----------+----------+
| |
YES NO
| |
Use SIN or COS Use TAN
(Opposite ==> sin) (Opposite / Adjacent)
(Adjacent ==> cos)
Category 1: The Broken Tree Problem
A celebrated classic problem involves a tree or pole broken by a windstorm such that its top touches the ground, forming a right-angled triangle with the ground:
The Broken Tree Geometry:
Original Tree of Total Height H = x + y:
+ Top (Originally at height H)
|
| Length y (Breaks and bends down!)
|
+ Break Point B (at height x)
| Height x | \ Hypotenuse y (Broken upper part)
| +-----+ Top touches ground at distance d
Base theta
Notice the crucial physical constraint: the total original height of the tree is the sum of the remaining upright trunk x and the fallen hypotenuse y:
H_total = x + y
In the right triangle: x = d * tan(theta) y = d * sec(theta) = d / cos(theta) Therefore: H_total = d * [tan(theta) + sec(theta)].
Detailed Worked Example: Broken Tree
A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30 degrees with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the total height of the tree.
Step 1: Identify components:
Angle with ground theta = 30°
Horizontal baseline d = 8 m
Let upright part = x
Let broken hypotenuse = y
Step 2: Find the upright trunk x using tangent:
tan(30°) = x / 8
1 / sqrt(3) = x / 8
x = 8 / sqrt(3) m
Step 3: Find the broken hypotenuse y using cosine:
cos(30°) = 8 / y
sqrt(3) / 2 = 8 / y
y * sqrt(3) = 16
y = 16 / sqrt(3) m
Step 4: Calculate total original height H:
H = x + y = 8 / sqrt(3) + 16 / sqrt(3)
= 24 / sqrt(3)
Step 5: Rationalize the denominator:
H = (24 * sqrt(3)) / (sqrt(3) * sqrt(3)) = 24*sqrt(3) / 3 = 8*sqrt(3) m.
Since sqrt(3) is approximately 1.732:
H = 8 * 1.732 = 13.856 m.
The total original height of the tree was 8*sqrt(3) meters (approximately 13.86 m).
Category 2: Kite Altitude and Tether Tension
When a kite is flown on a taut string of length L at an inclination angle theta: The string forms the hypotenuse: h = L. The vertical height above the ground is the opposite side: y = L * sin(theta). The horizontal ground span is the adjacent side: x = L * cos(theta).
Chapter Exercises
Exercise 1: A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60 degrees. Find the height of the tower.
Exercise 2: A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60 degrees. Find the length of the string, assuming that there is no slack in the string.
Exercise 3: The shadow of a vertical pole of height h on level ground is found to be h * sqrt(3) meters long. Find the angle of elevation of the sun at that moment.
Solutions and Explanations
Solution 1: Let AB be the vertical tower of height h standing on ground point B. Let C be the point on the ground such that BC = 15 m. The angle of elevation is Angle ACB = 60 degrees. In right-angled triangle ABC, the leg AB is opposite to Angle C, and leg BC is adjacent. Using the tangent ratio: tan(60°) = AB / BC ==> sqrt(3) = h / 15 ==> h = 15 * sqrt(3) m. Using the approximation sqrt(3) = 1.732, h = 15 * 1.732 = 25.98 m. Thus, the height of the tower is 15*sqrt(3) meters (approximately 25.98 meters).
Solution 2: Let K be the position of the kite, and let G be the point on the ground directly below the kite, so KG = 60 m. Let P be the point on the ground where the string is tied. The string KP forms the hypotenuse of right-angled triangle KGP, with Angle KPG = 60 degrees. The opposite side is KG = 60 m, and the string length is KP = L. Using the sine ratio: sin(60°) = Opposite / Hypotenuse = KG / KP ==> sqrt(3) / 2 = 60 / L. Cross-multiplying: L * sqrt(3) = 120 ==> L = 120 / sqrt(3). Rationalizing the denominator: L = (120 * sqrt(3)) / 3 = 40 * sqrt(3) m. Using sqrt(3) = 1.732, L = 40 * 1.732 = 69.28 m. Thus, the length of the string is 40*sqrt(3) meters (approximately 69.28 meters).
Solution 3: Let AB be the vertical pole of height h, and let BC be its horizontal shadow of length h*sqrt(3). Let theta be the angle of elevation of the sun. The shadow meets the pole at a right angle: Angle B = 90 degrees. In right triangle ABC: tan(theta) = Opposite / Adjacent = AB / BC = h / (h * sqrt(3)) = 1 / sqrt(3). We know from standard trigonometric values that tan(30°) = 1/sqrt(3). Therefore: theta = 30 degrees. The angle of elevation of the sun is 30 degrees.