Tenth Grade Mathematics: Two-Triangle Systems: Coplanar Observers
In real-world surveying, inaccessible obstacles prevent surveyors from walking directly to the base of a monument, cliff, or tower to measure a baseline. In such situations, observers employ two-triangle systems, measuring angles of elevation or depression from two distinct coplanar observation stations separated by a known baseline distance. By setting up simultaneous trigonometric equations across two overlapping right-angled triangles, the common unknown height or distance can be eliminated, allowing the inaccessible dimensions to be solved with complete precision.
The Classical Configurations of Two-Triangle Systems
Two-triangle heights and distances problems fall into three major structural geometric configurations:
Configuration A: Both Observers on the SAME SIDE of the Structure
P (Top of Tower)
| | Height h | | +---------+----+----+
B (Base) D C A
<-- x ---><-- d --->
Angle beta Angle alpha (alpha < beta)
Configuration B: Observers on OPPOSITE SIDES of the Structure
P (Top of Tower)
/| / | / | \ Height h
/ | +----+----+----+----+
A B C
<-- d1 --><-- d2 -->
Angle alpha Angle beta
Configuration C: Two Heights Observed from a SINGLE Station
P (Top of Flagstaff)
|
Q (Top of Building / Pedestal)
|
| Building Height h
|
+-------------------+ Station S
B (Base)
Mathematical Derivation for Same-Side Observers (Configuration A)
Let a vertical tower PB of height h stand on a horizontal plane. Two observers at points C and D, in a direct line with the base B, observe the top P with angles of elevation alpha and beta respectively (where beta > alpha, since D is closer to the base than C). Let the distance between the observers be CD = d, and let BD = x:
In Right Triangle PBD (Closer Station):
tan(beta) = h / x ===> x = h / tan(beta) = h * cot(beta) --- (Equation 1)
In Right Triangle PBC (Farther Station):
tan(alpha) = h / (x + d) ===> x + d = h / tan(alpha) = h * cot(alpha) --- (Equation 2)
Subtract Equation 1 from Equation 2 to eliminate the inaccessible distance x:
(x + d) - x = h * cot(alpha) - h * cot(beta)
d = h * [cot(alpha) - cot(beta)]
Isolating the unknown height h:
Theorem (Same-Side Height Formula):
d
h = ---------------------
cot(alpha) - cot(beta)
Notice how the inaccessible horizontal ground distance x has vanished entirely from the formula!
Detailed Worked Example: Opposite Sides
A statue stands in the center of a courtyard. Two observers standing on opposite sides of the statue, in line with its base, measure the angles of elevation of the top of the statue as 30 degrees and 45 degrees. If the distance between the two observers is 100 m, find the height of the statue.
Formulation:
Let statue height = h meters.
Let distance from first observer to base = x meters.
Then distance from second observer to base = (100 - x) meters.
Triangle 1 (Angle = 45°):
tan(45°) = h / x
1 = h / x ===> x = h
Triangle 2 (Angle = 30°):
tan(30°) = h / (100 - x)
1 / sqrt(3) = h / (100 - x)
100 - x = h * sqrt(3)
Substitute x = h into the equation:
100 - h = h * sqrt(3)
h * sqrt(3) + h = 100
h * (sqrt(3) + 1) = 100
h = 100 / (sqrt(3) + 1)
Rationalize the denominator by multiplying by (sqrt(3) - 1):
h = [100 * (sqrt(3) - 1)] / [(sqrt(3) + 1)*(sqrt(3) - 1)]
h = [100 * (sqrt(3) - 1)] / (3 - 1)
h = [100 * (sqrt(3) - 1)] / 2 = 50 * (sqrt(3) - 1) m.
Numerical Evaluation:
sqrt(3) is approximately 1.732:
h = 50 * (1.732 - 1) = 50 * 0.732 = 36.6 m.
The height of the statue is 50*(sqrt(3) - 1) meters, or approximately 36.6 meters.
Complementary Angles of Elevation
A famous theoretical case occurs when the angles of elevation from two points at distances a and b from the base of a tower are complementary (summing to 90 degrees):
Theorem (Complementary Heights Theorem):
If the angles of elevation of the top of a tower from two points on the ground
at distances a and b from its base are complementary, then the height h of the
tower is the geometric mean of the distances:
h = sqrt(a * b)
Proof: Let the angles be theta and (90° - theta). From first station: tan(theta) = h / a. From second station: tan(90° - theta) = h / b ===> cot(theta) = h / b. Multiply the two equations: tan(theta) * cot(theta) = (h / a) * (h / b). Since tan(theta) * cot(theta) = 1: 1 = h^2 / (a * b) ===> h^2 = a * b ===> h = sqrt(a * b).
Chapter Exercises
Exercise 1: As observed from the top of a 75 m high lighthouse from sea level, the angles of depression of two ships are 30 degrees and 45 degrees. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Exercise 2: A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60 degrees. After some time, the angle of elevation reduces to 30 degrees. Find the distance travelled by the balloon during the interval.
Exercise 3: The angles of elevation of the top of a tower from two points on the ground at distances of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solutions and Explanations
Solution 1: Let AB be the lighthouse of height AB = 75 m. Let C and D be the two ships on the same side of the lighthouse, with D closer to the lighthouse than C. The angle of depression of ship D is 45°, so Angle ADB = 45°. The angle of depression of ship C is 30°, so Angle ACB = 30°. In right-angled triangle ABD: tan(45°) = AB / BD ==> 1 = 75 / BD ==> BD = 75 m. In right-angled triangle ABC: tan(30°) = AB / BC ==> 1/sqrt(3) = 75 / BC ==> BC = 75sqrt(3) m. The distance between the two ships is CD = BC - BD = 75sqrt(3) - 75 = 75(sqrt(3) - 1) m. Using sqrt(3) = 1.732: CD = 75(1.732 - 1) = 75 * 0.732 = 54.9 m. Thus, the distance between the ships is 75*(sqrt(3) - 1) meters (approximately 54.9 meters).
Solution 2: The balloon flies at a constant height of 88.2 m from the ground. The girl is 1.2 m tall, so the vertical height of the balloon above her eye level is h = 88.2 - 1.2 = 87 m. Let A be the initial position of the balloon and B be the final position. In the first position, angle of elevation is 60°: tan(60°) = h / d1 ==> sqrt(3) = 87 / d1 ==> d1 = 87 / sqrt(3) = 87sqrt(3)/3 = 29sqrt(3) m. In the second position, angle of elevation is 30°: tan(30°) = h / d2 ==> 1/sqrt(3) = 87 / d2 ==> d2 = 87sqrt(3) m. The distance traveled by the balloon is the horizontal displacement: d = d2 - d1 = 87sqrt(3) - 29sqrt(3) = 58sqrt(3) m. Using sqrt(3) = 1.732: d = 58 * 1.732 = 100.456 m. Thus, the balloon traveled 58*sqrt(3) meters (approximately 100.46 meters).
Solution 3: Let AB be the tower of height h standing on ground base B. Let C and D be the two observation points at distances BC = 4 m and BD = 9 m from the base. Let the angle of elevation at C be theta; then because the angles are complementary, the angle of elevation at D is 90° - theta. In right triangle ABC: tan(theta) = AB / BC = h / 4 (Equation 1). In right triangle ABD: tan(90° - theta) = AB / BD ==> cot(theta) = h / 9 (Equation 2). Multiply Equation 1 and Equation 2: tan(theta) * cot(theta) = (h / 4) * (h / 9). Since tan(theta) * cot(theta) = 1: 1 = h^2 / 36 ==> h^2 = 36 ==> h = sqrt(36) = 6 m (taking the positive root since height must be positive). This proves that the height of the tower is 6 meters.