Unit 6: Introduction to Trigonometry
Exercise 6.6 Solutions
Real-world applications of angles of elevation and depression — flag posts, lighthouses, ladders, kites, cliffs, rivers, and the sun.
Question 1
Angle of Elevation
The angle of elevation of the top of a flag post from a point on the ground level 40 m away from the flag post is 60°. Find the height of the post.
Solution
Given: Distance AB = 40 m, Angle of elevation θ = 60°
Find: Height BC = ?
Find: Height BC = ?
In right-angled ▵ABC:
tan 60° = mBCmAB
√3 = mBC40
mBC = 40 × √3
mBC = 40 × 1.732 = 69.28 m
tan 60° = mBCmAB
√3 = mBC40
mBC = 40 × √3
mBC = 40 × 1.732 = 69.28 m
✓ Height of the flag post = 69.28 m
Question 2
Isosceles Triangle
An isosceles triangle has a vertical angle of 120° and a base 10 cm long. Find the length of its altitude.
Solution
The altitude AC bisects both the vertical angle and the base BD.
⇒ mBC = mCD = 5 cm
Each half of vertical angle = 120°2 = 60°
In right-angled ▵ABC (mZACB = 60°, mZABC = 90°):
Wait — the altitude stands from apex A to base BD. The angle at A is 120°, so the angle ACB = 60° and angle ABC = 90° because the altitude is perpendicular to the base. Actually, the right angle is at C.
Let the altitude = AC = h.
In right-angled ▵ABC (right angle at C):
mZA = 120°2 = 60° (half of vertical angle)
tan 60° = mBCmAC = 5h
√3 = 5h
h = 5√3 = 5√33
Altitude = 5/√3 ≈ 2.89 cm
⇒ mBC = mCD = 5 cm
Each half of vertical angle = 120°2 = 60°
In right-angled ▵ABC (mZACB = 60°, mZABC = 90°):
Wait — the altitude stands from apex A to base BD. The angle at A is 120°, so the angle ACB = 60° and angle ABC = 90° because the altitude is perpendicular to the base. Actually, the right angle is at C.
Let the altitude = AC = h.
In right-angled ▵ABC (right angle at C):
mZA = 120°2 = 60° (half of vertical angle)
tan 60° = mBCmAC = 5h
√3 = 5h
h = 5√3 = 5√33
Altitude = 5/√3 ≈ 2.89 cm
✓ Length of altitude = 2.89 cm
Question 3
Angle of Elevation
A tree is 72 m high. Find the angle of elevation of its top from a point 100 m away on the ground.
Solution
Height BC = 72 m, Distance AB = 100 m.
tan θ = mBCmAB = 72100 = 0.72
θ = tan−1(0.72) = 35.75°
tan θ = mBCmAB = 72100 = 0.72
θ = tan−1(0.72) = 35.75°
✓ Angle of elevation = 35.75°
Question 4
Angle of Elevation
A ladder makes an angle of 60° with the ground and reaches a height of 10 m along the wall. Find the length of the ladder.
Solution
Height of wall BC = 10 m, Angle with ground = 60°, Ladder length AC = ?
sin 60° = mBCmAC
√32 = 10mAC
mAC = 10 × 2√3 = 20√3 = 20√33
Length of ladder = 20/√3 ≈ 11.55 m
sin 60° = mBCmAC
√32 = 10mAC
mAC = 10 × 2√3 = 20√3 = 20√33
Length of ladder = 20/√3 ≈ 11.55 m
✓ Length of ladder = 11.55 m
Question 5
Angle of Depression
A lighthouse tower is 150 m high from the sea level. The angle of depression from the top of the tower to a ship is 60°. Find the distance between the ship and the tower.
Solution
Height BC = 150 m, Angle of depression = 60°, Distance AB = ?
By alternate interior angles (horizontal line is parallel to sea level):
Angle of elevation from ship to top of tower = 60°
tan 60° = mBCmAB
√3 = 150mAB
mAB = 150√3 = 150√33 = 50√3
Distance = 50√3 ≈ 86.60 m
By alternate interior angles (horizontal line is parallel to sea level):
Angle of elevation from ship to top of tower = 60°
tan 60° = mBCmAB
√3 = 150mAB
mAB = 150√3 = 150√33 = 50√3
Distance = 50√3 ≈ 86.60 m
✓ Distance between ship and tower = 86.60 m
Question 6
Two Angles
The measure of angle of elevation of the top of a pole is 15° from a point on the ground. Walking 100 m towards the pole, the angle of elevation becomes 30°. Find the height of the pole.
Solution
Let h = height of pole CD, mBC = x, mAB = 100 m.
mAC = x + 100 m
From triangle BCD (angle at B = 30°):
tan 30° = hx ⇒ 1√3 = hx ⇒ x = √3h …(i)
From triangle ACD (angle at A = 15°):
tan 15° = hx + 100
tan 15° ≈ 0.2679
0.2679 = hx + 100 …(ii)
Substituting x = √3h into (ii):
0.2679 = h√3h + 100
0.2679(√3h + 100) = h
0.2679 × 1.732 × h + 26.79 = h
0.4641h + 26.79 = h
26.79 = h − 0.4641h = 0.5359h
h = 26.790.5359 ≈ 50 m
mAC = x + 100 m
From triangle BCD (angle at B = 30°):
tan 30° = hx ⇒ 1√3 = hx ⇒ x = √3h …(i)
From triangle ACD (angle at A = 15°):
tan 15° = hx + 100
tan 15° ≈ 0.2679
0.2679 = hx + 100 …(ii)
Substituting x = √3h into (ii):
0.2679 = h√3h + 100
0.2679(√3h + 100) = h
0.2679 × 1.732 × h + 26.79 = h
0.4641h + 26.79 = h
26.79 = h − 0.4641h = 0.5359h
h = 26.790.5359 ≈ 50 m
✓ Height of the pole = 50 m
Question 7
Shadow Problem
Find the measure of the angle of elevation of the Sun, if a tower 300 m high casts a shadow 450 m long.
Solution
Height of tower BC = 300 m, Shadow length AB = 450 m.
tan θ = mBCmAB = 300450 = 23 ≈ 0.6667
θ = tan−1(23) = 33.69°
tan θ = mBCmAB = 300450 = 23 ≈ 0.6667
θ = tan−1(23) = 33.69°
✓ Angle of elevation of the Sun = 33.69°
Question 8
Two Angles
The angle of elevation of the top of a cliff is 25°. On walking 100 m towards the cliff, the angle of elevation becomes 45°. Find the height of the cliff.
Solution
Let h = height of cliff CD, mBC = x, mAB = 100 m.
mAC = x + 100 m
From triangle BCD (angle at B = 45°):
tan 45° = hx ⇒ 1 = hx ⇒ x = h …(i)
From triangle ACD (angle at A = 25°):
tan 25° = hx + 100
tan 25° ≈ 0.4663
0.4663 = hh + 100 [substituting x = h from (i)]
0.4663(h + 100) = h
0.4663h + 46.63 = h
46.63 = h − 0.4663h = 0.5337h
h = 46.630.5337 ≈ 87.37 m
mAC = x + 100 m
From triangle BCD (angle at B = 45°):
tan 45° = hx ⇒ 1 = hx ⇒ x = h …(i)
From triangle ACD (angle at A = 25°):
tan 25° = hx + 100
tan 25° ≈ 0.4663
0.4663 = hh + 100 [substituting x = h from (i)]
0.4663(h + 100) = h
0.4663h + 46.63 = h
46.63 = h − 0.4663h = 0.5337h
h = 46.630.5337 ≈ 87.37 m
Mathematical Note – Textbook Rounding (Q.8)
The textbook uses tan 25° ≈ 0.466 (3 decimal places only), which gives h = 46.6 / 0.534 ≈ 87.27 m. Using the more precise value tan 25° = 0.4663, the correct answer is h ≈ 87.37 m. Both are accepted.
✓ Height of the cliff ≈ 87.27 m (textbook) or 87.37 m (precise)
Question 9
Angle of Depression
From the top of a hill 300 m high, the angle of depression of a point on the nearer shore of a river is 70° and of a point directly across the river is 50°. Find the width of the river and the distance from the hill to the near shore.
Solution
Height CD = 300 m, Near shore angle of depression = 70°, Far shore = 50°.
Let BC = y (distance from foot of hill to near shore) and AB = x (width of river).
Near shore (right ▵BCD, angle at B = 70°):
tan 70° = 300y
y = 300tan 70° = 3002.747 = 109.19 m
Far shore (right ▵ACD, angle at A = 50°):
tan 50° = 300x + y
x + y = 300tan 50° = 3001.1918 = 251.7 m
x = 251.7 − 109.19 = 142.51 m
Let BC = y (distance from foot of hill to near shore) and AB = x (width of river).
Near shore (right ▵BCD, angle at B = 70°):
tan 70° = 300y
y = 300tan 70° = 3002.747 = 109.19 m
Far shore (right ▵ACD, angle at A = 50°):
tan 50° = 300x + y
x + y = 300tan 50° = 3001.1918 = 251.7 m
x = 251.7 − 109.19 = 142.51 m
✓ Width of river = 142.51 m, Distance from hill to near shore = 109.19 m
Question 10
Kite Problem
A kite has 120 m of string attached to it when at an elevation of 50°. How far is it above the hand holding it? (Assume the string is taut.)
Solution
Length of string AC = 120 m, Angle of elevation θ = 50°, Height BC = h = ?
sin 50° = BCAC = h120
h = 120 × sin 50°
h = 120 × 0.766
h = 91.92 m
sin 50° = BCAC = h120
h = 120 × sin 50°
h = 120 × 0.766
h = 91.92 m
✓ Height of kite above ground = 91.92 m
🎲 Elevation & Depression Physics Visualizer
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