Exercise 9.1 answers: Some Applications of Trigonometry

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1. The rope, the pole and the ground make a right triangle: the pole is vertical, the ground is horizontal.
2. The rope is the hypotenuse (20 m), and the pole's height is the side opposite the 30° angle.
3. Opposite and hypotenuse is the sine ratio: sin 30° = height ÷ 20.
4. sin 30° = 1/2, so height = 20 × 1/2 = 10 m.

AnswerThe pole is 10 m tall.

Watch this explained “Sighted at ground level”, 7:40 into The sightline, and the horizontal it gets measured against

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1. The standing stump is vertical, the ground is horizontal, and the broken (bent) part is the hypotenuse of a right triangle.
2. The 8 m distance is the side adjacent to the 30° angle at the point where the top touches the ground.
3. Stump height = adjacent × tan 30° = 8 × (1/√3) = 8/√3 m.
4. Broken part (hypotenuse) = adjacent ÷ cos 30° = 8 ÷ (√3/2) = 16/√3 m.
5. The tree's original height is the stump plus the broken part: 8/√3 + 16/√3 = 24/√3 = 8√3 m.

AnswerThe tree was 8√3 m (about 13.86 m) tall.

Watch this explained “When the ratio is only part of the answer”, 9:29 into Picking the one ratio that links what you know to what you want

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younger children

1. Height 1.5 m is opposite the 30° angle; the slide is the hypotenuse.
2. sin 30° = 1.5 ÷ length, so length = 1.5 × 2 = 3 m.

Answer3 m

elder children

1. Height 3 m is opposite the 60° angle; the slide is the hypotenuse.
2. sin 60° = 3 ÷ length, so length = 3 ÷ (√3/2) = 2√3 m.

Answer2√3 m (about 3.46 m)

Watch this explained “The ladder, where the tangent is the wrong reach”, 7:55 into Picking the one ratio that links what you know to what you want

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1. The 30 m ground distance is adjacent to the 30° angle; the tower's height is the opposite side.
2. Adjacent and opposite is the tangent ratio: tan 30° = height ÷ 30.
3. tan 30° = 1/√3, so height = 30 × 1/√3 = 30/√3 = 10√3 m.

AnswerThe tower is 10√3 m (about 17.32 m) tall.

Watch this explained “The tower: adjacent known, opposite wanted”, 5:57 into Picking the one ratio that links what you know to what you want

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1. The kite's height, 60 m, is opposite the 60° angle at the ground; the string is the hypotenuse.
2. Opposite and hypotenuse is the sine ratio: sin 60° = 60 ÷ length.
3. sin 60° = √3/2, so length = 60 ÷ (√3/2) = 120/√3 = 40√3 m.

AnswerThe string is 40√3 m (about 69.28 m) long.

Watch this explained “A ten-second checklist, run on a new figure”, 12:21 into Picking the one ratio that links what you know to what you want

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1. Since the boy's eyes are 1.5 m up, the triangle only sees the part of the building above his eyes: 30 − 1.5 = 28.5 m.
2. At the far point, tan 30° = 28.5 ÷ (far distance), so far distance = 28.5√3 m.
3. At the near point, tan 60° = 28.5 ÷ (near distance), so near distance = 28.5/√3 m.
4. Distance walked = far distance − near distance = 28.5√3 − 28.5/√3 = 28.5 × (2/√3) = 57/√3 = 19√3 m.

AnswerThe boy walked 19√3 m (about 32.91 m).

Watch this explained “The two storeys”, 5:38 into The sightline, and the horizontal it gets measured against

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1. The bottom of the tower is the top of the building: tan 45° = 20 ÷ distance, so distance = 20 m.
2. The top of the tower is higher still: tan 60° = (20 + tower height) ÷ 20.
3. tan 60° = √3, so 20 + tower height = 20√3, giving tower height = 20√3 − 20 m.

AnswerThe transmission tower is (20√3 − 20) m, about 14.64 m, tall.

Watch this explained “Two targets from one station”, 5:01 into Two angles in one figure, and the pair of equations they hand you

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1. Let the pedestal height be h and the ground distance be d.
2. Top of pedestal: tan 45° = h ÷ d, so d = h (since tan 45° = 1).
3. Top of statue: tan 60° = (h + 1.6) ÷ d = (h + 1.6) ÷ h.
4. √3 = (h + 1.6)/h, so h(√3 − 1) = 1.6, giving h = 1.6 ÷ (√3 − 1).
5. Multiplying top and bottom by (√3 + 1) removes the surd from the bottom: h = 0.8(√3 + 1) = (4√3 + 4)/5 m.

AnswerThe pedestal is (4√3 + 4)/5 m, about 2.19 m, tall.

Watch this explained “Two targets from one station”, 5:01 into Two angles in one figure, and the pair of equations they hand you

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1. The two structures stand on the same ground, so the horizontal distance between their feet is one shared unknown, d.
2. From the building's foot, tan 60° = 50 ÷ d, so d = 50/√3.
3. From the tower's foot, tan 30° = (building height) ÷ d.
4. Building height = d × tan 30° = (50/√3) × (1/√3) = 50/3 m.

AnswerThe building is 50/3 m, about 16.67 m, tall.

Watch this explained “Two targets from one station”, 5:01 into Two angles in one figure, and the pair of equations they hand you

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1. Let the point be x m from the pole seen at 60°, so it is (80 − x) m from the pole seen at 30°.
2. tan 60° = h ÷ x, so h = x√3.
3. tan 30° = h ÷ (80 − x), so h = (80 − x)/√3.
4. Setting these equal: x√3 = (80 − x)/√3, so 3x = 80 − x, giving 4x = 80, x = 20.
5. So the point is 20 m from one pole and 60 m from the other, and h = 20√3 m.

AnswerEach pole is 20√3 m (about 34.64 m) tall; the point is 20 m from one pole and 60 m from the other.

Watch this explained “Two equations, one substitution”, 2:57 into Two angles in one figure, and the pair of equations they hand you

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1. Let the canal width be w and the tower height be h.
2. From directly opposite: tan 60° = h ÷ w, so h = w√3.
3. From 20 m further back: tan 30° = h ÷ (w + 20).
4. Substituting h = w√3: w√3 = (w + 20)/√3, so 3w = w + 20, giving w = 10.
5. So h = 10√3 m.

AnswerThe canal is 10 m wide, and the tower is 10√3 m (about 17.32 m) tall.

Watch this explained “What a second look actually buys”, 0:41 into Two angles in one figure, and the pair of equations they hand you

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1. The depression of 45° to the tower's foot equals the elevation of 45° from the foot back up, so the horizontal distance d satisfies tan 45° = 7 ÷ d, giving d = 7 m (the building's own height).
2. The rise of the tower above the building's roof level: tan 60° = rise ÷ d, so rise = 7 × √3 = 7√3 m.
3. The tower's total height is this rise plus the 7 m the roof already sits above the tower's foot: 7√3 + 7 m.

AnswerThe cable tower is (7√3 + 7) m, about 19.12 m, tall.

Watch this explained “Three more of the same skeleton”, 10:39 into Looking up versus looking down: elevation and depression

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1. Swap each depression angle for the equal elevation angle from sea level back up to the lighthouse top.
2. Nearer ship (45°): distance = 75 ÷ tan 45° = 75 m.
3. Farther ship (30°): distance = 75 ÷ tan 30° = 75√3 m.
4. The gap between the ships is the difference: 75√3 − 75 m.

AnswerThe ships are (75√3 − 75) m, about 54.90 m, apart.

Watch this explained “Three more of the same skeleton”, 10:39 into Looking up versus looking down: elevation and depression

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1. The girl's eyes are 1.2 m up, so the triangle only sees the part of the balloon's height above her eyes: 88.2 − 1.2 = 87 m.
2. At 60°: tan 60° = 87 ÷ (near distance), so near distance = 87/√3 m.
3. At 30°: tan 30° = 87 ÷ (far distance), so far distance = 87√3 m.
4. Distance travelled = far distance − near distance = 87√3 − 87/√3 = 87 × (2/√3) = 174/√3 = 58√3 m.

AnswerThe balloon travelled 58√3 m (about 100.46 m).

Watch this explained “The two storeys”, 5:38 into The sightline, and the horizontal it gets measured against

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1. Let the tower's height be h. At 30°, the car's distance is h/tan 30° = h√3. At 60°, its distance is h/tan 60° = h/√3.
2. The car covered the gap between these in 6 seconds: h√3 − h/√3 = 2h/√3, so its speed is (2h/√3) ÷ 6 = h/(3√3).
3. The remaining distance to the foot of the tower is h/√3.
4. Time = remaining distance ÷ speed = (h/√3) ÷ (h/(3√3)) = 3 seconds. The height h cancels out of the answer entirely.

AnswerThe car takes 3 more seconds to reach the foot of the tower.

Watch this explained “The height walks out”, 13:27 into Two angles in one figure, and the pair of equations they hand you