3.P — Worked Problems: Geometry & Trigonometry

Fifteen problems on shapes, angles and waves. Every formula used is derived in the chapter that introduces it.

Problem 1 — A ladder, a wall, and a right triangle

A 6 m ladder leans against a vertical wall with its foot 1.8 m from the base. How far up the wall does it reach, and what angle does it make with the ground?

Solution

Step 1 — draw the triangle and name the parts. The ladder is the hypotenuse (c = 6), the distance from the wall is the adjacent side (b = 1.8), and the height up the wall is the opposite side (a, unknown).

Step 2 — height, by Pythagoras.

a^2 + b^2 = c^2 \quad \Rightarrow \quad a^2 = 6^2 - 1.8^2 = 36 - 3.24 = 32.76

a = \sqrt{32.76} = 5.724 \text{ m}

Step 3 — the angle. We know the adjacent side and the hypotenuse, so cosine is the ratio to use:

\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1.8}{6} = 0.3

\theta = \arccos(0.3) = 72.54°

Answer: it reaches 5.72 m up the wall, at 72.5° to the ground.

Check by a second route. Using the height we found: \sin\theta = \frac{5.724}{6} = 0.954, and \arcsin(0.954) = 72.5° ✓. Two different ratios agreeing is the strongest check available.

The practical note. Ladder safety guidance says one unit out for every four units up, which is \arctan(4) = 76°. At 72.5° this ladder is slightly too shallow, and the number that matters is exactly the one we computed.

Problem 2 — Heron, with no height in sight

A triangular plot has sides 13 m, 14 m and 15 m. Find its area, and then the height from the 14 m side.

Solution

Step 1 — the semi-perimeter.

s = \frac{13+14+15}{2} = \frac{42}{2} = 21

Step 2 — Heron's formula.

A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21 \times (21-13) \times (21-14) \times (21-15)}

= \sqrt{21 \times 8 \times 7 \times 6}

Multiply carefully: 21 \times 8 = 168, 7 \times 6 = 42, and 168 \times 42 = 7056.

A = \sqrt{7056} = 84 \text{ m}^2

That \sqrt{7056} is exactly 84, which is why this particular triangle appears in every textbook.

Step 3 — the height from the 14 m side. Use the ordinary area formula backwards:

A = \tfrac12 \times \text{base} \times h \quad \Rightarrow \quad 84 = \tfrac12 \times 14 \times h \quad \Rightarrow \quad 84 = 7h \quad \Rightarrow \quad h = 12 \text{ m}

Answer: area 84 m², height 12 m from the 14 m side.

Check with the law of cosines and \frac12 ab\sin C. The angle between the 13 and 15 sides: \cos C = \frac{13^2+15^2-14^2}{2(13)(15)} = \frac{169+225-196}{390} = \frac{198}{390} = 0.50769, so C = 59.49° and \sin C = 0.86154. Then A = \frac12(13)(15)(0.86154) = 84.0 ✓

Problem 3 — A circle sliced up

A circle of radius 10 cm has a chord subtending an angle of 120° at the centre. Find the arc length, the sector area, the chord length and the segment area.

Solution

Step 0 — convert to radians, because every one of the formulas needs them:

\theta = 120° \times \frac{\pi}{180} = \frac{2\pi}{3} = 2.0944 \text{ rad}

Step 1 — arc length.

L = r\theta = 10 \times 2.0944 = 20.94 \text{ cm}

Step 2 — sector area.

A_{\text{sector}} = \tfrac12 r^2\theta = \tfrac12 (100)(2.0944) = 104.72 \text{ cm}^2

Step 3 — chord length, by the law of cosines on the triangle made of two radii and the chord:

c^2 = 10^2 + 10^2 - 2(10)(10)\cos120°

Since \cos120° = -0.5, the subtraction of a negative becomes an addition:

c^2 = 100 + 100 + 100 = 300 \quad \Rightarrow \quad c = \sqrt{300} = 17.32 \text{ cm}

Step 4 — segment area is the sector minus the triangle:

A_{\text{triangle}} = \tfrac12 r^2\sin\theta = \tfrac12(100)(\sin120°) = 50 \times 0.86603 = 43.30 \text{ cm}^2

A_{\text{segment}} = 104.72 - 43.30 = 61.42 \text{ cm}^2

Or in one step with the combined formula: \frac12 r^2(\theta - \sin\theta) = 50(2.0944 - 0.86603) = 61.42 ✓

Answers: arc 20.94 cm, sector 104.72 cm², chord 17.32 cm, segment 61.42 cm².

Sanity check. 120° is one third of the circle, so the sector should be one third of \pi(10)^2 = 314.16, which is 104.72 ✓

Problem 4 — Proving an identity

Prove that \dfrac{1-\cos2\theta}{\sin2\theta} = \tan\theta.

Solution

The strategy for any identity proof: work on one side only, and push everything towards the simplest functions available — usually \sin\theta and \cos\theta of the single angle.

Step 1 — start with the left side and expand both double angles. Choose the version of \cos2\theta that will make the numerator simplify. Since the numerator is 1 - \cos2\theta, pick \cos2\theta = 1 - 2\sin^2\theta, because then the 1's cancel:

\frac{1 - \left(1-2\sin^2\theta\right)}{2\sin\theta\cos\theta}

Step 2 — simplify the top.

1 - 1 + 2\sin^2\theta = 2\sin^2\theta

= \frac{2\sin^2\theta}{2\sin\theta\cos\theta}

Step 3 — cancel. The 2's go, and one \sin\theta cancels from top and bottom:

= \frac{\sin\theta}{\cos\theta} = \tan\theta

Which is the right-hand side. \blacksquare

What made it work. Step 1 was the whole problem. Picking \cos2\theta = \cos^2\theta - \sin^2\theta instead would have given \frac{1 - \cos^2\theta+\sin^2\theta}{2\sin\theta\cos\theta}, which still works after using 1-\cos^2\theta = \sin^2\theta, but takes an extra step. Choose the form of the identity that cancels something you already have. Here we had a bare 1, so we chose the version containing a bare 1.

The domain footnote. The identity holds wherever both sides exist, so not at \theta = 0, 90°, 180°, \ldots, where \sin2\theta = 0 or \tan\theta is undefined.

Problem 5 — Solving a trigonometric equation completely

Solve 2\sin^2 x - \sin x - 1 = 0 for 0° \le x < 360°.

Solution

Step 1 — recognise the quadratic. Let u = \sin x:

2u^2 - u - 1 = 0

Step 2 — factorise. Two numbers multiplying to 2 \times (-1) = -2 and adding to -1 are -2 and +1:

2u^2 - 2u + u - 1 = 0 \quad \Rightarrow \quad 2u(u-1) + 1(u-1) = 0 \quad \Rightarrow \quad (2u+1)(u-1) = 0

u = -\tfrac12 \quad \text{or} \quad u = 1

Step 3 — undo the substitution, one case at a time.

Case \sin x = 1. On the unit circle the height reaches 1 at exactly one place in a full turn:

x = 90°

Case \sin x = -\frac12. The calculator gives \arcsin(-0.5) = -30°, which is outside our range, so we must find the two solutions that are inside it. The height is negative in the third and fourth quadrants. The reference angle — the acute angle whose sine is \frac12 — is 30°. So:

• third quadrant: x = 180° + 30° = 210°
• fourth quadrant: x = 360° - 30° = 330°

Answer: x = 90°,\ 210°,\ 330°.

Check each. 2(1)^2 - 1 - 1 = 0 ✓. At 210°, \sin = -0.5: 2(0.25) + 0.5 - 1 = 0 ✓. At 330°, same ✓.

What to notice. The calculator gives one answer; the circle gives all of them. Sketching the unit circle and marking where the height equals the value you want takes five seconds and is the only reliable way not to lose solutions. Every trigonometric equation needs this step.

Problem 6 — The ambiguous case, met head on

In triangle ABC, a = 8, b = 11, and A = 32°. Find angle B.

Solution

Step 1 — set up the law of sines. We have a side and its opposite angle, plus one more side — the classic ambiguous configuration.

\frac{a}{\sin A} = \frac{b}{\sin B} \quad \Rightarrow \quad \sin B = \frac{b\sin A}{a}

Step 2 — compute.

\sin B = \frac{11 \times \sin 32°}{8} = \frac{11 \times 0.52992}{8} = \frac{5.8291}{8} = 0.72864

Step 3 — find both angles with that sine. The calculator gives one:

B_1 = \arcsin(0.72864) = 46.77°

But \sin\theta = \sin(180°-\theta), so there is a second candidate:

B_2 = 180° - 46.77° = 133.23°

Step 4 — test whether each candidate actually makes a triangle. The three angles must total 180°.

• With B_1 = 46.77°: C = 180 - 32 - 46.77 = 101.23°. Positive, so this triangle exists.
• With B_2 = 133.23°: C = 180 - 32 - 133.23 = 14.77°. Also positive, so this one exists too.

Answer: both. B = 46.77° or B = 133.23° — two genuinely different triangles satisfy the given data.

Why this happens, seen physically. Fix the 32° angle at A and draw the side b = 11 along one arm. Now swing the side a = 8 from the far end like a compass. Because 8 is longer than the perpendicular distance (11\sin32° = 5.83) but shorter than 11, the arc crosses the base line in two places. Hence two triangles.

When it does not happen. If a were less than 5.83 the arc would not reach the line at all — no triangle. If a were greater than 11 the arc would cross only once — one triangle. This is why the law of cosines is preferred whenever the data allows it: it has no ambiguity.

Problem 7 — A real navigation problem

A ship sails 40 km on a bearing of 065°, then 55 km on a bearing of 130°. How far is it from the start, and on what bearing?

Solution

Bearings are measured clockwise from north, which is why they need care.

Step 1 — find the angle inside the triangle at the turning point. The ship arrives travelling on 065° and leaves on 130°, so it turned through 130 - 65 = 65°. The interior angle of the triangle at that vertex is the supplement of the turn:

B = 180° - 65° = 115°

Drawing this is worth the thirty seconds; the interior angle is between the incoming leg extended backwards and the outgoing leg.

Step 2 — the distance, by the law of cosines (two sides and the angle between them — no ambiguity):

b^2 = 40^2 + 55^2 - 2(40)(55)\cos115°

= 1600 + 3025 - 4400 \times (-0.42262)

The cosine of an obtuse angle is negative, so this term adds:

= 4625 + 1859.5 = 6484.5

b = \sqrt{6484.5} = 80.53 \text{ km}

Step 3 — the bearing. First find the angle at the start, using the law of sines:

\frac{\sin A}{55} = \frac{\sin 115°}{80.53} \quad \Rightarrow \quad \sin A = \frac{55 \times 0.90631}{80.53} = 0.61896

A = 38.24°

No ambiguity here: A must be acute, since the triangle already has an obtuse angle and cannot have two.

Step 4 — turn that into a bearing. The first leg was on 065°, and the direct line to the finish is a further 38.24° clockwise from it:

\text{bearing} = 65 + 38.24 = 103.24°

Answer: 80.5 km away, on a bearing of 103°.

Sanity check. The two legs total 95 km of sailing, and the direct distance is 80.5 km — shorter, as it must be, but not much shorter, because the course change was modest. Both facts pass inspection.

Problem 8 — Two waves adding to one

Express 3\sin\theta + 4\cos\theta as a single sine wave, then find its maximum value and where it occurs.

Solution

Step 1 — write the target form and expand it.

3\sin\theta + 4\cos\theta = R\sin(\theta+\alpha) = R\cos\alpha\sin\theta + R\sin\alpha\cos\theta

Step 2 — match the coefficients of \sin\theta and of \cos\theta on each side:

R\cos\alpha = 3, \qquad R\sin\alpha = 4

Step 3 — find R by squaring and adding. The Pythagorean identity makes the \alpha vanish:

R^2\cos^2\alpha + R^2\sin^2\alpha = 9+16 \quad \Rightarrow \quad R^2 = 25 \quad \Rightarrow \quad R = 5

Step 4 — find \alpha by dividing.

\frac{R\sin\alpha}{R\cos\alpha} = \tan\alpha = \frac43 \quad \Rightarrow \quad \alpha = 53.13°

Both \cos\alpha and \sin\alpha came out positive, so \alpha is in the first quadrant and 53.13° is correct as it stands.

3\sin\theta+4\cos\theta = 5\sin(\theta + 53.13°)

Step 5 — the maximum. A sine never exceeds 1, so the largest the whole thing can be is 5, reached when

\sin(\theta+53.13°) = 1 \quad \Rightarrow \quad \theta + 53.13° = 90° \quad \Rightarrow \quad \theta = 36.87°

Answer: 5\sin(\theta+53.13°), maximum value 5 at \theta = 36.87°.

Check. At \theta = 36.87°: 3\sin(36.87°) + 4\cos(36.87°) = 3(0.6) + 4(0.8) = 1.8 + 3.2 = 5 ✓

What to notice. The amplitude is 5, not 7. Adding a wave of size 3 to a wave of size 4 gives 5 because they peak at different moments, and the combination is exactly the Pythagorean \sqrt{3^2+4^2}. This is why alternating-current calculations use \sqrt{a^2+b^2} everywhere and never a+b, and it is the same reason a 3-4-5 triangle keeps appearing.

Problem 9 — Coordinate geometry with a circle

A circle passes through A(1,2), B(5,2) and C(5,6). Find its centre, radius and equation.

Solution

Step 1 — spot the right angle. AB runs horizontally (both y's are 2) and BC runs vertically (both x's are 5). So the angle at B is a right angle.

Step 2 — use the theorem that saves all the work. An angle in a semicircle is a right angle, and its converse holds too: if \angle ABC = 90°, then AC must be a diameter. So the centre is the midpoint of AC.

\text{centre} = \left(\frac{1+5}{2}, \frac{2+6}{2}\right) = (3, 4)

Step 3 — the radius is half the diameter.

AC = \sqrt{(5-1)^2 + (6-2)^2} = \sqrt{16+16} = \sqrt{32} = 5.657

r = \frac{\sqrt{32}}{2} = \sqrt{8} = 2.828

Step 4 — the equation.

(x-3)^2 + (y-4)^2 = 8

Answer: centre (3,4), radius 2\sqrt2 \approx 2.83, equation (x-3)^2+(y-4)^2 = 8.

Check all three points.

• A(1,2): (1-3)^2+(2-4)^2 = 4+4 = 8 ✓
• B(5,2): (5-3)^2+(2-4)^2 = 4+4 = 8 ✓
• C(5,6): (5-3)^2+(6-4)^2 = 4+4 = 8 ✓

The general method, for when there is no right angle. Substitute all three points into x^2+y^2+Dx+Ey+F = 0 and solve the three linear equations for D, E, F. Slower, but it always works.

Problem 10 — Volume and surface area together

A solid is a cylinder of radius 5 cm and height 12 cm with a hemisphere of the same radius on top. Find the total volume and the total exposed surface area.

Solution

Step 1 — volume of the cylinder.

V_{\text{cyl}} = \pi r^2 h = \pi(25)(12) = 300\pi = 942.48 \text{ cm}^3

Step 2 — volume of the hemisphere, half of \frac43\pi r^3:

V_{\text{hemi}} = \tfrac12 \times \tfrac43\pi(125) = \tfrac23\pi(125) = \frac{250\pi}{3} = 261.80 \text{ cm}^3

Step 3 — total volume.

V = 942.48 + 261.80 = 1204.28 \text{ cm}^3

Step 4 — surface area, counting only surfaces you can actually touch. This is where the marks are lost. The pieces are:

• the flat base of the cylinder: \pi r^2 = 25\pi
• the curved side of the cylinder: 2\pi rh = 2\pi(5)(12) = 120\pi
• the curved top of the hemisphere: half of 4\pi r^2, which is 2\pi r^2 = 50\pi

The circular top face of the cylinder is not counted, because the hemisphere is sitting on it. Neither is the hemisphere's flat face, for the same reason.

A = 25\pi + 120\pi + 50\pi = 195\pi = 612.61 \text{ cm}^2

Answers: volume 1204.3 cm³, surface area 612.6 cm².

Rough check. A plain cylinder of the same radius and total height 12+5 = 17 would hold \pi(25)(17) = 1335 cm³. Ours is less, because a dome holds less than a flat-topped cylinder of the same height — and 1204 < 1335 ✓

Problem 11 — Where the hanging cable actually hangs

A power line hangs as y = 20\cosh(x/20), with x and y in metres and the lowest point at x=0. The poles are at x = \pm 30. How much lower is the middle than the ends, and how long is the cable?

Solution

Step 1 — the height at the lowest point. Put x=0. Since \cosh 0 = \frac{e^0+e^0}{2} = \frac{1+1}{2} = 1:

y(0) = 20 \times 1 = 20 \text{ m}

That constant 20 is not decoration: for the catenary y = a\cosh(x/a), the parameter a is exactly the height of the lowest point above the origin.

Step 2 — the height at a pole. Put x = 30, so x/a = 1.5:

\cosh(1.5) = \frac{e^{1.5}+e^{-1.5}}{2} = \frac{4.48169 + 0.22313}{2} = \frac{4.70482}{2} = 2.35241

y(30) = 20 \times 2.35241 = 47.05 \text{ m}

Step 3 — the sag.

47.05 - 20 = 27.05 \text{ m}

Step 4 — the cable's length. For a catenary there is an exact formula, and it is unusually clean:

L = 2a\sinh\frac{x}{a}

\sinh(1.5) = \frac{4.48169 - 0.22313}{2} = \frac{4.25856}{2} = 2.12928

L = 2 \times 20 \times 2.12928 = 85.17 \text{ m}

Answers: it sags 27.05 m, and the cable is 85.17 m long.

Check against the straight-line distance. The two poles are 60 m apart, and the cable is 85.17 m — longer, necessarily, since it hangs. And the straight distance from a pole to the lowest point is \sqrt{30^2 + 27.05^2} = 40.4 m, so two of those give 80.8 m, which is a little less than the true curved length of 85.17 m ✓ — a curve is always longer than the chord.

Where the length formula comes from. It is the arc-length integral \int\sqrt{1+(y')^2}\,dx from 5.5. Here y' = \sinh(x/a), so the thing under the root is 1+\sinh^2 = \cosh^2 by the hyperbolic identity, and its square root is simply \cosh(x/a) — which integrates to a\sinh(x/a). The catenary is the one curve whose arc length integral simplifies perfectly, and that is not a coincidence: it is the property that defines it.

Problem 12 — Small angles doing real work

The Moon's diameter is 3,474 km and it is 384,400 km away. What angle does it span in the sky, in degrees? Compare with the Sun, diameter 1,392,000 km at 149,600,000 km.

Solution

Step 1 — set up the geometry. The diameter is the "opposite" side and the distance is the "adjacent", so

\tan\theta = \frac{3474}{384400}

But for a tiny angle the small-angle rule says \tan\theta \approx \theta in radians, and this is a case where the approximation is spectacular rather than merely acceptable.

Step 2 — the Moon.

\theta \approx \frac{3474}{384400} = 0.0090375 \text{ rad}

Convert to degrees:

0.0090375 \times \frac{180}{\pi} = 0.5178°

Step 3 — the Sun.

\theta \approx \frac{1392000}{149600000} = 0.0093048 \text{ rad} = 0.5331°

Answers: the Moon spans 0.518°, the Sun 0.533° — both almost exactly half a degree.

How good was the approximation? \arctan(0.0090375) = 0.00903748 rad. The approximation was right to eight decimal places. At angles this small the tangent, the sine and the angle itself are indistinguishable.

The remarkable part. The Sun is 400 times wider than the Moon and happens to sit 400 times further away, so the two look the same size from Earth. That coincidence — and nothing else — is why a total solar eclipse works: the Moon covers the Sun's disc almost exactly, leaving the corona visible. No other planet in the solar system gets this. The Moon is drifting away at about 3.8 cm a year, so in around 600 million years total eclipses will stop happening.

Problem 13 — Deriving an identity you were not given

Show that \sin 3\theta = 3\sin\theta - 4\sin^3\theta, using only the addition and double-angle formulas.

Solution

Step 1 — split the angle into pieces you have formulas for.

\sin3\theta = \sin(2\theta + \theta)

Step 2 — apply the addition formula \sin(A+B) = \sin A\cos B + \cos A\sin B with A = 2\theta, B = \theta:

= \sin2\theta\cos\theta + \cos2\theta\sin\theta

Step 3 — expand each double angle. For \sin2\theta use 2\sin\theta\cos\theta. For \cos2\theta choose the version in sines only, 1-2\sin^2\theta, because the target answer is in sines only:

= \left(2\sin\theta\cos\theta\right)\cos\theta + \left(1-2\sin^2\theta\right)\sin\theta

Step 4 — tidy.

= 2\sin\theta\cos^2\theta + \sin\theta - 2\sin^3\theta

Step 5 — remove the last cosine with \cos^2\theta = 1-\sin^2\theta:

= 2\sin\theta\left(1-\sin^2\theta\right) + \sin\theta - 2\sin^3\theta

= 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta

Step 6 — collect like terms.

= 3\sin\theta - 4\sin^3\theta \qquad \blacksquare

Check at \theta = 30°, where \sin30° = 0.5 and \sin90° = 1:

3(0.5) - 4(0.125) = 1.5 - 0.5 = 1 \; ✓

What to notice. Step 3 was the decision point again: choosing the right form of \cos2\theta made the rest automatic. And compare this with Problem 13 of 2.P, which got the same class of result from De Moivre in half the steps. Both routes are worth knowing — this one needs no complex numbers, that one needs no cleverness.

Problem 14 — A three-dimensional angle

A cuboid room is 8 m long, 6 m wide and 3 m high. Find the length of the diagonal from one bottom corner to the opposite top corner, and the angle that diagonal makes with the floor.

Solution

Step 1 — the diagonal of the floor first. The floor is a rectangle 8 by 6, so by Pythagoras:

d_{\text{floor}} = \sqrt{8^2+6^2} = \sqrt{100} = 10 \text{ m}

Step 2 — the space diagonal, using a second right triangle standing vertically: its base is the floor diagonal (10 m), its height is the room's height (3 m), and its hypotenuse is what we want.

d = \sqrt{10^2 + 3^2} = \sqrt{109} = 10.44 \text{ m}

The general formula, which this derives, is d = \sqrt{l^2+w^2+h^2} — Pythagoras applied twice, once in the plane and once out of it.

Step 3 — the angle with the floor. The space diagonal, the floor diagonal and the vertical edge form a right triangle. The angle at the bottom has opposite side 3 and adjacent side 10:

\tan\theta = \frac{3}{10} = 0.3 \quad \Rightarrow \quad \theta = 16.70°

Answers: the diagonal is 10.44 m, at 16.7° to the floor.

Check. \sin\theta should be \frac{3}{10.44} = 0.2874, and \arcsin(0.2874) = 16.70° ✓

What to notice. Every three-dimensional angle problem reduces to finding the right two-dimensional triangle inside the solid. The skill is identifying which flat triangle contains the angle you want, and the reliable method is to name the three points that define it before doing any arithmetic.

Problem 15 — Everything at once

A Ferris wheel of radius 25 m has its centre 28 m above the ground and completes one turn every 40 seconds, anticlockwise, with you starting at the lowest point. Write your height as a function of time, then find when you are first 40 m up, and how long you spend above 40 m on each turn.

Solution

Step 1 — build the model from the circle. At time t you have turned through an angle \theta = \frac{2\pi}{40}t = \frac{\pi t}{20} radians, because one full turn of 2\pi takes 40 seconds.

Measuring the angle from the bottom of the wheel, your height above the centre is -25\cos\theta: at t=0 that gives -25, putting you 25 m below the centre, which is the lowest point ✓.

h(t) = 28 - 25\cos\left(\frac{\pi t}{20}\right)

Step 2 — check the model at the obvious moments.

• t = 0: h = 28 - 25(1) = 3 m — the lowest point, 3 m off the ground ✓
• t = 20: h = 28 - 25(-1) = 53 m — the top, 28+25 ✓
• t = 10: h = 28 - 0 = 28 m — level with the centre ✓

Step 3 — solve h = 40.

28 - 25\cos\left(\frac{\pi t}{20}\right) = 40

-25\cos\left(\frac{\pi t}{20}\right) = 12

\cos\left(\frac{\pi t}{20}\right) = -0.48

Step 4 — find every solution in one turn, not just the calculator's.

\frac{\pi t}{20} = \arccos(-0.48) = 2.0714 \text{ rad}

Cosine takes the same value at 2\pi - 2.0714 = 4.2118 rad, so there are two crossings per turn — as there must be, since you pass 40 m once going up and once coming down.

Step 5 — convert both back to time.

t_1 = \frac{20 \times 2.0714}{\pi} = 13.19 \text{ s}, \qquad t_2 = \frac{20\times4.2118}{\pi} = 26.81 \text{ s}

Answers: you first reach 40 m at t = 13.2 s, and you are above 40 m from 13.2 s to 26.8 s — that is 13.6 seconds of every 40.

Check for reasonableness. The high point is at t=20 s, which sits exactly halfway between 13.19 and 26.81 ✓ — the interval must be symmetric about the top, since the motion is.

What to notice. Every periodic real-world quantity gets modelled this way: tides, daylight hours, engine pistons, alternating current. The recipe is always the same four numbers — the midline (28), the amplitude (25), the period (40 s, which sets the coefficient \frac{2\pi}{40} inside), and the starting phase (a minus sign here, because we began at the bottom rather than the middle). Get those four right and the model is right.

Next: Part 4 — Linear Algebra, where the coordinate geometry of this Part becomes a machine that handles any number of dimensions at once.