Mathematics • Year 10 • Unit 3 • Lesson 3

Finding Angles in the Real World

Apply inverse trig to real Australian contexts — the pitch of a roof, the steepness of a Blue Mountains hiking track, the angle of a satellite dish on a Brisbane rooftop. Each problem asks for an unknown angle given two sides. Round to the nearest degree unless told otherwise. Calculator in DEG mode.

Apply · Real-World Maths

1. Word problems

For each problem: (i) sketch the right-angled triangle, (ii) identify the two given sides, (iii) write the ratio, (iv) apply the matching inverse function, (v) round.

1.1 — Roof pitch in Penrith. A pitched gable roof has rafters of 5.2 m and a horizontal half-span (from the wall to directly under the ridge) of 4.5 m. The ridge sits above one end of the half-span. Find the angle the rafter makes with the horizontal (the "roof pitch"). Round to the nearest degree. 3 marks

Stuck? Rafter = hypotenuse = 5.2 m. Half-span = adjacent = 4.5 m. A + H means cosine. cos θ = 4.5 / 5.2.

1.2 — Blue Mountains track. A short straight section of a hiking track in the Blue Mountains rises 80 m vertically over a horizontal distance of 350 m. Find the angle of the slope to the horizontal, to 1 decimal place. 3 marks

Stuck? Vertical rise = opposite. Horizontal = adjacent. O + A means tangent. tan θ = 80 / 350.

1.3 — Satellite dish. A satellite dish on a Brisbane rooftop is angled so the centre of the dish sits 1.4 m above the centre of its base. The straight line from the rooftop directly below the centre of the dish to the centre of the dish itself is 2.1 m long (the hypotenuse). Find the elevation angle of the dish above the horizontal rooftop. 3 marks

Stuck? Height above rooftop = opposite = 1.4. Slope distance = hypotenuse = 2.1. O + H means sine. sin θ = 1.4 / 2.1.

1.4 — Skateboard ramp. A skateboard ramp at a Marrickville skate park is 1.2 m high at its highest point and 1.8 m long along the ground from the bottom of the ramp to directly under the highest point. Find the angle the ramp surface makes with the ground. 3 marks

Stuck? Height = opposite, ground distance = adjacent, ramp surface = hypotenuse (unknown). tan θ = 1.2 / 1.8.

1.5 — A leaning pole. A 3 m flagpole leans slightly. Its tip is 2.85 m directly above the ground (vertical distance from the base of the pole). Find the angle the pole makes with the vertical. (Round to the nearest degree.) 3 marks

Stuck? The pole itself is the hypotenuse (3 m). The vertical 2.85 m is the adjacent (alongside the vertical line, between the base and the tip's vertical drop). cos θ = 2.85 / 3.

2. Explain your thinking

This question is about communication, not just numbers. Use full sentences. 4 marks

2.1 A friend types sin⁻¹(2) into a calculator and gets the answer ERROR (or NaN). They are confused. Using your understanding of trig ratios from Lessons 1-3, explain in 4-6 sentences (i) why the calculator returns an error, (ii) what range of values the input to sin⁻¹ or cos⁻¹ must lie in, (iii) why tan⁻¹ does not have the same restriction, and (iv) how you could re-frame the question to make it valid. Include the words "ratio" and "hypotenuse".

Stuck? sin θ = O / H. The opposite can never be longer than the hypotenuse, so the ratio can never exceed 1.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Penrith roof pitch

cos θ = 4.5 / 5.2 = 0.8654. θ = cos⁻¹(0.8654) = 30.07° ≈ 30°.
A 30° pitch is a typical Australian gable roof.

1.2 — Blue Mountains track slope

tan θ = 80 / 350 = 0.2286. θ = tan⁻¹(0.2286) = 12.876° ≈ 12.9° (to 1 d.p.).
A gradient of about 23% — moderately steep walking.

1.3 — Satellite dish elevation

sin θ = 1.4 / 2.1 = 0.6667. θ = sin⁻¹(0.6667) = 41.81° ≈ 42°.
A reasonable elevation for a geostationary-satellite dish in Brisbane.

1.4 — Skateboard ramp angle

tan θ = 1.2 / 1.8 = 0.6667. θ = tan⁻¹(0.6667) = 33.69° ≈ 34°.
This is steep enough for a quarter-pipe.

1.5 — Leaning pole

cos θ = 2.85 / 3 = 0.95. θ = cos⁻¹(0.95) = 18.19° ≈ 18°.
The pole is leaning 18° away from vertical — surprisingly noticeable for just a 15 cm gap.

2.1 — Why sin⁻¹(2) errors (sample response)

The calculator returns an error because sin θ is defined as opposite divided by hypotenuse in a right-angled triangle, and the hypotenuse is always the longest side — so the ratio can never exceed 1 in size. The valid inputs for sin⁻¹ and cos⁻¹ therefore lie between −1 and 1 inclusive. tan θ does not have this restriction because tangent is opposite ÷ adjacent (no hypotenuse involved), and the opposite can be far larger than the adjacent — for example, in a very steep triangle, tan θ can be 100 or 1000. To re-frame the question, my friend could try a valid value such as sin⁻¹(2/3), which gives an angle of about 41.8°, or recognise that they meant tan⁻¹(2), which gives about 63.4°.

Marking: 1 for the ratio / hypotenuse argument, 1 for the −1 to 1 range, 1 for the tan exception, 1 for a valid re-framing.