Mathematics Advanced • Year 11 • Module 2 • Lesson 8
Complementary Angle Relationships
Apply co-function identities to surveying triangles, roof pitch, lens optics, ramp design, and identity-driven simplification.
Problem 1 — Surveying right triangle (geometric)
A surveyor establishes a right-angled triangle between three markers A, B, C, with the right angle at B. The angle at A is measured as $\angle A = 32^\circ$. The opposite side $a$ (across from A) is 17.5 m and the hypotenuse $c$ (across from C, the right angle) is 33.0 m.
Set up: What are we solving for?
(i) Find $\angle C$ exactly using the complementary-angle fact ($\angle A + \angle C = 90^\circ$). 1 mark
(ii) Show, by writing both as ratios of sides, that $\sin A = \cos C$. Then verify this numerically using the given side lengths (compute $\sin 32^\circ$ and $\cos 58^\circ$ to 3 dp using a calculator if needed). 3 marks
(iii) Without recomputing, use the co-function identity to write $\tan 32^\circ$ as a cotangent. 2 marks
Stuck on (ii)? In the triangle, $\sin A = \frac{a}{c}$ and $\cos C = \frac{a}{c}$ as well (the side opposite A is adjacent to C).Problem 2 — Roof pitch (construction)
A roof truss has a rafter inclined at $\theta = 35^\circ$ to the horizontal. The complementary angle at the apex is $\phi = 55^\circ$ (since $35 + 55 = 90$). A carpenter needs to know both $\sin \theta$ (vertical rise per unit rafter) and $\sin \phi$ (a quantity used by a second tradesperson's worksheet).
Set up: What are we solving for?
(i) Without a calculator, express $\sin 55^\circ$ as the cosine of a complementary angle. 2 marks
(ii) Using your answer to (i) and the identity $\sin^2 \theta + \cos^2 \theta = 1$, show that $\sin^2 35^\circ + \sin^2 55^\circ = 1$ without a calculator. 3 marks
(iii) A different worksheet computes $\sin 35^\circ + \sin 55^\circ$ rather than the squared sum. Without a calculator, can you conclude this equals 1? Justify your answer in one sentence. 2 marks
Problem 3 — Lens incidence angles (optics)
For light hitting a glass lens at incidence angle $\alpha$, the refracted ray makes angle $\beta$ with the normal. At a critical configuration, the incident and refracted rays inside the glass are complementary: $\alpha + \beta = 90^\circ$. Engineers compare $\tan \alpha$ (used in the optical solver) with $\cot \beta$ (used in a reference table).
Set up: What are we solving for?
(i) Show that $\tan \alpha = \cot \beta$ whenever $\alpha + \beta = 90^\circ$. Cite the co-function identity. 2 marks
(ii) For $\alpha = 25^\circ$, write $\cot \beta$ in the form $\tan(\text{angle})$ and state the angle. 2 marks
(iii) A junior engineer writes $\sin \alpha = \sin \beta$ on the basis that $\alpha + \beta = 90^\circ$. Identify the trap this commits (referencing the lesson by name) and write the correct identity. 2 marks
Stuck on (iii)? Lesson Trap 1: $\sin \theta = \sin(90^\circ - \theta)$ holds only when $\theta = 45^\circ$.Problem 4 — Ramp design with complementary slopes
Two ramps meet at a platform. Ramp 1 has angle $\theta$ to the horizontal; Ramp 2 has angle $(90^\circ - \theta)$. (Together they form a right angle at the platform edge.)
Set up: What are we solving for?
(i) The mechanical advantage of a ramp is $\csc(\text{slope angle})$. Show that $\text{MA}_1 \cdot \text{MA}_2 = \sec \theta \cdot \csc \theta$ for these two ramps. (You'll need a co-function identity.) 3 marks
(ii) Take $\theta = \frac{\pi}{6}$. Find $\text{MA}_1$ and $\text{MA}_2$ exactly. Then verify your product formula from (i). 3 marks
(iii) Explain in 1-2 sentences why making both ramps "easier" (larger MA) is impossible — what trade-off does the complementary relationship enforce? 2 marks
Problem 5 — Compound-expression detective
Simplify each compound expression using a co-function identity and (where helpful) a Pythagorean identity.
Set up: What are we solving for?
(i) Simplify $\sin 68^\circ \cdot \cos 22^\circ + \cos 68^\circ \cdot \sin 22^\circ$. Hint: rewrite each cos as a sin and group. (Alternatively, recognise the sum-of-angles pattern, but the cleaner route here uses complementary rewrites.) 3 marks
(ii) Without a calculator, evaluate $\cos^2 25^\circ + \cos^2 65^\circ$. 2 marks
(iii) A textbook asserts $\frac{\tan 80^\circ}{\cot 10^\circ} = 1$. Verify or refute by rewriting one side using a co-function identity. 2 marks
Stuck on (ii)? Convert $\cos 65^\circ$ to $\sin 25^\circ$ and use Identity 1.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Surveying triangle
Set up. We're applying complementary-angle and co-function identities to a real triangle, then rewriting tan as cot.
(i) $\angle A + \angle C = 90^\circ$, so $\angle C = 90^\circ - 32^\circ = \mathbf{58^\circ}$.
(ii) By definition, $\sin A = \frac{\text{opposite to A}}{\text{hypotenuse}} = \frac{a}{c}$. The side opposite A (length $a$) is also adjacent to C, so $\cos C = \frac{\text{adjacent to C}}{\text{hypotenuse}} = \frac{a}{c}$. Hence $\sin A = \cos C$. Numerically: $\sin 32^\circ \approx 0.530$, $\cos 58^\circ \approx 0.530$. ✓
(iii) $\tan 32^\circ = \cot(90^\circ - 32^\circ) = \mathbf{\cot 58^\circ}$.
Problem 2 — Roof pitch
Set up. We're converting a sine to a cosine via the complement, then using the Pythagorean identity to evaluate a sum of squares without a calculator.
(i) $\sin 55^\circ = \cos(90^\circ - 55^\circ) = \mathbf{\cos 35^\circ}$.
(ii) Substitute the rewrite into the sum: $\sin^2 35^\circ + \sin^2 55^\circ = \sin^2 35^\circ + \cos^2 35^\circ = \mathbf{1}$ (Pythagorean Identity 1). No calculator needed.
(iii) No: $\sin 35^\circ + \sin 55^\circ \approx 0.574 + 0.819 = 1.393$, not 1. The identity $\sin^2 + \cos^2 = 1$ applies to the squared sum, not the linear sum.
Problem 3 — Lens incidence
Set up. We're applying the tan-cot co-function identity, then critiquing a common Trap 1 misuse.
(i) $\beta = 90^\circ - \alpha$, so $\cot \beta = \cot(90^\circ - \alpha) = \tan \alpha$ (co-function identity for tan-cot). Hence $\tan \alpha = \cot \beta$.
(ii) $\cot \beta = \tan \alpha = \mathbf{\tan 25^\circ}$.
(iii) This commits Trap 1 in the lesson: writing $\sin \theta = \sin(90^\circ - \theta)$ is only true at $\theta = 45^\circ$. The correct identity is $\sin \alpha = \cos \beta$ (since $\beta = 90^\circ - \alpha$).
Problem 4 — Ramp design
Set up. We're applying the csc-sec co-function pair to express two MA values in different forms, then exploring a numerical case and the design trade-off.
(i) $\text{MA}_1 = \csc \theta$, $\text{MA}_2 = \csc(90^\circ - \theta) = \sec \theta$ (co-function: $\csc(90^\circ - \theta) = \sec \theta$). Product: $\csc \theta \cdot \sec \theta = \mathbf{\sec \theta \cdot \csc \theta}$. ✓
(ii) $\text{MA}_1 = \csc \frac{\pi}{6} = \frac{1}{1/2} = 2$. $\text{MA}_2 = \sec \frac{\pi}{6} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$. Product: $2 \cdot \frac{2\sqrt{3}}{3} = \mathbf{\frac{4\sqrt{3}}{3}}$. Verify via formula: $\sec \frac{\pi}{6} \cdot \csc \frac{\pi}{6} = \frac{2\sqrt{3}}{3} \cdot 2 = \frac{4\sqrt{3}}{3}$. ✓
(iii) Increasing $\theta$ to make Ramp 1 gentler ($\csc \theta$ larger) automatically increases $90^\circ - \theta$ toward 0, making Ramp 2 steeper ($\sec \theta = \csc(90^\circ - \theta)$ smaller). The two MAs trade off — a fixed right angle at the platform forces this constraint.
Problem 5 — Compound expressions
Set up. We're using co-function identities to simplify expressions and to verify a textbook claim.
(i) $\cos 22^\circ = \sin 68^\circ$ and $\cos 68^\circ = \sin 22^\circ$. Substituting: $\sin 68^\circ \cdot \sin 68^\circ + \sin 22^\circ \cdot \sin 22^\circ = \sin^2 68^\circ + \sin^2 22^\circ$. Now $\sin 22^\circ = \cos 68^\circ$, so the sum becomes $\sin^2 68^\circ + \cos^2 68^\circ = \mathbf{1}$ (Identity 1).
(ii) $\cos 65^\circ = \sin 25^\circ$, so $\cos^2 65^\circ = \sin^2 25^\circ$. Sum: $\cos^2 25^\circ + \sin^2 25^\circ = \mathbf{1}$.
(iii) $\cot 10^\circ = \tan(90^\circ - 10^\circ) = \tan 80^\circ$. So $\frac{\tan 80^\circ}{\cot 10^\circ} = \frac{\tan 80^\circ}{\tan 80^\circ} = \mathbf{1}$. ✓ The textbook claim is correct.