Sine rule and a percentage-error interval

Triangle PQR: sine-rule geometryP; Q; R; π/6 + θ; 2π/3; a. Source triangle labels P,Q,R,a,2π/3,π/6+θ retained; schematic a=1.Triangle PQR: sine-rule geometryPQRπ/6 + θ2π/3a
Illustration of the stated givens.
Question diagram 1
Triangle PQR: sine-rule geometryP; Q; R; π/6 + θ; 2π/3; a. Source triangle labels P,Q,R,a,2π/3,π/6+θ retained; schematic a=1.Triangle PQR: sine-rule geometryPQRπ/6 + θ2π/3a

Triangle PQR with PR=a, angle PRQ=2π/3 and angle RPQ=π/6+θ.

In triangle PQR, ∠PRQ=2π3\angle\mathrm{PRQ}=\frac{2\pi}3, ∠RPQ=π6+θ\angle\mathrm{RPQ}=\frac\pi6+\theta, and PR = aa units, where aa is a positive real constant.

(i) Show that PQ=3acos⁡θ−3sin⁡θ\mathrm{PQ}=\frac{\sqrt3a}{\cos\theta-\sqrt3\sin\theta}. marks

(ii) Given that θ\theta is sufficiently small, show that PQ≈3a(1+bθ+cθ2)\mathrm{PQ}\approx\sqrt3a(1+b\theta+c\theta^2), where b,cb,c are real constants. marks

(iii) Given that θ>0\theta>0, find the range of values of θ\theta such that the percentage error of the approximation in (ii) is less than 5%. marks

[2023/SAJC/Prelim/P2/2]

Total marks: 9

Solution

(i)

Sine rule

In triangle PQRPQR, ∠PRQ=2π3\angle PRQ=\frac{2\pi}3, ∠RPQ=π6+θ\angle RPQ=\frac\pi6+\theta, PR=aPR=a. Show that PQ=3acos⁡θ−3sin⁡θPQ=\dfrac{\sqrt3a}{\cos\theta-\sqrt3\sin\theta}.

Hint

  1. Find ∠PQR\angle PQR from the angle sum.

  2. Apply the sine rule: PQsin⁡∠PRQ=PRsin⁡∠PQR\frac{PQ}{\sin\angle PRQ}=\frac{PR}{\sin\angle PQR}.

Answer · (i)

PQ=3a/(cos⁡θ−3sin⁡θ)PQ=\sqrt3a/(\cos\theta-\sqrt3\sin\theta).

(ii)

Reciprocal series

For small θ\theta, show PQ≈3a(1+bθ+cθ2)PQ\approx\sqrt3a(1+b\theta+c\theta^2) and find bb and cc.

Explore this idea

Hint

  1. Use cos⁡θ≈1−θ22\cos\theta\approx1-\frac{\theta^2}2 and sin⁡θ≈θ\sin\theta\approx\theta.

  2. Expand (1−u)−1(1-u)^{-1}.

Answer · (ii)

b=3, c=7/2b=\sqrt3,\ c=7/2.

Explore the idea

1/(cos θ − √3 sin θ) and its series

Compare the function with a finite polynomial. Change the highest included power and test x, then use the error plot to see how the approximation changes at that x.

Highest included powerxⁿ, n = 2
Test value xx = 0.15
Reset the controls
Function
1.36998
P₂(x)
1.33856
Error
-0.03142

Inside the stated open interval. A finite truncation can still have a noticeable error.

Teal marks the function, amber marks the polynomial, and the red segment joins their values at the selected x.

Does keeping more powers help at this x?

Highest included power n runs horizontally; signed error Pₙ(x) − f(x) runs vertically. Zero means agreement at this x. A missing power leaves the polynomial unchanged; individual steps need not improve the error.

  • Function
  • Series truncated at the chosen power
  • Stated open interval. Endpoint convergence requires a separate check
Key idea

The series is only an approximation for small θ\theta; it is poor near the pole at θ=π6\theta=\frac\pi6.

Ideas to try
Slide θ\theta towards π6≈0.524\frac\pi6\approx0.524 where the denominator is zero.
(iii)

Percentage error

For θ>0\theta>0, find the range of θ\theta such that the percentage error of the approximation in (ii) is less than 5%.

Explore this idea

Hint

  1. Percentage error =100∣approximate−actualactual∣=100\left|\dfrac{\text{approximate}-\text{actual}}{\text{actual}}\right|.

  2. Use a graphing calculator to solve error <5<5.

Answer · (iii)

0<θ<θ∗0<\theta<\theta_* radians, where θ∗≈0.194141\theta_*\approx0.194141 (approximately 0.194).

Physical domain 0<θ<π/60<\theta<\pi/6; the exact positive boundary has percentage error 5%, and the required error is strictly smaller. Rounded values are illustrative.

Explore the idea

Percentage error of the approximation

The error grows from 0 as θ increases. The approximation is within 5% until the curve crosses the line at θ ≈ 0.194.

Height of the horizontal linek = 5
Try an example
Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct crossings
1

Each positive tolerance here gives one boundary in 0 < θ < π/6; for 5% it is approximately 0.194141.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Pointxy
0.1941415
  • Percentage error of the approximation
  • Tolerance
  • Boundary θ
  • Intersection selected in the coordinate table
Key idea

The strict 5% condition is 0<θ<θ∗0<\theta<\theta_*, with f(θ∗)=5f(\theta_*)=5 and θ∗≈0.194141\theta_*\approx0.194141. Rounded values are not exact boundaries; the error increases towards the pole at θ=π/6\theta=\pi/6.

Ideas to try
What angle gives a 1% error?