The diagram shows two chords AB and CD intersect at X such that angle CXB = 90∘90^\circ. Given that AC=5 cmAC = 5\ \text{cm}, BD=12 cmBD = 12\ \text{cm} and angle BDC=θBDC = \theta.

  1. (a)

    Prove that AB=5cos⁡θ+12sin⁡θAB = 5\cos\theta + 12\sin\theta.

  2. (b)

    Express ABAB in the form Rcos⁡(θ−α)R\cos(\theta - \alpha) where R>0R > 0 and α\alpha is an acute angle.

  3. (c)

    Find the maximum length of ABAB and the corresponding value of θ\theta.

    Hence state the radius of the circle.

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Worked solution

4(a)

1. Find angle BAC using circle properties.

Angles subtended by the same arc BCBC at the circumference are equal.
∠BAC=∠BDC=θ\angle BAC = \angle BDC = \theta

2. Express the length of AX in terms of theta.

In △AXC\triangle AXC, ∠AXC=90∘\angle AXC = 90^\circ (angles on a straight line with ∠CXB=90∘\angle CXB = 90^\circ).
cos⁡(∠CAX)=AXAC\cos(\angle CAX) = \frac{AX}{AC}
cos⁡θ=AX5\cos\theta = \frac{AX}{5}
AX=5cos⁡θAX = 5\cos\theta

3. Express the length of BX in terms of theta and sum them to find AB.

In △BXD\triangle BXD, ∠BXD=90∘\angle BXD = 90^\circ (angles on a straight line with ∠CXB=90∘\angle CXB = 90^\circ).
sin⁡(∠BDX)=BXBD\sin(\angle BDX) = \frac{BX}{BD}
sin⁡θ=BX12\sin\theta = \frac{BX}{12}
BX=12sin⁡θBX = 12\sin\theta

Since XX lies on the chord ABAB,
AB=AX+BXAB = AX + BX
AB=5cos⁡θ+12sin⁡θAB = 5\cos\theta + 12\sin\theta (Shown)

Answer

AB=5cos⁡θ+12sin⁡θAB = 5\cos\theta + 12\sin\theta (Shown)

4(b)

1. Expand the target form and compare coefficients with the given expression

Let 5cos⁡θ+12sin⁡θ=Rcos⁡(θ−α)5\cos\theta + 12\sin\theta = R\cos(\theta - \alpha).
Expanding the right side:
Rcos⁡(θ−α)=Rcos⁡θcos⁡α+Rsin⁡θsin⁡αR\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha
Comparing coefficients of cos⁡θ\cos\theta and sin⁡θ\sin\theta:
Rcos⁡α=5R\cos\alpha = 5
Rsin⁡α=12R\sin\alpha = 12

2. Calculate the value of R

R=52+122R = \sqrt{5^2 + 12^2}
R=25+144=169R = \sqrt{25 + 144} = \sqrt{169}
R=13R = 13

3. Calculate the value of alpha

tan⁡α=Rsin⁡αRcos⁡α=125\tan\alpha = \frac{R\sin\alpha}{R\cos\alpha} = \frac{12}{5}
α=tan⁡−1(125)\alpha = \tan^{-1}\left(\frac{12}{5}\right)
α≈67.380∘≈67.4∘\alpha \approx 67.380^\circ \approx 67.4^\circ (to 1 d.p.)

4. State the final expression

AB=13cos⁡(θ−67.4∘)AB = 13\cos(\theta - 67.4^\circ)

Answer

13cos⁡(θ−67.4∘)13\cos(\theta - 67.4^\circ)

4(c)

1. State the maximum length of AB using the R-formula expression

From part (b), AB=13cos⁡(θ−67.38∘)AB = 13\cos(\theta - 67.38^\circ).
The maximum value of cos⁡(θ−67.38∘)\cos(\theta - 67.38^\circ) is 11.
Maximum length of AB=13(1)=13 cmAB = 13(1) = 13\text{ cm}.

2. Find the corresponding value of $\theta$

For maximum ABAB, cos⁡(θ−67.38∘)=1\cos(\theta - 67.38^\circ) = 1.
θ−67.38∘=0∘\theta - 67.38^\circ = 0^\circ
θ=67.38∘≈67.4∘\theta = 67.38^\circ \approx 67.4^\circ (to 1 d.p.)

3. State the radius of the circle

The maximum length of a chord in a circle is its diameter.
Diameter =13 cm= 13\text{ cm}
Radius =132=6.5 cm= \frac{13}{2} = 6.5\text{ cm}

Answer

Maximum length =13 cm= 13\text{ cm}, θ=67.4∘\theta = 67.4^\circ, Radius =6.5 cm= 6.5\text{ cm}

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