The diagram below shows a circle with chords PQPQ and RSRS intersect at the point XX. P,Q,RP, Q, R and SS are points on the circle. Given that PQ=7 cmPQ = 7\text{ cm}, RS=24 cmRS = 24\text{ cm}, angle PXQ=90∘PXQ = 90^\circ and angle PRS=θPRS = \theta, where θ\theta varies.

  1. (i)

    Prove that PR=24cos⁡θ+7sin⁡θPR = 24\cos\theta + 7\sin\theta.

  2. (ii)

    Express PRPR in the form Rcos⁡(θ−α)R\cos(\theta - \alpha).

  3. (iii)

    Hence, find the value of θ\theta for which PR=15 cm.PR = 15\text{ cm}.

  4. (iv)

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

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Worked solution

11(ii)

1. Find the value of $R$

R=242+72=576+49=625=25R = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25

2. Find the value of $\alpha$

tan⁡α=724\tan \alpha = \frac{7}{24}
α=tan⁡−1(724)≈16.260∘\alpha = \tan^{-1}\left(\frac{7}{24}\right) \approx 16.260^\circ

3. Express $PR$ in the required form

PR=25cos⁡(θ−16.3∘)PR = 25\cos(\theta - 16.3^\circ)

Answer

25cos⁡(θ−16.3∘)25\cos(\theta - 16.3^\circ)

11(iii)

1. Find the values of $R$ and $\alpha$ from part (ii) to express $PR$ in the form $R\cos(\theta - \alpha)$

R=242+72=25R = \sqrt{24^2 + 7^2} = 25
tan⁡α=724  ⟹  α=16.260∘\tan\alpha = \frac{7}{24} \implies \alpha = 16.260^\circ
So, PR=25cos⁡(θ−16.260∘)PR = 25\cos(\theta - 16.260^\circ)

2. Set up the equation for $PR = 15$ and isolate the cosine term

25cos⁡(θ−16.260∘)=1525\cos(\theta - 16.260^\circ) = 15
cos⁡(θ−16.260∘)=1525=0.6\cos(\theta - 16.260^\circ) = \frac{15}{25} = 0.6

3. Solve for $\theta$

θ−16.260∘=cos⁡−1(0.6)\theta - 16.260^\circ = \cos^{-1}(0.6)
θ−16.260∘=53.130∘\theta - 16.260^\circ = 53.130^\circ
θ=53.130∘+16.260∘=69.390∘≈69.4∘\theta = 53.130^\circ + 16.260^\circ = 69.390^\circ \approx 69.4^\circ (to 1 d.p.)

Answer

69.4∘69.4^\circ

11(iv)

1. State the maximum length of PR using the amplitude from the R-formula

From part (ii), PR=25cos⁡(θ−16.26∘)PR = 25\cos(\theta - 16.26^\circ).
The maximum value of cos⁡(θ−16.26∘)\cos(\theta - 16.26^\circ) is 11.
Therefore, the maximum length of PRPR is 25 cm25\text{ cm}.

2. Find the corresponding value of theta

The maximum occurs when cos⁡(θ−16.26∘)=1\cos(\theta - 16.26^\circ) = 1.
θ−16.26∘=0∘\theta - 16.26^\circ = 0^\circ
θ=16.26∘≈16.3∘\theta = 16.26^\circ \approx 16.3^\circ (to 1 d.p.)

Answer

Maximum length =25 cm= 25\text{ cm}, θ=16.3∘\theta = 16.3^\circ

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