Math  /  Geometry

QuestionQUESTION 7 In the diagram below, TQRT Q R represents three points in a horizontal plane on a sportsfield. PQ represents a vertical flagpole.
The angle of elevation of the top of the pole from R is equal to θTundefined=θTQ=TR=y\theta \cdot \widehat{\mathrm{T}}=\theta \cdot \mathrm{TQ}=\mathrm{TR}=y. QR=x\mathrm{QR}=x units. TQR=α\mathrm{TQ} \mathrm{R}=\alpha. 7.1 Express θ\theta in terms of α\alpha and show that sinθ=sin2α\sin \theta=\sin 2 \alpha 7.2 Hence, prove that in PQR:PR=2ycosαcosθ\triangle \mathrm{PQR}: \mathrm{PR}=\frac{2 y \cos \alpha}{\cos \theta} 7.3 If α=49;x=20 m\alpha=49^{\circ} ; x=20 \mathrm{~m} and y=15 my=15 \mathrm{~m}, calculate the area of ΔTQR\Delta \mathrm{TQR}.

Studdy Solution
Calculate sin(49)0.7547 \sin(49^\circ) \approx 0.7547 : Area=12×15×15×0.7547 \text{Area} = \frac{1}{2} \times 15 \times 15 \times 0.7547 Area=12×225×0.7547 \text{Area} = \frac{1}{2} \times 225 \times 0.7547 Area84.41m2 \text{Area} \approx 84.41 \, \text{m}^2
Solution: The area of ΔTQR \Delta TQR is approximately 84.41m2 84.41 \, \text{m}^2 .

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