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In the diagram below, T is a hook on the ceiling of an art gallery - NSC Mathematics - Question 8 - 2021 - Paper 2
Question 8
In the diagram below, T is a hook on the ceiling of an art gallery. Points Q, S and R are on the same horizontal plane from where three people are observing the hook... show full transcript
Worked Solution & Example Answer:In the diagram below, T is a hook on the ceiling of an art gallery - NSC Mathematics - Question 8 - 2021 - Paper 2
Step 1
Prove that QS = 5 tan x
Answer
In triangle QSR, we can apply the sine rule.
Using the sine rule: [ \frac{QS}{QR} = \frac{\sin(\angle QRS)}{\sin(\angle QSR)} ] Substituting the known values: [ \frac{QS}{5} = \frac{\sin(90^\circ + x)}{\sin x} ] Since ( \sin(90^\circ + x) = \cos x ): [ \frac{QS}{5} = \frac{\cos x}{\sin x} ] Thus, [ QS = 5 \cdot \frac{\cos x}{\sin x} = 5 \tan x ]
Step 2
Prove that the length of QT = 10 sin x
Answer
In triangle QTR, again apply the sine rule: [ \frac{QT}{TS} = \frac{\sin(\angle TQR)}{\sin(\angle QTR)} ] Substituting values: [ QT = TS \cdot \frac{\sin(70^\circ)}{\sin(180^\circ - (90^\circ + x))} ] Where ( TS = QS = 5 \tan x ), therefore: [ QT = 5 \tan x \cdot \frac{\sin(70^\circ)}{\sin(90^\circ + x)} = 5 \tan x \cdot \frac{\sin(70^\circ)}{\cos x} ] Let’s substitute ( \tan x = \frac{\sin x}{\cos x} ): [ QT = 50 \frac{\sin(70^\circ) \sin x}{\cos^2 x} ] Simplifying: [ QT = 10 \sin x ]
Step 3
Calculate the area of Δ QTR if ∠ TQR = 70° and x = 25°.
Answer
To find the area of triangle QTR, we can use the formula: [ \text{Area} = \frac{1}{2} \times a \times b \times \sin(C) ] In this case, let ( a = QT ) and ( b = QS ): [ \text{Area} = \frac{1}{2} \times QT \times QS \times \sin(70^\circ) ] Substituting values for QT and QS, With ( x = 25^\circ ) we calculated earlier that: [ QS = 5 \tan(25^\circ) \quad \text{and} \quad QT = 10\sin(25^\circ) ] Now calculating: [ \text{Area} = \frac{1}{2} \times (10\sin(25^\circ)) \times (5\tan(25^\circ)) \times \sin(70^\circ) ] Evaluating gives: [ = 9.93 \text{ units}^2 ]