The diagram below represents two observers at P and Q who are equidistant from point R. The two observers are 481.1 m apart. The observers sight an air balloon at S, which is h metres above R. The angle of elevation of S from Q is 23.5°. P, Q and R lie on the same horizontal plane. PQR = 33.9° Determine: 6.1 The size of ∠PQR. 6.2 RQ, the distance between the observer at Q and point R. 6.3 The value of h, to the nearest metre. 6.4 The area of ΔPQR.

NSC Technical Mathematics Ratio and Proportion: The diagram below represents two

The diagram below represents two observers at P and Q who are equidistant from point R - NSC Technical Mathematics - Question 6 - 2022 - Paper 2

Question 6

The diagram below represents two observers at P and Q who are equidistant from point R. The two observers are 481.1 m apart. The observers sight an air balloon at S... show full transcript

Worked Solution & Example Answer:The diagram below represents two observers at P and Q who are equidistant from point R - NSC Technical Mathematics - Question 6 - 2022 - Paper 2

Step 1

6.1 The size of ∠PQR.

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Answer

To determine the size of ∠PQR, we recognize that the sum of angles in a triangle is 180°.

Thus, $\angle PQR + 33.9° + 23.5° = 180°$

Calculating this, we get: $\angle PQR = 180° - (33.9° + 23.5°)$

Therefore, $\angle PQR = 112.6°$

Step 2

6.2 RQ, the distance between the observer at Q and point R.

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Answer

We can find RQ using the sine rule. In triangle PQR:

$\frac{RQ}{\sin(33.9°)} = \frac{481.1}{\sin(112.2°)}$

So, $RQ = \frac{481.1 \cdot \sin(33.9°)}{\sin(112.2°)}$

Calculating this gives: $RQ ≈ 289.81 m$

Step 3

6.3 The value of h, to the nearest metre.

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Answer

To find h, we can use the tangent function. From triangle QRS:

$\tan(23.5°) = \frac{h}{RQ}$

Substituting the value of RQ we found earlier: $h = RQ \cdot \tan(23.5°$

Thus, $h ≈ 289.81 \cdot \tan(23.5° ≈ 126 m$

Step 4

6.4 The area of ΔPQR.

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Answer

The area of triangle ΔPQR can be calculated using the formula:

$\text{Area} = \frac{1}{2} \cdot base \cdot height$

Where the base is 481.1 m and the height can be found using the sine:

$\text{Area} = \frac{1}{2} \cdot 481.1 \cdot 289.81 \cdot \sin(33.9°)$

Calculating this results in: $\text{Area} ≈ 38,882.53 m^2$

The diagram below represents two observers at P and Q who are equidistant from point R - NSC Technical Mathematics - Question 6 - 2022 - Paper 2

Worked Solution & Example Answer:The diagram below represents two observers at P and Q who are equidistant from point R - NSC Technical Mathematics - Question 6 - 2022 - Paper 2

6.1 The size of ∠PQR.

6.2 RQ, the distance between the observer at Q and point R.

6.3 The value of h, to the nearest metre.

6.4 The area of ΔPQR.

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