How many different triangles can be drawn from the information given? (b) Calculate: (i) R̂ (ii) P̂ and (iii) QR.

Let the helicopter's position be C, and the two points on the horizontal plane be A and B. We are given:

• Distance from helicopter to A, $AC = 850$ m
• Distance from helicopter to B, $BC = 1200$ m
• Angle of elevation from B to the helicopter, $\angle B = 43^\circ$ We need to find the distance $AB$. Let $AB = x$.

The Law of Cosines states $a^2 = b^2 + c^2 - 2bc \cos A$. In our triangle ABC, we can write it as:

$$AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos B$$

Substitute the known values:

$$850^2 = x^2 + 1200^2 - 2(x)(1200) \cos 43^\circ$$ $$722500 = x^2 + 1440000 - 2400x \cos 43^\circ$$

First, calculate $\cos 43^\circ$:

$$\cos 43^\circ \approx 0.7313537$$

Substitute this value into the equation:

$$722500 = x^2 + 1440000 - 2400x (0.7313537)$$ $$722500 = x^2 + 1440000 - 1755.24888x$$

Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$:

$$x^2 - 1755.24888x + (1440000 - 722500) = 0$$ $$x^2 - 1755.24888x + 717500 = 0$$

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{1755.24888 \pm \sqrt{(-1755.24888)^2 - 4(1)(717500)}}{2(1)}$$ $$x = \frac{1755.24888 \pm \sqrt{3080898.10 - 2870000}}{2}$$ $$x = \frac{1755.24888 \pm \sqrt{210898.10}}{2}$$ $$x = \frac{1755.24888 \pm 459.23643}{2}$$

This gives two possible values for $x$:

$$x_1 = \frac{1755.24888 + 459.23643}{2} = \frac{2214.48531}{2} \approx 1107.24 \text{ m}$$ $$x_2 = \frac{1755.24888 - 459.23643}{2} = \frac{1296.01245}{2} \approx 648.01 \text{ m}$$

This is an ambiguous case (SSA) where two valid triangles can be formed.

• If $AB \approx 1107.24$ m, then $\angle A \approx 74.3^\circ$ (acute).
• If $AB \approx 648.01$ m, then $\angle A \approx 105.7^\circ$ (obtuse). The diagram provided shows $\angle A$ as an acute angle. Therefore, we select the solution that corresponds to an acute $\angle A$.

The distance $AB$ is approximately 1107.2 m.

The final answer is $\boxed{\text{1107.2 m}}$.