From ground level, a shooter is aiming at targets on a vertical brick wall. How far is the shooter from the wall?

Question

From ground level, a shooter is aiming at targets on a vertical brick wall. At the current angle of elevation of his rifle, he will hit a target 20 m above ground level. If he doubles the angle of elevation of the rifle, he will hit a target 45 m above ground level. How far is the shooter from the wall?

My current working out is as follows:

• tanθ = 20/x
• tan2θ = 45/x
• x = 20/tanθ
• x = 45/tan2θ

20/tanθ = 45/tan2θ

20/tanθ = 45/tanθ+tanθ/1-tanθ.tanθ

20/tanθ = 45/2tanθ/1-tanθ^2

My calculations could continue on from here, but I don't believe I am going down the right track as this website (which has had the working out deleted) suggests another process: http://mymathforum.com/trigonometry/28959-trig-againn.html

The answer to the calculation is 60m.

Answer 1

Here is a way without using trigonometry. The angle bisector theorem states that $\frac{BD}{DC} = \frac{AB}{AC}$. Let $AC$ be $x$. Here, $DC = 20$, $BD = 45 -20 = 25$, and then $AB = \sqrt{45^2 + x^2}$. Therefore, we have:

$$\frac{25}{20} = \frac{\sqrt{45^2 + x^2}}{x}$$ $$\frac{5}{4} = \frac{\sqrt{45^2 + x^2}}{x}$$

Cross-multiplying, we have:

$$5x = 4\sqrt{45^2 + x^2}$$

Squaring, we have:

$$25x^2 = 16 \cdot 45^2 + 16x^2$$ $$9x^2 = 32400$$ $$x^2 = 3600$$ $$x = +60, -60$$

However, since $x$ cannot be negative, the only solution is $x = 60$.

Answer 2

Everything you have done so far is correct:

$$\frac{20}{\tan \theta} = \frac{45}{(2 \tan \theta) / (1 - \tan^2 \theta)}$$

Now you can cross-multiply:

$$\frac{40 \tan \theta}{1 - \tan^2 \theta} = 45 \tan \theta$$ $$40 \tan \theta = 45 \tan \theta - 45 \tan^3 \theta$$ $$5 \tan \theta = \color{red}{+}45 \tan^3 \theta$$ $$\tan^2 \theta = \frac{1}{9} (\tan \theta \ne 0)$$

Can you continue? Not all of the solutions are valid.