Bird's Eye View (Mathathon Problem 7)

First of all, let's give some names to the known values!

Let $x_e$ be the position of the elephant on the ground and let $\epsilon=\frac{\pi}{3}$ be the elephant's angle of depression from the bird's eye view.

Let $x_l$ be the position of the lion on the ground and let $\lambda=\frac{\pi}{4}$ be the lion's angle of depression from the bird's eye view.

What we want to find is the distance $d=x_l-x_e$

Now we can make a drawing of the whole situation (not to scale).

One nice thing immediately springs to attention. The diagonal of a square has angle $\frac{\pi}{4}$ with respect to the horizontal, just like our lion! Since the sides of a square have equal length and the bird's location is perpendicular to the ground, we can immediately conclude that $x_l$ must also equal $45m$ .

Now we still need $x_e$ . It isn't too hard to see that the angle of depression of the elephant plus some angle must equal $\frac{\pi}{2}$ , let's call that angle $\overline{\epsilon}$ , so that we get $\epsilon + \overline{\epsilon} = \frac{\pi}{2} \iff \overline{\epsilon} = \frac{\pi}{2} -\epsilon .$ What's left is to note that we know the angle $\overline{\epsilon}$ , the height of the bird and want to know the length of the other leg of the right triangle which the three are part of - this just begs for some trigonometry! So let's take the tangent of $\overline\epsilon$ : $\tan\overline\epsilon = \frac{x_e}{45} \iff x_e = 45\cdot \tan\overline\epsilon$ That was already the hardest part, but further simplification can make the problem even easier. It can be proven that $\tan\overline\epsilon =\tan\left[\frac{\pi}{2}-\epsilon\right] = \cot(\epsilon) = \cot\left[\frac{\pi}{3}\right] = \frac{\sqrt{3}}{3}.$ All that's left to do is plugging in these values! $\begin{aligned} d = x_l -x_e &= 45 - x_e\\ d&= 45 - \frac{45\cdot\sqrt{3}}{3}\\[10pt] d &=45 - 15\sqrt{3} \end{aligned}$

$\frac{sin(A)}{a}$ = $\frac{sin(B)}{b}$ So, we start off with $\frac{sin(60)}{45}$ = $\frac{sin(90)}{x}$ We then cross multiply to get $45(sin (90))$ = $x(sin (60))$ $45=x(0.86602540378)$ So $x=51.961$ . We then repeat the process to obtain $\frac{sin(45)}{51.961}$ = $\frac{sin(15)}{y}$ . We again cross-multiply to obtain $51.961((\sqrt{6}-\sqrt{2}$ )/ $4)$ = y\(\frac{1}{\sqrt{2} We finally obtain by rearranging , \frac{y}{\sqrt{2} to get $y=19.019=45-15(\sqrt{3}$ )

In the graph above, we can see that the triangle that the elephant is a vertex of is a 30-60-90 triangle. Since the 45 meter side is the long leg, the shorter leg of a 30-60-90 triangle is always the $\text{longer leg of the triangle}\times \frac{\sqrt(3)}{3}$ , so the shorter side of the 30-60-90 triangle, which is the distance from the bird to the elephant, is, in this case, $15\sqrt(3)$ . The bigger triangle is a 45-45-90 triangle, and because the leg lengths are always the same in a 45-45-90 triangle, the distance from the tree to the lion is 45 meters. Taking the difference between out two distances, we get $\boxed{45-15\sqrt(3)}$ , which is our final answer.

Bird witnessed a crime

Let $\textcolor{#EC7300}{B1}$ be the distance from tree to 🦁

And $\textcolor{#0C6AC7}{B2}$ be the distance from tree to 🐘

$\textcolor{#EC7300}{tan45°}$ = $\frac{45}{B1}$

$\textcolor{#0C6AC7}{tan60°}$ = $\frac{45}{B2}$

$\textcolor{#EC7300}{tan45°}$ =1

$\textcolor{#0C6AC7}{tan60°}$ = $\frac {√3}{2}$

Hence $\textcolor{#EC7300}{B1}$ = $45*1$ and $\textcolor{#0C6AC7}{B2}$ = $\frac{45}{√3}$

Note:we can rationalise $\frac{45}{√3}$ * $\frac{√3}{√3}$ to get $15*√3$

To find the distance between the 🦁 and the 🐘: Subtract $\textcolor{#EC7300}{B1}$ and $\textcolor{#0C6AC7}{B2}$

Subtracting the two we get,

$\textcolor{limegreen}{ 45-15√3}$

Hence option $\boxed{\textcolor{#69047E}{B}}$ is correct!

$α = 30° (90° - 45° - 15°)$

$\text{tan θ} = \dfrac{\text{Opposite}}{\text{Adjacent}}$

$Y = 45 \text{ tan }α = 45 \text{ tan 30°} = 45 \cdot \dfrac{1}{\sqrt{3}} = \dfrac{15 \cdot 3}{\sqrt{3}} = 15 \sqrt{3}$

$Z = 45 \text{ tan }(α+15°) = 45 \text{ tan } (30° + 15°) = 45 \text{ tan 45°} = 45 \cdot 1 = 45$

$X = Z - Y = \boxed{45 - 15 \sqrt{3}}$

birds eye view

Look at the picture. It is given that $\angle ABE = 45\degree \text{(the yellow angle)}$ and $\angle ABD = 60\degree \text{(the cinnamon angle)}$

From this we get $\angle BEC = 45\degree$ and $\angle BDC = 60\degree \hspace{5 cm} \because AB \parallel CF \text{and they are alternate interior angles}$

We need to find $EC - DC$

BEC and BDC are two triangles right-angled at C and we know BC = 45m and also the measures of angles BEC and BDC.

We can use trigonometry for this. We will use $tan \theta = \dfrac{opposite}{hypotenuse}$ to get EC and DC.

For $\triangle BEC$ :

$tan 45\degree = \dfrac{BC}{EC}$

$EC = \dfrac{BC}{tan 45\degree}$

$EC = \dfrac{45}{1} = 45m$

$$

For $\triangle BDC$ :

$tan 60\degree = \dfrac{BC}{DC}$

$DC = \dfrac{BC}{tan 60\degree}$

$DC = \dfrac{45}{\sqrt{3}} = \dfrac{45\sqrt{3}}{3} = 15\sqrt{3}m$

Now $EC - DC = 45m - 15\sqrt{3}m$

Therefore, the distance between the lion and the elephant = $\boxed{45m - 15\sqrt{3}m}$

This is how I solved it. For the 45 degree angle, it creates an isosceles triangle. The leg along the ground is the same length as the vertical leg, so the lion is 45 meters away.

For the elephant, it is half of an equilateral right triangle, so the leg along the ground is exactly half of the straight line distance between the bird and the elephant. Let $x$ be the distance to the elephant. Using the pythagorian theorem:

• ${45}^2 + x^2 = (2x)^2$
• ${45}^2 + x^2 = 4x^2$
• ${45}^2 = 3x^2$
• $x = \frac{45}{\sqrt{3}}$
• The answer is $45 (lion) - \frac{45}{\sqrt{3}}$
• $45 - 15 \sqrt{3}$

Now we need to find DB - DC = BC

in $\Delta$ ABD, by using sin(45) = $\frac{AD}{AB}$ , we get

$\begin{aligned} \frac{1}{\sqrt{2}}&= \frac{AD}{AB}\\ AB&= AD×\sqrt{2}&= 45\sqrt{2} m \end{aligned}$

The above image indicates the positions of lion and elephant

Now by applying sine rule in $\Delta$ ABC, we get

$\begin{aligned} \frac{BC}{Sin(15)}&= \frac{AB}{Sin(120)}\\ BC&= AB × \frac{Sin(15)}{Sin(120)}\\ BC&= AB × \frac{Sin(15)}{Sin(60)} \Rightarrow using& Sin(180-\theta)&= Sin(\theta) \end{aligned}$

Now by substituting the values we get
$\begin{aligned} BC&= 45\sqrt{2} × \normalsize\frac{\frac{\sqrt{3}-1}{2\sqrt{2}}}{\frac{\sqrt{3}}{2}}\\ &= 45\frac{\sqrt{3}-1}{\sqrt{3}}\\ &= 45 - 15\sqrt{3} \end{aligned}$

Note:

• Sin(45) = $\frac{1}{\sqrt{2}}$
• Sin(15) = $\frac{\sqrt{3}-1}{2\sqrt{2}}$
• Sin(60) = $\frac{\sqrt{3}}{2}$

The Sine Rule

Let the Point where the bird is sitting be B, where the lion is be L, where the elephant is be E.

Simple calculations tell us that $\angle {ABE} = 30^{\circ}$ & $\angle {ABL} = 45^{\circ}$

In $\triangle ABL$ , $\angle {ABL} = 45^{\circ}$ & $\angle {A} = 90^{\circ}$ . So $\angle {ALB} = 45^{\circ}$ . Hence $\triangle ABL$ is an isosceles triangle.

Hence , $\text {AL = AB = 45m}$

We have to find EL = AL - AE = 45 - AE. So EL is less than 45.

Now we can find the answer by eliminating the options (consider $\sqrt {3} = 1.7$ )

Option 1 : $60 - 45 \sqrt{3} = 60-45\cdot 1.7 = -16.5$ As Distance can't be negative $\color{#D61F06}{option~1~is~eliminated}.$

Option 2 : $45 - 15 \sqrt {3} = 45 - 15 \cdot 1.7 = 19.5$ . Cannot be eliminated

Option 3 : $45 + 15 \sqrt{3}$ As d(EL) has to be less than d(AL) = 45 $\color{#D61F06}{option~3~is~eliminated}.$

Option 4 : $45 \sqrt{3} - 15 = 45*1.7 - 15 = 61.5$ $\color{#D61F06}{option~4~is~eliminated}.$ (same reason as for option 3)

Only remaining Option 2 = $45 - \sqrt{3}$ is the correct answer.

$Let\space the\space horizontal\space distance\space between\space the\space \space bird\space and$ $the\space lion\space be\space l\space and\space between\space the\space bird\space and\space the\space elephant\space be\space e$ $in\space \yellow{yellow\space\triangle}$ $\tan \red{60\degree}=\dfrac{45m}{e}=\sqrt{3}$ $\Rightarrow e=\dfrac{45m\red{\times\sqrt{3}}}{\sqrt{3}\red{\times\sqrt{3}}}=\dfrac{45\sqrt{3}m}{3}=15\sqrt{3}m$ $in\space \green{green\space\triangle}$ $\tan \blue{45\degree}=\dfrac{45m}{l}=1$ $\Rightarrow l=45m$ $Distance\space between\space the\space bird\space and\space the\space elephant = l-e=(45-15\sqrt{3})\red{m}$

Using basics of trigonometry, $45 \tan(45^{\circ}) - 45 \tan(30^{\circ}) =45 - \dfrac{45}{\sqrt3}=\boxed{45-15\sqrt3}$