mcr3u trigonometry assignment t /29\n4. jason is standing 8.7 km from town x and 11.5 km from town y. from…

Explanation:

Step1: Use the Law of Cosines to find the distance between town X and town Y ($d$)

The Law of Cosines formula is $d^{2}=x^{2}+y^{2}-2xy\cos Z$, where $x = 11.5$, $y=8.7$, and $Z = 37^{\circ}$. $$ \begin{align*} d^{2}&=11.5^{2}+8.7^{2}-2\times11.5\times8.7\times\cos(37^{\circ})\ d^{2}&=132.25 + 75.69-2\times11.5\times8.7\times0.7986\ d^{2}&=207.94-(200.1\times0.7986)\ d^{2}&=207.94 - 159.79\ d^{2}&=48.15\ d&=\sqrt{48.15}\approx6.94\mathrm{km} \end{align*} $$

Step2: Find the distance from Jason to the mid - point of the road ($D$)

Since the hotel is at the mid - point, $D=\frac{d}{2}$. Using the formula for the length of the median in a triangle (or we can also use the Law of Cosines in a more complex way, but an alternative is: Let the two sides be $a = 11.5$, $b = 8.7$, and the included angle $C = 37^{\circ}$. The formula for the length of the median $m$ is $m=\frac{1}{2}\sqrt{2a^{2}+2b^{2}-d^{2}}$. Since we already found $d^{2}\approx48.15$, $a = 11.5$, $b = 8.7$ $$ \begin{align*} m&=\frac{1}{2}\sqrt{2\times11.5^{2}+2\times8.7^{2}-48.15}\ &=\frac{1}{2}\sqrt{2\times132.25+2\times75.69 - 48.15}\ &=\frac{1}{2}\sqrt{264.5+151.38 - 48.15}\ &=\frac{1}{2}\sqrt{367.73}\ &\approx\frac{1}{2}\times19.18\approx9.59\mathrm{km} \end{align*} $$ Another way (using the Law of Cosines in two sub - triangles): Let the two sides be $x = 11.5$, $y = 8.7$, included angle $Z=37^{\circ}$. Let the median be $m$. We know that $\cos A=\frac{y^{2}+d^{2}-x^{2}}{2yd}$ (from $\triangle XYZ$) and also in the sub - triangle with sides $y$, $m$, $\frac{d}{2}$: $\cos A=\frac{y^{2}+(\frac{d}{2})^{2}-m^{2}}{2y\times\frac{d}{2}}$. But since we already have $d\approx6.94$, the first method (median formula) is more straightforward.

Step3: Use the tangent function to find the height of the hotel ($h$)

We know that $\tan\theta=\frac{h}{D}$, where $\theta = 1^{\circ}$ and $D\approx9.59\mathrm{km}=9590\mathrm{m}$ $$h = D\times\tan(1^{\circ})$$ $$h=9590\times\tan(1^{\circ})\approx9590\times0.017455\approx167.5\mathrm{m}$$

Answer:

The height of the hotel is approximately $167.5$ metres.