# [SOLVED] Sine Question

#### RogueDemon

On a sunny day, a tower casts a shadow 35.2m long. At the same time, a 1.3m parking meter that is nearby casts a shadow 1.8m long. How high is the tower to the nearest tenth of a metre?

Triangle formed by parking meter:
$$\displaystyle c^2 = a^2 + b^2$$
$$\displaystyle c^2 = 1.3^2 + 1.8^2$$
$$\displaystyle c^2 = \frac{493}{100}$$
$$\displaystyle c = \frac{\sqrt{493}}{10}$$

$$\displaystyle SinA = \frac{O}{H}$$
$$\displaystyle SinA = \frac{1.3}{\frac{\sqrt{493}}{10}}$$
$$\displaystyle Sin^{-1}\frac{1.3}{\frac{\sqrt{493}}{10}} = A$$
$$\displaystyle 35.84^{\circ} = A$$

Triangle formed by tower (Total triangle):
$$\displaystyle \frac{a}{SinA} = \frac{b}{SinB}$$
$$\displaystyle \frac{a}{Sin35.84} = \frac{37}{Sin54.16}$$
$$\displaystyle a = Sin35.84(\frac{37}{Sin54.16})$$
$$\displaystyle a = 26.7m$$

*The $$\displaystyle 37m$$ value was obtained by adding $$\displaystyle 35.2m$$ and $$\displaystyle 1.8m$$.
*The $$\displaystyle 54.16^{\circ}$$ value was obtained by subtracting $$\displaystyle 90deg$$ and $$\displaystyle 35.84deg$$ from $$\displaystyle 180deg$$.

Therefore, the height of the tower is $$\displaystyle 26.7m$$.

Textbook answer: $$\displaystyle 25.4m$$.

Where did I make the mistake?

#### masters

MHF Helper
Hi RogueDemon,

Actually, you made this way harder than it is.

You have two similar right triangles.

Set up a proportion using the ratios of the heights and shadows.

$$\displaystyle \frac{bldg \:\: ht.}{1.3}=\frac{35.2}{1.8}$$

• RogueDemon

#### RogueDemon

Thanks very much!