Question

1. how wide is this bay?
2. on june 30, 1859, jean françois gravelet crossed the niagara gorge on a tightrope. since he could not measure the distance across the gorge directly to determine the length of rope he would need, he used indirect measurement.

a) explain why $\triangle dec$ is similar to $\triangle abc$.
b) what ratio might gravelet have used to determine the scale factor of the two triangles?
c) calculate the distance across the gorge.

Explanation:

Part 5 (How wide is this bay?):

Step1: Identify Similar Triangles

$\triangle ABC$ and $\triangle EDC$ are right triangles ( $\angle B = \angle D = 90^\circ$ ) and share $\angle ACB = \angle ECD$ (vertical angles). By AA (Angle - Angle) similarity criterion, $\triangle ABC \sim \triangle EDC$.

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. So, $\frac{AB}{ED}=\frac{BC}{DC}$. Let $AB = x$ (width of the bay), $ED = 15.0\ m$, $BC = 30.0\ m$, $DC = 18.0\ m$.

Step3: Solve for $x$

Substitute the values into the proportion: $\frac{x}{15.0}=\frac{30.0}{18.0}$. Cross - multiply: $18.0x=15.0\times30.0$. Then $x = \frac{15.0\times30.0}{18.0}=\frac{450}{18}=25.0\ m$.

$\angle ABC$ and $\angle DEC$ are right angles (so $\angle ABC=\angle DEC = 90^\circ$). Also, $\angle ACB$ and $\angle DCE$ are vertical angles, so $\angle ACB=\angle DCE$. By the AA (Angle - Angle) similarity postulate, if two angles of one triangle are congruent to two angles of another triangle, the triangles are similar. Thus, $\triangle DEC\sim\triangle ABC$.

The scale factor of similar triangles is the ratio of corresponding sides. For $\triangle DEC$ and $\triangle ABC$, corresponding sides are $DE$ and $AB$, $EC$ and $BC$, or $DC$ and $AC$. A possible ratio is $\frac{DE}{AB}$ or $\frac{EC}{BC}$ or $\frac{DC}{AC}$. Since we know $DE = 26\ m$, $EC = 37\ m$, $BC = 740\ m$ (from the diagram), a practical ratio could be $\frac{EC}{BC}=\frac{37}{740}$ or $\frac{DE}{AB}$ (once $AB$ is found) or $\frac{DC}{AC}$ (but $DC$ and $AC$ are not directly given with numbers in the problem's diagram for part 6, but $DE$ and $EC$ are). The ratio of corresponding legs (e.g., $\frac{DE}{AB}$ or $\frac{EC}{BC}$) can be used. A likely ratio is $\frac{EC}{BC}=\frac{37}{740}$ (simplifies to $\frac{1}{20}$) or $\frac{DE}{AB}$ (when $AB$ is calculated).

Answer:

The width of the bay is $25.0\ m$.

Part 6:

a) Explain why $\triangle DEC$ is similar to $\triangle ABC$: