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- Homework Statement
- Two straight roads, which cross one another, meet a river with straight course at angles ##60^{\circ}## and ##30^{\circ}##, respectively. If it be 3 miles by the longer of the two roads, from the crossing to the river, how far is it by the ##\text{shortest}## (or the shorter of the two?)? If there be a foot-path which goes the shortest way from the crossing to the river, what is the distance by it?

- Relevant Equations
- 1. In a right-angled triangle, ##\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}##.

2. For a triangle ##\text{ABC}##, ##\frac{a}{\sin A} = \frac{b}{\sin B}##. (The Law of Sines)

**Problem statement :**I copy and paste the problem as it appears in the text. I hope am understanding its wording correctly.

**The river is shown in blue. The two roads start from the crossing C and end in A and B, making angles ##60^{\circ}## and ##30^{\circ}## resectively. The longer of the two roads CB is given to be 3 miles. Question is, how long is the shorter of the two roads AC, assuming my understanding of the problem is correct. Additionally, how long is the shortest way from C to the river, that is CD, where CD is ##\perp^r## to BA produced.
Diagram : **

**Solution :**Using the law of sines (see Relevant Equations above), ##\text{AC} = \text{BC}\times \frac{\sin 30^{\circ}}{\sin 60^{\circ}} = 3\times \frac{1}{2}\times \frac{2}{\sqrt 3} = \sqrt 3 = \boxed{1.73\,\, \text{miles}}##.

The distance CD = ##BC\, \sin 30^{\circ} = 3\times\frac{1}{2} = \boxed{1.5\,\text{miles}}##.

**Issue :**The answers don't match with those in the text. I copy and paste the text answers below.

A hint or help would be welcome. I am perplexed at the moment, given the simplicity of the problem. Or am I deceiving myself?