Question

A building lot in a city is shaped as a 30°-60°-90° triangle, like the figure shown. The side opposite the 30° angle measures 41 feet. a. Find the length of the side of the lot opposite the 60° angle. Show how you know. b. Find the length of the hypotenuse of the triangular lot. Show how you know.

Answer 1

You haven't shared the figure mentioned. However, we can still solve this problem.

If you have a 30-60-90 triangle, then the sides opposite these angles have basic lengths 1, sqrt(3) and 2. In other words, the side with length 2 is the hypotenuse and is opposite the right angle (90 deg).

Let's apply the Law of Sines:

a b c
------- = --------- = -----------
sin A sin B sin C

If a = 41 ft is opposite the 30 degre angle, then

41 ft x
---------- = -----------
sin 30 sin 60

Then x(sin 30) = (41 ft)(sin 60), or
41sqrt(3)
x = --------------- = 82sqrt(3)
0.5

The hypotenuse could be found using a similar approach:

41 ft x
---------- = -----------
sin 30 sin 90

In this case x will represent the length of the hypotenuse.

Answer 2

a. 41√3 unit

b. 82 unit

Explanation:

Let ABC is a triangle,

In which m∠B = 90°,

m∠A= 30°, m∠C = 60°,

Also, BC = 41 feet,

a. By the law of sine,

`(sin A)/(BC)=(sin C)/(BA)`

By substituting the value,

`(sin 30^(\circ))/(41)=(sin 60^(\circ))/(BA)`

`(1)/(82)=(√(3))/(2BA)`

`\implies BA=(82√(3))/(2)=41√(3)\text{ unit}`

b. AC is the hypotenuse of the triangle ABC,

By the pythagoras theorem,

`AC=√(BC^2+BA^2)=\sqrt{41^2+(41√(3))^2}=√(1681+5043)=82\text{ unit}`