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

Answer:

a. 41√3 unit

b. 82 unit

Step-by-step 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,

$\frac{sin A}{BC}=\frac{sin C}{BA}$

By substituting the value,

$\frac{sin 30^{\circ}}{41}=\frac{sin 60^{\circ}}{BA}$

$\frac{1}{82}=\frac{\sqrt{3}}{2BA}$

$\implies BA=\frac{82\sqrt{3}}{2}=41\sqrt{3}\text{ unit}$

b. AC is the hypotenuse of the triangle ABC,

By the pythagoras theorem,

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