Find the area of each triangle.

Question

b=50in, c=75in, C=15degrees

Answer 1

Let's call the remaining side "a" and the remaining vertex "A".

 

There are two triangles with the given description. To find their areas do the following.

 

1. Law of sines.

 

(sin B)/b = (sin C)/c

 

(sin B)/50 = (sin 15°)/75 = (√6 - √2)/300

 

sin B = (√6 - √2)/6

 

2. Altitude dropped from vertex A

 

h = b sin B = 50 sin 15° = 25(√6 - √2)/2

 

This is a common leg of the two right triangles. Drawing a picture will help here.

 

3. The bases and areas of the right triangles

 

The bigger one has 50 cos 15° = 25(√6 + √2)/2 for the other leg for an area

 

A_big = (1/2)[25(√6 + √2)/2][25(√6 - √2)/2] = 625/2

 

The smaller one has a base

 

h·cot B = h(sin B)/√(1-sin² B)

 

= [25(√6 - √2)/2][(√6 - √2)/6]/√[1-[(√6 - √2)/6]²]

 

= 25√(7+√3)   ---- Simplified using Wolfram-Alpha online calculator.

 

and area

 

A_small = (1/2)[25√(7+√3)][25(√6 - √2)/2] = (625/4)(√6 - √2)√(7+√3)

 

4. Final answer

 

The two triangles with the description in your problem have areas

 

A_big ± A_small = (625/2) [2±(√6 - √2)√(7+√3)]