Answer:
Option A.
The closest area of triangle ABC is [tex]17\ units^{2}[/tex]
Step-by-step explanation:
In this problem
If sin(a)=cos(b)
then
Angles a and b are complementary
so
[tex]a+b=90\°[/tex]-----> equation A
therefore
The triangle ABC is a right triangle
step 1
Find the value of x
substitute the given values of a and b in the equation A
[tex](2x-15)\°+(5x-21)\°=90\°\\(7x-36)\°=90\°\\ 7x=90+36\\ x=18\°\\ a=(2x-15)\°=2(18)-15=21\°\\ b=(5x-21)\°=5(18)-21=69\°[/tex]
step 2
Find the length of side AC
we know that
In the right triangle ABC
[tex]cos(a)=AC/BC\\ AC=(BC)cos(a)[/tex]
substitute the given values
[tex]AC=(10)cos(21\°)=9.34\ units[/tex]
step 3
Find the length of side AB
we know that
In the right triangle ABC
[tex]sin(a)=AB/BC\\ AB=(BC)sin(a)[/tex]
substitute the given values
[tex]AB=(10)sin(21\°)=3.58\ units[/tex]
step 4
Find the area of triangle ABC
The area is equal to
[tex]A=(1/2)(AB)(AC)[/tex]
substitute
[tex]A=(1/2)(3.58)(9.34)=16.7\ units^{2}[/tex]
therefore
The closest area of triangle ABC is [tex]17\ units^{2}[/tex]
