Solve the following triangles:
a = 21 degrees a = 10 b = 6
a = 48 degrees a = 32 b = 6

find the area and perimeter of the two trianlgles formed in question 1.

1.1) B=12.42°, C=146.58°, c=15.37  

1.2) B=8°, C=124°, c=35.7  

2.1) Perimeter = 31.37 units, Area = 16.5 square units  

2.2) Perimeter = 73.70 units, Area = 75.6 square units  

Step-by-step explanation:

2.1) Perimeter: P = a+b+c → P = 10+6+15.37 → P = 31.37  

Area: A = a b sinC / 2

Replacing the known values:

A = (10)(6) sin 146.58325418° / 2 → A = 60(0.55072472) / 2 →

A = 16.52174152 → A = 16.52 square units

2.2) Perimeter: P = a+b+c → P = 32+6+35.70 → P=73.70

Area: A = a b sin C /2

Replacing the known values:

A = (32)(6) sin 123.99036327° / 2 → A = 192(0.82913161) / 2 →

A = 75.59663485 → A = 75.6 square units

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

150.6

all i did was i multiplied 5.02 by 30