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

19.

step-by-step explanation:

you add the 9 in the tens column since it is a single digit and 9 plus zero is 9 so it is 19

correct answers below: