Answer:
It's Ok
Step-by-step explanation:
Mark me brainliest :>
Answer:
The distance from A to B is 736.2 to the nearest tenth foot
Step-by-step explanation:
In ΔCAB
∵ m∠CAD = 30° ⇒ exterior angle of Δ at vertex A
∴ m∠CAD = m∠ACB + m∠ABC
∵ m∠ABC = 20°
∴ m∠ACB = 30° - 20° = 10°
We will use the sin rule to find the distance AB
∵ 
∴ 
≅ 736.2 to the nearest tenth foot
Answer:
26.2 units
Step-by-step explanation:
We are given the points/vertices
A(6, 3),
B(6, - 2) , and
C(- 4, 3)
Step two
Let us find the distances between the given points/vertices
A-B =A(6, 3) to B(6,-2)
d=√((x2-x1)²+(y2-y1)²)
Substitute
d=√((6-6)²+(-2-3)²)
d=√(-2-3)²)
d=√(-5)²)
d=5 units
B-C=B(6, - 2) to C(-4, 3)
d=√((x2-x1)²+(y2-y1)²)
Substitute
d=√((-4-6)²+(3+2)²)
d=√(-10)²+(5)²)
d=√100+25
d=√125
d=11.2 units
C-A=C(-4, 3) to A(6, 3)
d=√((x2-x1)²+(y2-y1)²)
Substitute
d=√((6+4)²+(3-3)²)
d=√(10)²
d=√100
d=10 units
Hence the perimeter is 5+11.2+10
P=26.2 units
