H = 68
g = 112
k = 82
m = 98
Angle A of triangle ABC will be ...
∠A = θ₂ -θ₃ = 77° -26° = 51°
Angle B of triangle ABC will be ...
∠B = 180° -θ₂ -θ₁ = 180° -77° -44° = 59°
Then angle C will be ...
∠C = 180° -∠A -∠B = 180° -51° -59° = 70°
By the Law of Sines,
a/sin(∠A) = c/sin(∠C)
a = c×sin(∠A)/sin(∠C)
a = (57 yd)×sin(51°)/sin(70°) ≈ 47.14 yd
The distance from B to C is 47.14 yd.
I would say B in all honesty. Others probably would want A
Answer:
15
Step-by-step explanation:
Note: Take note , absolute value will give the magnitude of the value. (Which means ignoring the + / - signs.)
Eg. | - 19 | = 19
Therefore,
Step-by-step explanation:
AB=BC
8x+6=6x+18
2x=12
x=6
so
AB= 48+6=54cm
BC=36+18=54cm
AD=CD
8y+20=12y-8
-4y=-28
y=7
so
AD= 84-8=76cm
CD=56+20=76cm
AE=2x+4y
AE=2×6+4×7
AE=40cm
AC=2×40=80cm
