Someone help me pls, will give brainiest and high points

Step-by-step explanation:

Using sine rule on triangle ABC as re - shown in the attachment below to get side AC first we have;

∠A = 34°, BC = 164cm, ∠B = 27°, AC = ?

AC/sin 27° = BC/sin 34°

AC/sin 27° = 164/sin 34°

AC = 164 * sin 27° / sin 34°

AC = 74.45/0.5592

AC = 133.14 cm

To get the side 'a', we will use the trig ratio SOH CAH TOA on the right angled ACD.

According to SOH,

sin 61° = a/AC

a = ACsin61°

a = 133.14 * sin 61°

a = 116.45

a ≈ 117 cm

The marked side is approximately 117 cm

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!