Atriangular lot bounded by three streets has a length of 300 feet on one street, 250 feet on the second, and 420 feet on the third. find the measure of the largest angle formed by these streets. which of the following equations can be used to solve the problem? 2502 = 3002 + 4202 - (2)(300)(420)cosx 3002 = 2502 + 4202 - (2)(250)(420)cosx 4202 = 3002 + 2502 - (2)(300)(250)cos

420^2 = 300^2 + 250^2 - 2(300)(250)cosC

Step-by-step explanation:

a = 300

b= 250

c = 420

C= ??

Using Cosine rule

Cos C = (a^2 + b^2 - c^2) / 2ab

Cos C = (300^2 + 250^2 - 420^2) / 2*300*250

= (90000 + 62500 - 176400) / 150000

= -23900/150000

= -0.159

C = cos^-1(-0.159)

C = 80.85°

To find angle A

Cos A = (b^2 + c^2 - a^2) / 2bc

Cos A = (250^2 + 420^2 - 300^2) / 2*250*420

= (62500 + 176400 - 90000) / 210000

= 0.71

A = cos^-1(0.71)

A = 44.77°

A+B+C = 180°

B = 180 - (A+C)

B = 180 + (80.85 + 44.77)

B = 54.38°

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180