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°

this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

hope this !

~just a girl in love with shawn mendes

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.