0 0 A man wants to build a triangular lot. What would Angle A, B, and C? (Side lengths in description)

Question

P.s sorry can't draw the triangle it would have made it easier.
AB is 211.
BC is 142.
AC is 185.
Thanks :)

Answer 1

 

Since you have the measures for all the sides, You would use the law of cosines to find any 1 of the 3 angles.

Note that in the typical diagram, the angles are named using upper case letters, and the sides opposite those angles will have lower case letters.

 

let a = 142, b = 185, c = 211

 

Law of Cosines:  1 of 3 formulas

 

c² = a²  + b²  - 2ab (Cos C) 

 

Plug in the side measures ... then solve for Cos C

 

211² = 142² + 185² - 2(142)(185)(Cos C)

44,521 = 20,164 + 34,225 -52,540(Cos C)

44,521 = 54,389 - 52,540(Cos C)

 

Subtract 54,389 from both sides

 

-9,868 = -522,540(Cos C)

 

Divide both sides by -522,540

 

,01888468 = Cos C

 

Use the inverse trig function (in degree mode) to find the measure of angle A

 

Cos^-1 .01888468 = 88.92 degrees

 

*** Angle C is 88.92 degrees

 

Now use the law of sines to find a second angle

 

(Sine C)/c = (Sine B)/b

 

(Sine 88.92)/211 = (Sine B)/185

 

Solve for the Sine B by multiplying both sides by 185

 

185(.99982235)/211 = Sine B

.8766214917 = Sine B

 

Use inverse Sine function (in degree mode) to find the measure of angle A

 

sine (.8766214817) = A

 

*** Angle  A = 61.24

 

Since we have 2 of the angles, we subtract their sum from 180 to get the third angle ... The sum of the internal angles of all triangles is 180 degrees

 

Angle B = 180 - (61.24 + 88.92)

 

*** Angle B = 29.84