Question ABCD is a regular tetrahedron (right triangular pyramid). If M is the midpoint of CD, then what is cos ABM

If M is the midpoint of CD, then the is cos ABM = 8 / (235) sqrt What is regular tetrahedron? The Platonic solid known as the regular tetrahedron has four polyhedron vertices, six polyhedron edges, and four equal equilateral triangular faces. It is frequently referred to as “the” tetrahedron. Also included are the Wenninger model and uniform polyhedron. What is Law of Cosines? 1: According to a rule of trigonometry, the square of a side in a plane triangle is equal to the sum of the squares of the other sides minus twice that much of the product of those sides and the cosine of the angle between them. According to the given information: To make things easier, set the tetrahedron’s side equal to 1. A = ( -1/2 , 0 , 0) B = ( 1/2, 0, 0 ) C = (0, sqrt (3)/2, 0) D = (0, sqrt (3)/4 , 6/sqrt (3) ) M = (0 , (3/8) sqrt (3) , 3/sqrt (3) ) = (0, (3/8)sqrt (3) , sqrt (3) ) AB = 1 AM = BM = sqrt [ (-1/2)^2 + (3sqrt (3) / 8)^2 + (sqrt 3)^2 = sqrt [ 1/4 + 27/64 + 3 ] = sqrt (235) / 8 By the Law of Cosines AM^2 = AB^2 + BM^2 – 2 (AB) (BM)cos (ABM) 0 = 1 – 2 (1) (235 squared / 8) cos ABM 0 = 1 – [sqrt ( 235) / 8] since ABM cos ABM = -1 / – [ sqrt (235) / 8] 8 / (235) sqrt = cos ABM If M is the midpoint of CD, then the is cos ABM = 8 / (235) sqrt To know more about regular tetrahedron visit: https://brainly.com/question/11097227 #SPJ4 Reply

midpointofCD, then the iscos ABM= 8 / (235) sqrt## What is regular tetrahedron?

Platonicsolid known as theregular tetrahedronhas four polyhedron vertices, six polyhedron edges, and four equalequilateraltriangularfaces. It is frequently referred to as “the” tetrahedron. Also included are the Wenninger model and uniform polyhedron.## What is Law of Cosines?

trigonometry, the square of a side in a plane triangle is equal to the sum of the squares of the other sides minus twice that much of the product of those sides and the cosine of the angle between them.## According to the given information:

midpointof CD, then the iscos ABM = 8 / (235)sqrtregular tetrahedronvisit:#SPJ4