Question

А(1; 1), В(4; 1) және С(4; 5) нүктелері ABC үшбұрышының төбелері болса, онда

cos∠A табыңыз.
Срочно көмек керек
15 бал беремін​

Answer 1

Given :  А(1; 1), В(4; 1) and С(4; 5)  are vertices of Triangle

To Find :  cos∠A

Solution:

А(1; 1), В(4; 1) and С(4; 5)

Slope of AB =  0

Slope of BC  = 1/0

Hence AB ⊥ BC

=> ABC is right angled triangle at B

Cos ∠A    =  AB/ AC

AB  =  3

BC  = 4

AC = √(4 -1)² + (5 - 1)²  = 5

Cos ∠A    =  3/5

Cos ∠A    =  3/5

Learn More:

find the area of triangle having vertices at (8,1)

https://brainly.in/question/12350448

If two vertices of an isosceles right triangle areA(–1, –7) and B(–7 ...

https://brainly.in/question/22705923