Question

abc is an equilateral triangle with base bc on x axis and the vertex b is the origin. find the co ordinates of the vertices of the triangle abc if the sides are equal to 5 cm.​

Answer 1

Answer:

A(12/5,root of 481/5)B(0,0)C(5,0)

Step-by-step explanation:

Since Point B is on origin it's coordinates are (0,0),BC is on x axis the coordinates of point C are (5,0)

Let coordinates of A be (x,y)

AB=

$\sqrt{x { }^{2} + y {}^{2} }$

AC=

$5 = \sqrt{(5 - x) ^{2} + (0 - y) {}^{2} } \\ 5 = \sqrt{25 + x {}^{2} - 10x + y ^{2} } \\ 25 = 25 + x {}^{2} + y {}^{2} - 10x \\ 1 = 25 - 10x \\ x = \frac{24 }{10}$

further you can get Y coordinate