Question

ABCD is a rhombus. The diagonals intersect at point E. If angle CDB= 6x and angle ACD= 2x+10, find x.

Answer 1

x =
`10^(o)`

Explanation:

The diagonals of a rhombus are perpendicular to each other, thus the angle at point E is
`90^(o)`. So that ΔCED is a right triangle, and that;

<CDB = <CDE

<ACD = <ECD

<CDE + <ACD + <CED =
`180^(o)` (sum of angles in a triangle)

6x + (2x + 10) +
`90^(o)` =
`180^(o)`

8x + 100 =
`180^(o)`

8x =
`180^(o)` - 100

8x = 80

x =
`(80)/(8)`

=
`10^(o)`

Thus, the value of x is
`10^(o)`.