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

ABCD is a rhombus. The diagonals intersect at point

ainhoagmz903

13.06.2021

x = $10^{o}$

Step-by-step 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 = $\frac{80}{8}$

= $10^{o}$

Thus, the value of x is $10^{o}$ .

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19thomasar

20.05.2022

-cosx+c is your answer

Step-by-step explanation:

$\displaystyle \int \: \tan(x) \cos(x) \: dx$

$\displaystyle \int \: \sin \: x \div \cos(x) \times \cos(x) dx$

$\displaystyle \int \: sinx \: dx \\ - \cos(x) + c$

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