Find the fifth roots of 32(cos 280° + i sin 280°)

The fifth root is 2[cos(56°) + i sin(56°)]

Step-by-step explanation:

* To solve this problem we must revise De Moiver's rule

- In the complex number with polar form

∵ z = r(cosФ + i sinФ)

∴ z^n = r^n(cos(nФ) + i sin(nФ))

* In the problem

- The fifth root means z^(1/5)

- We can put 32 as a form a^n

∵ 32 = 2 × 2 × 2 × 2 × 2 = 2^5

∴ z = 2^5[cos(280°) + i sin(280°)]

* Lets find z^(1/5)

∴ z^(1/5) = 2[cos(56) + i sin(56)]

* The fifth root of 32[cos(280°) + i sin(280°)] is 2[cos(56°) + i sin(56°)]

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.