Find angle ABC this is for geometry i need help please

Mathematics

2 answers:

zmey [24]3 years ago

8 0

2x+5x+5=180

7x+5=180

7x=175

x=25

angle A,B,C=130

hope this helps

P.S can you give me a brainiest pls!!!!!!

Levart [38]3 years ago

3 0

A straight line is also called a supplementary angle which equals 180 degrees. Angle DBC is a supplementary angle and angles ADB and ABC equals 180 degrees.

Your starting equation would be : 2x + 5x + 5 = 180

From here, combine like terms : 7x + 5 = 180

Subtract 5 from both sides : 7x = 175

Divide 7 on both sides : x = 25

From here, you need to insert the value of x (which is 25) into the equation for angle ABC (which is 5x + 5)

5(25) + 5

Angle ABC = 130 degrees

To ensure you have the correct answer, you can also find angle ADB (50) and add both values. If they add up to 180, you know for sure that you have the correct answer.

50 + 130 = 180 The answer is correct!

Just to be clear, angle ABC = 130 degrees

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Answer and step-by-step explanation:

The polar form of a complex number $a+ib$ is the number $re^{i\theta}$ where $r = \sqrt{a^2+b^2}$ is called the modulus and $\theta = tan^-^1 (\frac ba)$ is called the argument. You can switch back and forth between the two forms by either remembering the definitions or by graphing the number on Gauss plane. The advantage of using polar form is that when you multiply, divide or raise complex numbers in polar form you just multiply modules and add arguments.

(a) let's first calculate moduli and arguments

$r_1 = \sqrt{(-2\sqrt3)^2+2^2}=\sqrt{12+4} = 4\\ \theta_1 = tan^-^1(\frac{2}{-2\sqrt3}) =-\pi/6\\r_2=\sqrt{1^2+1^2}=\sqrt2\\ \theta_2 = tan^-^1(\frac 11)= \pi/4$

now we can write the two numbers as

$z_1=4e^{-i\frac \pi6}; z_2=e^{i\frac\pi4}$

(b) As noted above, the argument of the product is the sum of the arguments of the two numbers:

$Arg(z_1\cdot z_2) = Arg(z_1)+Arg(z_2) = -\frac \pi6 + \frac \pi4 = \frac\pi{12}$

(c) Similarly, when raising a complex number to any power, you raise the modulus to that power, and then multiply the argument for that value.

$(z_1)^1^2=[4e^{-i\frac \pi6}]^1^2=4^1^2\cdot (e^{-i\frac \pi6})^1^2=2^2^4\cdot e^{-i(12)\frac\pi6}\\=2^2^4 e^{-i\cdot2\pi}=2^2^4$

Now, in the last step I've used the fact that $e^{i(2k\pi+x)} = e^i^x ; k\in \mathbb Z$ , or in other words, the complex exponential is periodic with $2\pi$ as a period, same as sine and cosine. You can further compute that power of two with the help of a calculator, it is around 16 million, or leave it as is.