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3 years ago

Solve for x (Angle relationship in Circles)

Mathematics

1 answer:

makkiz [27]3 years ago

4 0

Step-by-step explanation:

$180 \degree - 102 \degree = \frac{1}{2} (x + 95\degree) \\ \\ \therefore \: 72 \degree = \frac{1}{2} (x + 95\degree) \\ \\ \therefore \: 72 \degree \times 2= (x + 95\degree) \\ \\ \therefore \: 144 \degree = x + 95\degree \\ \\ \therefore \:x = 144 \degree - 95\degree \\ \\ \huge \red{ \boxed{\therefore \:x = 49\degree}}$

You might be interested in

Diego is solving the equation x^2-12x = 21

uysha [10]

Answer:

The solutions to the quadratic equations will be:

$x=\sqrt{57}+6,\:x=-\sqrt{57}+6$

Step-by-step explanation:

Given the expression

$x^2-12x\:=\:21$

Let us solve the equation by completing the square

$x^2-12x\:=\:21$

Add (-6)² to both sides

$x^2-12x+\left(-6\right)^2=21+\left(-6\right)^2$

simplify

$x^2-12x+\left(-6\right)^2=57$

Apply perfect square formula: (a-b)² = a²-2ab+b²

i.e.

$x^2-12x+\left(-6\right)^2=\left(x-6\right)^2$

so the expression becomes

$\left(x-6\right)^2=57$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solve

$x-6=\sqrt{57}$

add 6 to both sides

$x-6+6=\sqrt{57}+6$

Simplify

$x=\sqrt{57}+6$

also solving

$x-6=-\sqrt{57}$

add 6 to both sides

$x-6+6=-\sqrt{57}+6$

Simplify

$x=-\sqrt{57}+6$

Therefore, the solutions to the quadratic equation will be:

$x=\sqrt{57}+6,\:x=-\sqrt{57}+6$

5 0

2 years ago

Please show work and thank youuu

Assoli18 [71]

Answer: $6\sqrt{3}$

======================================================

Explanation:

Method 1

We can use the pythagorean theorem to find x.

$a^2+b^2 = c^2\\\\6^2+x^2 = 12^2\\\\36+x^2 = 144\\\\x^2 = 144-36\\\\x^2 = 108\\\\x = \sqrt{108}\\\\x = \sqrt{36*3}\\\\x = \sqrt{36}*\sqrt{3}\\\\x = 6\sqrt{3}\\\\$

-----------------------------------

Method 2

Use the sine ratio to find x. You'll need a reference sheet or the unit circle, or simply memorize that sin(60) = sqrt(3)/2

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(60^{\circ}) = \frac{x}{12}\\\\\frac{\sqrt{3}}{2} = \frac{x}{12}\\\\x = 12*\frac{\sqrt{3}}{2}\\\\x = 6\sqrt{3}\\\\$