All categories

1 year ago

Help the question is on the image.

Mathematics

1 answer:

Phantasy [73]1 year ago

5 0

Answer:

$\boxed {tan\theta = \frac{2}{3}}$

Step-by-step explanation:

Let's derive a formula :

⇒ Area (triangle) × 3 = Area (square)

⇒ 1/2(bh) × 3 = a²

∴ But a = h (shown in diagram)

⇒ 3/2(bh) = h²

⇒ h = 3b/2

Now, taking tan θ :

⇒ tan θ = b/h

⇒ tan θ = b/(3b/2)

⇒ tan θ = 2b/3b

⇒ tan θ = 2/3

You might be interested in

How many distinguishable a arrangements are there of the letters in "REPRESENTATION"?

Anika [276]

Answer: There are 1,816,214,400 ways for arrangements.

Step-by-step explanation:

Since we have given that

"REPRESENTATION"

Here, number of letters = 14

There are 2 R's, 3 E's, 2 T's, 2 N's

So, number of permutations would be

$\dfrac{14!}{2!\times 3!\times 2!\times 2!}\\\\=1,816,214,400$

Hence, there are 1,816,214,400 ways for arrangements.

5 0

3 years ago

What is the relationship between the two variables represented in the table?

Zina [86]

I think that it's - positive linear association with one deviation

5 0

2 years ago

Read 2 more answers

X2 - 6x + 12 and y = 2x - 4, algebraically are

olga_2 [115]

The solution is (4, 4)

Solution:

Given system of equations are:

$y = x^2 - 6x + 12 ------ eqn\ 1\\\\y = 2x - 4 ---------- eqn\ 2$

Substitute eqn 2 in eqn 1

$x^2 - 6x + 12 = 2x - 4$

Make the right side of equation 0

$x^2 - 6x + 12 - 2x + 4 = 0\\\\x^2 -8x + 16 = 0$

Solve by quadratic equation

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=-8,\:c=16:\\\\x=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:16}}{2\cdot \:1}\\\\x=\frac{-\left(-8\right)\pm \sqrt{0}}{2\cdot \:1}\\\\x = \frac{8}{2}\\\\x = 4$