Step-by-step explanation: Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

$\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},$

and from the right-angled triangle WAB, we have'

$\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.$

Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.

Thus, the height of the building across the street is 381.85 feet.

8 0

3 years ago

Taiga wants to make a circular loop that she can twirl around her body for exercise. He will use a tube that is 2.5 meters long.

Travka [436]

Answer: 0.796 meters

Step-by-step explanation:

Circumference of circle = $2\pi r$ , where r=radius

Here, Circumference of loop = 2.5 meters

i.e. $2\pi r=2.5$

$\Rightarrow\ r=\dfrac{2.5}{2\pi}$

$\Rightarrow\ r=\dfrac{2.5}{2\times\dfrac{22}{7}}\\\\\Rightarrow\ r=\dfrac{2.5\times7}{2\times22}\\\\\Righttarrow\ r=0.398\ m$

Diameter = 2r = 2(0.398) = 0.796 meters

Hence, The diameter of Taiga's exercise hoop = 0.796 meters

7 0

3 years ago

The quotient of (x^3+3x^2-4x-12)/(x^2+5x+6)

Elis [28]

Answer: The quotient is (x-2).

Step-by-step explanation:

Since we have given that

$f(x)=(x^3+3x^2-4x-12)\\\\and\\\\g(x)=x^2+5x+6\\\\So,\ \frac{\left(x^3+3x^2-4x-12\right)}{\left(x^2+5x+6\right)}$

Now, we have to find the quotient of the above expression.

So, here we go:

$Factorise\ (x^3+3x^2-4x-12)\\\\=\left(x^3+3x^2\right)+\left(-4x-12\right)\\\\=-4\left(x+3\right)+x^2\left(x+3\right)\\\\=\left(x+3\right)\left(x^2-4\right)$

Now, we will divide the above simplest form with g(x):

$\frac{\left(x+3\right)\left(x^2-4\right)}{\left(x+2\right)\left(x+3\right)}\\\\=\frac{x^2-4}{x+2}\\\\=\frac{\left(x+2\right)\left(x-2\right)}{x+2}\ using\ (a^2-b^2)=(a+b)(a-b)\\\\=x-2$