Question 18 (5 points)

A flag staff is fixed at one corner of a rectangular room of side 20×10m.The angle of elevation on the top of the flag from oppo

Vedmedyk [2.9K]

Answer:

The height is 12.9m

Step-by-step explanation:

First we have to find the distance from the corner of the flag to the opposite corner, for this we will use Pythagoras

In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides

h² = s1² + s2²

h² = 10² + 20²

h² = 100 + 400

h² = 500

h = √500

h = 22.36

Now that we know this measurement we can calculate the height of the flagpole

well to start we have to know the relationship between angles, legs and the hypotenuse

α = 30

a: adjacent = 22.36

o: opposite = ?

h: hypotenuse

sin α = o/h

cos α= a/h

tan α = o/a

we see that it has (angle, adjacent, opposite)

is the tangent

tan α = o/a

tan 30 = o/22.36

tan30 * 22.36 = o

12.9 = o

The height is 12.9m

7 0

3 years ago

I WILL GIVE BRAINIEST

SOVA2 [1]

Answer:

The flight took 34 minutes to arrive Bakersfield wich is 85 miles far from Santa Barbara, at a speed of 150 \ mi/hr150 mi/hr

4 0

2 years ago

Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

Question 1 a. i) Find the value of x.

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

Question 1a ii) Find the volume of the frustrum

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

Question 1b. Calculate the volume of the bin

i) Upper frustrum

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

ii) Lower frustrum

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

iii) Add the volume of the two frustrums

$Volume=27.6m^3+38.3m^3=65.9m^3$

6 0

3 years ago

The points shown on a graph represent the numbers in a geometric sequence.

WARRIOR [948]

Answer:

2

Step-by-step explanation:

6 0

3 years ago

Frank invests $5100 for 8 years at a simple interest rate of 5%. How much interest will he earn? What will his total value be?

Westkost [7]

Answer:

A = P(1 + r)t

Step-by-step explanation:

Or this formula:

$A = P(1 + \frac{r}{n} )$

3 0

3 years ago