A flag pole and a building stand on the same horizontal level. From the point p at the bottom of a building the angle of the top T of the flagpole is 65°. From the point Q of the building the angle of elevation of the point T is 25°. If the building is 20m high. calvulate the distance PT, The heigt of the flagpole, The distance QT​

PT is approximately 28.2 m

The height of the flagpole is approximately 25.56 m

QT is approximately 13.15 m

Step-by-step explanation:

Part A

From the attached drawing of the building and the flagpole, we have;

∠TQP = 25° + 90° = 115°

∠TPQ = 90° - 65° = 25°

∴ ∠QTP = 180° - (115° + 25°) = 40°

By sine rule, we have;

∴ PT = sin(115°) × 20 m/(sin(40°)) = 28.1992924 ≈ 28.2

PT ≈ 28.2 m

Part B

The height of the flagpole, Ta = PT × sin(∠TPa)

∴ Ta ≈ 28.2 × sin(65°) ≈ 25.56

The height of the flagpole, Ta ≈ 25.56 m

Part C

The distance QT= sin(25°) × 20 m/(sin(40°)) ≈ 13.15

QT ≈ 13.15 m

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!