Question

can anyone please help me with this ?

Answer 1

Answer:

The distance from point A to the top of the building is 5$\sqrt{2}$ feet

The Height of the Skyscraper is 5 feet

Step-by-step explanation:

Given

Let Top point of building be point C

Also Let Base of the building be point  B

Distance from point A to base B of the building AB= 5 feet

∴ It makes a Right angle triangle

Also ∠ACB = 45°

Also tan 45° = 1

Now tan ∠ACB =$\frac{AB}{BC}$

∴ AB= BC =5 feet

The Height of the Skyscraper is 5 feet

Now Triangle ABC is right angle triangle with right angled at B

So by Pythagoras theorem

AC= $\sqrt{AB^2+BC^2}=\sqrt{5^2+5^2} =\sqrt{50} =\sqrt{25\times2} =\sqrt{5^2\times2}=5\sqrt{2}$

The distance from point A to the top of the building is 5$\sqrt{2}$ feet

Answer 2

Answer:

Distance from point A to the top of the building(hypotenuse)= $5\sqrt{2}feet$

Height of the landscaper= $5 feet$

Step-by-step explanation:

Given:

Distance from point A to bottom of building = 5 feet

Angle of depression from the top of building = 45°

We see that the triangle formed is a special 45-45-90 right triangle

The sides of such triangle is given by:

Leg1 = $x$

Leg2=$x$

Hypotenuse = $x\sqrt{2}$

We know the  $x=5 feet$

So we can find all the sides of the triangle:

Leg1 =$5 feet$

Leg 2 =$5 feet$

Hypotenuse= $5\sqrt{2}feet$

It is shown in figure attached.

Distance from point A to the top of the building(hypotenuse)= $5\sqrt{2}feet$

Height of the landscaper= $5 feet$