Domenic and Adriana are observing a hot-air balloon from two tracking stations on the ground.

Mathematics

1 answer:

mel-nik [20]4 years ago

8 0

Answer:

height of the balloon = 6.32 km

Step-by-step explanation:

The question wants us to find the height of the balloon .

They are observing the balloon from two tracking stations on the ground. The tracking stations are 6 km apart . From Dominic point of view the hair balloon is at an angle of elevation of 72° and from Adriana point of view the elevation is 58° . The illustration forms a triangle containing 2 right angles with a similar height.

To find the height we must find the hypotenuse of one of the right angle triangle. WE are given 2 angle and a side of 6 km. The third angle is 180 - 72 - 58 = 50°.

Using sin rule

6/sin 50° = b/sin 72°

cross multiply

6 sin 72° = b sin 50°

divide both sides by sin 50°

b = 6 sin 72°/sin 50°

b = 6 × 0.95105651629 /0.76604444311

b = 5.70633909777 /0.76604444311

b = 7.44909665372

b ≈ 7.45 km

The height can be found since we know the hypotenuse of one side of one of the right angle triangle.

sin 58° = opposite/hypotenuse

sin 58° = h/7.45

cross multiply

h = 7.45 × sin 58°

h = 7.45 × 0.84804809615

h = 6.31795831637

h ≈ 6.32 km

You might be interested in

Write these equations of the line in point-slope form given the slope

Mice21 [21]

Answer: #1) y=5x-73

#2) y= 2/5x - 47/5

Step-by-step explanation:

7 0

3 years ago

1!+2(2!)+3(3!)+...+n(n!)=(n+1)!-1

anyanavicka [17]

Step-by-step explanation:

OK, let's assume it this way:

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!=(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+...+((n-1)n!-n!)=(2!-1!)+(3!-2!)+(4!-3!)+

Sn=1.1!+2.2!+3.3!+...+n.n!=(2‐1).1!+(3-1).2!+(4-1)3!+...+((n+1)-1).n!=(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+...+((n-1)n!-n!)=(2!-1!)+(3!-2!)+(4!-3!)+...+(n+1)!-n!=(n+1)!-1!=(n+1)!-1

and boom problem solved

7 0

2 years ago

What is the net force on a 1,000 kg object accelerating at 3 m/s2 ?

anzhelika [568]

3 $3 \leq m+1.4$