Two skyscrapers are located 25m apart. A 62.3m cable links them together. The taller one is 200m. What's the height of the smaller building?

Apply pythagorean theorem..

a = √(c^2 - b^2)

a = √[(62.3)^2 - (25)^2]

a = 57.06 m

solving the height of the smaller building is to get the difference between the upper part of the taller building from height of the taller building..

so

Height of short building = 200m - 57.06 m = 142.94 m or 143 m answer//

You have a right angled triangle with base 25 and hypotenuse 62.3.

The difference in height is the third side.

You can use Pythagoras:

The square on the hypotenuse = sum of squares on the other two sides.

25^2 + x^2 = 62.3^2

625 + x^2 = 3881.29

x^2 = 3256.29

x = 57.06m (less than 200m)

200m - 57.06m = 142.94m

(62.3 x 62.3) - (25 x 25) = 3256.29

sq rt = 57

Smaller tower = 200-57

= 143m

A right triangle is formed. The base is the distance between the buildings. The height is the difference in height of the buildings. The length of the cable is the hypotenuse. Tall building Height A, Short Building Height B.

A=200

62.3 ² = 25 ² + (A-B) ²

62.3 ² = 25 ² + (200-B) ²

62.3 ² = 25 ² + 200 ² -400B +B ²

0 = 25 ² + 200 ² -400B +B ² - 62.3 ²

0 = B ² - 400B +40000 + 625 - 3881.29

0 = B ² - 400B + 36743.71

solve for B as the height difference

or

25/62.3 is cosin of the angle at the shorter building

0.401284109

66.341521 degrees, and 23.658479 degrees at the top of the taller building

(200-B) / 25 = tangent 66.341521 degrees

200-B = 57.063911311

B = 200 - 57.063911311

B = 142.9361 m

0 = B ² - 400B + 36743.71

0 = 20430.73 - 57174.44 + 36743.71

0=0

Smaller building is 142.9361 meters

143m is close enough.

Assume the cable is straight.

Let the height of the smaller building be s.

The cable is the hypotenuse of a right angled triangle with the difference in height between the taller and smaller building and 25 m, the horizontal distance between the two skyscrapers, as the adjacent sides.

Hence,

(200 - s)² + 25² = 62.3² (Pythagorean theorem)

200² - 400s + s² + 25² - 62.3² = 0

s² - 400s + 36743.71 = 0

Using the quadratic formula

x = (- b ± √(b² - 4ac))/2a

for

ax² + bx + c,

s = (400 ± √((- 400)² - 4*1*36743.71))/2*1

s = (400 ± √(13025.16))/2

s = 142.94 or 257.06 (reject because greater than 200)

Height of the smaller building = 142.94 m

h = difference in height

h^2 = 62.3^2-25^2

h = 57.1 m

200-57.1 = 142.9 m

Answer 142.9 m

Using Pythagoras; theorem the height of the smaller building is 200 - 57.06391154 = 142.9360885 or about 143m

Presuming the cable is attached to the tops since the distance difference between the length of the cable and the distance between the buildings.

The cable becomes the hypotenuse of a right triangle with the distance between the buildings being the base. The hight is unknown.

Once we solve for the height we can subtract that from 200m to find the height of the smaller building:

a² + b² = c²

25² + b² = 62.3²

625 + b² = 3881.29

b² = 3256.29

b = 57.064 m (rounded to 3 DP)

So the height of the smaller building is:

200 - 57.064 = 142.936 m tall.

The cable links them together at the top of each skyscraper

Not enough info. Does the cable go from top of one to top of other? Is the cable taut? Does the cable run between them on the ground?