The diagram shown below represents the height of a blimp flying over a football stadium. find the height of the blimp if the angle of elevation at the southern end zone, point a, is 70∘, the angle of elevation from the northern end zone, point b, is 62∘, and the distance between the viewing points of the two end zones is 145 yards. round your answer to the nearest tenth and do not include units.

The height of the blimp over the football stadium = 161.9 yards

Step-by-step explanation:

* We will consider that the 3 points construct ΔABC

- To find the height of the blimp over the football stadium,

  draw CD perpendicular to AB and intersect it at D

- CD is the height of the blimp over the football stadium

* Now lets think how we will solve the problem

- We have two right triangles ADC and BDC

- The height CD is opposite to angle A of measure 70°

 and to angle B of measure 62°

∵ AD + DC = 145 yards

- We can split them ⇒ let AD = x, then BD = 145 - x

∴ AD = x , BD = 145 -x

∵ tanФ = opposite/adjacent

∴ tan(70) = CD/AD and tan(62) = CD/BD

* Make CD as a subject using cross multiplication

∴ CD = AD tan(70) and CD = BD tan(62)

* Now we can equate them

∴ AD tan(70) = BD tan(62)

* Substitute AD and BD by their values

∴ xtan(70) = (145 - x)tan(62) ⇒ open the bracket

∴ xtan(70) = 145 tan(62) - xtan(62) ⇒ collect like terms

∴ xtan(70) + xtan(62) = 145 tan(62) ⇒ take x as a common factor

∴ x[tan(70) + tan(62)] = 145 tan(62) ⇒ divide 2 sides by [tan(70) + tan(62)]

∴ x = [145 tan(62)]/[tan(70) + tan(62)] = 58.922498

∵ AD = x

∴ AD = 58.922498

* Use this value to find the height CD

∵ CD = x tan(70)

∴ CD = 58.922498 × tan(70) = 161.888234

- Proximate the value to the nearest tenth

∵ The hundredth digit is 9

∴ Add the tenth digit by 1

∴ CD = 161.9 ⇒ to the nearest tenth

* The height of the blimp over the football stadium = 161.9 yards