Question

URGENT!! 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.

Answer 1

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

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