Question

Question #deeb3

Answer 1

In figure
AB = h ft = Height of the building
C is the point on the ground from which angle of elevation of building `/_ACB=30^@20'=(30+20/60)^@=(91/3)^@`

D is the point on the ground(50 ft closer to the building from C) from which angle of elevation of building `/_ADB=45^@`
Let BD = x ft

Now `(AB)/(BD)=h/x=tan/_ADB=tan45^@=1`

`=>h/x=1=>x=h`

Again`(AB)/(BC)=(AB)/(BD+DC)=h/(x+50)=tan/_ACB=tan(91/3)^@`

`=>h/(x+50)=0.585`

`=>h/(h+50)=0.585`

`=>h=(h+50)xx0.585`

`=>h(1-0.585)=50xx0.585~~29.25`

`h=29.25/0.415~~70.5 ft`

Answer 2

Height of building`=70.5ft`

Explanation:

Let line AB be the building with A the top.
Let C be the first point with ascending angle 30°20'
Let D be the second point with ascending angle 45°

In triangle ACD `angle D = 180°-45°=135°`

`:.180°-(135°+30°20')=14°40'= angle A`

In triangle ACD:

`(AD)/(Sin30°20')=50/(sin14°40')`

Multiply both sides by `sin30°20'`

`:.cancel(sin30°20')/1 xx ( AD)/cancel(sin30°20')=(50 xx sin30°20')/(sin14°40')`

`:.AD=(50 xx 0.505029841)/(0.253195168)`

`:.AD=(25.25149208)/(0.253195168)`

`:.AD=99.73133484=hypotenuse`

In triangle ADB `angle D = 45°`

AB=height of building=opposite side

`:.sin45°=(opposite)/(hypotenuse)`

`:.AB=sin45 xx 99.73133484`

`:.AB=70.5ft`