Plane trigonometry, by S.L. Loney.

42

TRIGONOMETRY.

Let P be the top of the spire and A and B the two points at which
the angles of elevation are taken. Draw
PMl perpendicular to AB produced and p
let MP be x.
We are given AB= 100 feet,
MAP= 450, x
and z LBP=60~.
We then have
45" ~ /60
AM 45 \ /60 __
=- cot 450, A 100 B M
x
BM 1
and = cot 60=x ^/3
Hence AM=x, and BM=-3.
100=AM -BM=x- =zax /3 -100/3 100/3 (/3 + 1) 50 (3 3)
x-/3- 1 3-1
=50 [3 + 1'73205...] = 236-6... feet.
Also AM1=x, so that both of the required distances are equal to
236-6... feet.
Ex. 3. From the top of a cliff, 200 feet high, the angles of depression
of the top and bottom of a tower are observed to be 30~ and 60~; find the
height of the tower.
Let A be the point of observation and BA the height of the cliff and
let CD be the tower.
Draw AE horizontally, so that L EAC= 30~ and E A
EAD = 60~. -"" 3O —
Let x feet be the height of the tower and produce
DC to meet AE in E, so that CE=AB- x=200- x. C
Since / ADB= DAE=60~ (Euc. I. 29), 200
200.DB = AB cot ADB = 200 cot 60~ = -. x
t3.
200 - x CE 1
Also - - tan 30 = D B
DB EA 3' D B