P,Q,R on same horizontal plane.S is vertically abv R wif h m.On de west of S is P,s. east is Q.PQ=300M.elev. a

Question

The 3 points P,Q,R lies on same horizontal plane. S, the pinnacle of a tower is h m vertically abv R. P is to the west of the tower whereas Q is to its southeast. The angle of elevation of the tower from both P & Q are thirty degree respectively.PQ=300m. Calculate h m & find the largest possible angle of elevation of S from any point on PQ.


ans:h=93.74m, 56 degree 27 minute

can any 1 help me? thanks

Answer 1

PQR will form an isosceles triangle with PR=QR as they both subtend same angle of elevation. Also angle PRQ is 135 degrees as per the directions mentioned.
Also each PR and QR=hcot30=hsqrt3
drop a perpendicular at PQ from R which meets PR at T.
In triangle PTQ. PT=150m, PR=hsqrt3 and angle PRT is 135/2=67.5 degrees.
sin 67.5=150/h sqrt3
so h=93.74m

Largest possible elevation will be from T
which will be equal to
tan(angle)=h/RT
now RT=150cos67.5=57.40 m
so tan(angle)=93.74/57.40=1.633
or angle = 58.52degrees

Answer 2

You have an isosceles triangle PRQ with:

PQ = 300 m
m Point S is directly above point R by a distance h
T is the closest point on PQ to R
m
We have:

By symmetry PT = TQ = 300/2 = 150.

Looking at triangle PRS we have:

tan(30°) = h/PR
h = PR*tan(30°) = PR(1/√3) = PR/√3

Looking at triangle PTR we have:

m
cos(22.5°) = PT/PR = 150/PR
PR = 150/cos(22.5°) = 150/[√(2 + √2) / 2] = 300/√(2 + √2)

Plug back into the formula for h.

h = PR/√3 = [300/√(2 + √2)] / √3 = 300√3 / [3√(2 + √2)]
h = 100√3 / √(2 + √2) ≈ 93.74 m
__________

T is the closest point on PQ to R and hence has the largest possible angle of elevation α, to S from any point on line segment PQ.

sin(22.5°) = RT/PR
RT = PR*sin(22.5°) = [150/cos(22.5°)]*sin(22.5°)
RT = 150*tan(22.5°) = 150(√2 - 1)

Now look at triangle TRS.

tanα = RS/RT = h / RT = [100√3 / √(2 + √2)] / [150(√2 - 1)]
tanα = 100√3 / {150[√(2 + √2)]*(√2 - 1)}
tanα = 2√3 / {3[√(2 + √2)]*(√2 - 1)}
tanα = 2(√2 + 1) / [√3*√(2 + √2)]
tanα = 2(√2 + 1) / √(6 + 3√2) ≈ 1.508689

α = arctan[2(√2 + 1) / √(6 + 3√2)] ≈ 56.462502°

Converting to degrees, minutes and seconds we have:

α ≈ 56° 27' 45"

Answer 3

Consider the pinnacle as S
In triangle PRS

tan(30)=h/PR

Similarly in triangle QRS

tan(30)=h/QR

=>PR=sqrt(3)h=QR

Consider triangle PQR apply cosine rule in it taking PQ as resultant and angle as 135 deg (90 + 45) draw figure for clarity.

PQ^2=QR^2 + RP^2 - 2(QR)(RP)Cos(135)
=>90000=3h^2 + 3h^2 + 3sqrt(2)h^2

=>h^2=8789.0625

=>h=93.75 m


For the angle tan(A)=h/d
where d is the distance of the point from base of tower to the point on line PQ. For maximum A d should be minimum.
So it will lie at mid point of PQ.

Consider triangle PQR, let a point on the mid-point of PQ be T. In Triangle RPT, Angle RPT=22.5 deg as triangle PQR is isosceles and angle PRQ is 135 deg.

Now Sin(angle RPT= 22.5 deg)=d/RP
=>d=62.14

The distance of the point from the foot of tower will be 62.14m

=>tan(A)=93.75/62.14

=>A=56.46 deg