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1.

A tower subtends an angle of `30^@` at a point on the same level as its foot. At a second point h metres above the first, the depression of the foot of the tower is `60^@` . The height of the tower is:

Answer» With the given details, we can draw the figure.
Please refer to video for he figure.
In the figure, `CD` is the height of the tower and `A` and `B` are the two given points such that `AB = h`
Let `AB = h, BC = y and CD = x`
Then, `h/y = tan60^@ => y = h/sqrt3`
Also, `x/y = tan 30^@ => x = y/sqrt3`
`=> x = (h/sqrt3)/(sqrt3) = h/3`
`:. ` Height of the tower is `h/3` metres.
Discussion
2.

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is `30^@`.

Answer» We can create a right angle triangle with the given details.
Please refer to video for the figure.
From the figure,
`AC` = Length of rope `=20m`
Height of pole will be `AB`.
Now,`(AB)/(AC) = sin30^@ =>AB = 1/2**20 = 10m`
Discussion
3.

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is `60^@`. Find the length of the string, assuming that there is no slack in the string.

Answer» let the point on the ground be C making angle`60^@`
A be the point at where kite is
B be the distance of kite from ground
so we have to find AC
`/_ ABC ` is a right angled triangle
`/_ C = 60^@`
`sin /_C = (AB)/(AC)`
`sqrt 3/2 = sin 60^@ = (60)/(AC)`
`AC(sqrt3/2) = 60`m
AC`= 60*2/sqrt3`
`= 120/sqrt3`
`120 sqrt3/3`
`=40sqrt3`m
answer
Discussion
4.

From the bottom of a pole of height h, the angle of elevation of the top of a tower is `alpha`. The pole subtends an angle `beta` at the top of the tower. find the height of the tower.

Answer» 1)`tanalpha=4/x`
`90-(90-alpha+beta)`
`90-90+alpha-beta`
2)`tan(alpha-beta)=(4-h)/x`
`tan(apha-beta)=(H-h)/(H/tanalpha)`
`tan(alpha-beta)=(H-h)/Htanalpha`
`h/H=1-tan(alpha-beta)/tanalpha`
`h/H=(tanalpha-tan(alpha-beta))/tanalpha`
`h/H=1-tan(alpha-beta)/tanalpha`
`=1-sin(alpha-beta)/cos(alpha-beta)*cosalpha/sinalpha`
`h/H=(sinalphacos(alpha-beta)-sin(alpha-beta)cosalpha)/(sinalphacos(alpha-beta)`
`h/H=sinn(alpha-alpha+beta)/sinalphacos(alpha-beta)`
`H=(hsinalphacos(alpha-beta))/sinbeta`.
Discussion
5.

An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is `45o`.What is the height of the chimney?

Answer» We can create a right angle triangle `ADE` with the given details.
Please refer to video for the figure.
Here, `CD = BE= 1.5m, BC = DE = 28.5m`
Now, from figure,
`(AE)/(DE) = tan45^@ =>AE=DE=28.5m`
So, height of chimney `= AB = AE+BE = 28.5+1.5 = 30m`
Discussion
6.

A tree standing on horizontal plane is leaning towards east. At two points situated at distances a and b exactly due west on it, angles of elevation of the top are respectively `alpha` and `beta`. Prove that height of the top from the ground is `((b-a).tanalpha.tanbeta)/(tanalpha-tanbeta)`

Answer» We can draw a diagram with the given details.
Please refer to video for the diagram.
Here, height of top from the ground ` = OB = h`
`AC = a, AD = b, OB =x`
`/_OCB = alpha, /_ODB = beta`
`:.tan alpha = (OB)/(OC) `
`=> tan alpha = h/(x+a)`
`h = tan alpha(x+a)->(1)`
`tan beta = (OB)/(OD) `
`=> tan beta = h/(x+b)`
`=> tan beta(x+b) = h`
Putting value of `h` from (1),
`=>tan alpha (x+a) = tan beta(x+b)`
`=>x(tan alpha- tan beta) = btanbeta - atanalpha`
`=>x = ( btanbeta - atanalpha)/(tan alpha- tan beta)`
`:. h = tan alpha(( btanbeta - atanalpha)/(tan alpha- tan beta) +a)`
`=> h = tan alpha(( btanbeta - atanalpha+atanalpha-atanbeta)/(tan alpha- tan beta))`
`=> h =( (b-a)tan beta tanalpha)/(tan alpha- tan beta)`
Discussion
7.

A person standing between two posts finds that angle subtended at his eyes by the tops of the post is right angle.If height of the post is two times and four times , the height of the person and distance between two post is equal to the length of the longer post. Find the ratio of distance of the person from shorter to longer post.

Answer» `In/_DCH`
`tantheta=(DH)/(HC)=h/x`
`h/x=tantheta-(1)`
`In/_BCG`
`tan(90-theta)=(BG)/(GC)`
`cottheta=(3h)/(4h-x)-(2)`
from equation 1 and 2
`(3h)/(4h-x)=x/h`
`x^2-4hx+3h^2=0`
`x^2-3hx-hx+3h^2=0`
`x(x-3h)-h(x-3h)=0`
`x=h,3h`
When x=h
`AF=4h-x=4h-h=3h`
`FE=x=h`
Ratio`=(EF)/(AF)=h/(3h)=1/3`
Ratio=1/3.
Discussion
8.

The shadow of a tower standing on a level ground is found to be 40 m longer when the Suns altitude is `30o`than when it is `60o`. Find the height of the tower.

Answer» let h be the height of th tower
D is the point where it is forming angle `30^@`
& C is the point where it is forming angle`60^@`
now `tan 60^@ = h/(BC)`
`BC= h/tan 60^@ = h/sqrt3`
now, in `/_ ABD`
`In 30^@ = (AB)/(BD)= h/(BD)`
`BD= h/(tan 30^@) = h/(1/sqrt3)= hsqrt3`
as we can see, `BD-BC=CD`
`(sqrt3h) -(1/sqrt3)h = 40`
`h(sqrt3 - 1/sqrt3) = 40`
`h(3-1)/sqrt3 = 40`
`h(2/sqrt3) = 40`
`h=40*sqrt3/2 = 20sqrt3`m
answer
Discussion
9.

A bridge across the river makes an angle of `30^@` with the river bank. If the length of the bridge across the river is 60m, Find the width of the river.

Answer» Let width of river is `x` in meters.We can create a triangle with the given details.
Please refer to video for the diagram.
From the figure,`(AB)/(AC) = cos30^@`
`=>x/60 = cos30^@`
`=> x = 60**sqrt3/2`
`=>x = 30sqrt3 m`
So, width of river is `30sqrt3 m`.
Discussion
10.

From a point on a cricket ground ,the angle of elevation of the top a lower is found to `30^@` at a distance of 225m from the tower. On walking 150m towards the tower, again the angle of elevation is found Find the new angle of elevation and the height of the tower

Answer» `In/_ABD`
`tan30=(AB)/(BD)=h/225`
`h=225tan30`
`h=225/sqrt3m`
`h=75sqrt3m`
`In/_ABC`
`tantheta=h/75`
`tantheta=sqrt3`
`theta=60`.
Discussion
11.

A boy 1.5 m tall standing from a lamp at a distance of 3m. The lamp casts a shadow of 4.5 m. Find the height of the lamp.

Answer» `/_ABE & /_CDE`are similar
`(DE)/(AD+DE)=(CD)/(AB)`
`4.5/(3+4.5)=1.5/h`
`4.5/7.5=1.5/h`
`hxx10=25`
`h=2.5m`
Discussion
12.

The angle of elevation of the top of a tower at a point on the line through the foot of the tower is `45^@`. After walking a distance towards the foot of the tower along the same horizontal line elevation of the top of the tower changes to `60^@`. Find the height of tower.

Answer» With the given details, we can create a diagram.
Please refer to video to see the diagram.
From the digram,
In `Delta OBC`,
`h/x = tan60^@ = sqrt3`
`=>x = h/sqrt3->(1)`
In `Delta OAC`,
`h/(80+x) = tan45^@ = 1`
`=> h = 80+x`
From (1),
`=>h = 80+h/sqrt3`
`=>sqrt3h-h = 80sqrt3`
`=>h = (80sqrt3)/(sqrt3-1)**(sqrt3+1)/(sqrt3+1)`
`=>h = 40sqrt3(sqr3+1)m`
So,the height of the tower is `40sqrt3(sqrt3+1)m`.
Discussion
13.

A person walking along a straight road observes that at two consecutive kilometre stones the angles of elevation of a hill in front of him are `30^@` and `45^@` find the height of the hill

Answer» In`/_ACD=tan30^@=h/(x+1)`
`1/sqrt3=h/(x+1)`
`x+1=sqrt2h`
`x=sqrt3h-1`
In`/_BCD`
`tan45^@=h/x`
`x=h`
`sqrt3h-h=1`
`h(sqrt3-1)=1`
`h=1/(sqrt3-1)km`
`h=(sqrt3+1)/2 km`
Discussion