Question

Question #632ef

Answer 1

`color(red)[34^o]`

Explanation:

We can construct this diagram from the given information.

Let length of `AB` be `l_1` and length of `BC` be `l_2`.

It is given that `l_1/l_2 = 2/3` `->` 1.

Also, observe that `BD = FE` and `BF=DE`. `->` 2.

We have to find `angle CAE` which will give us elevation of C from A.

Let `angle CAE` be `theta`.

From `triangle ABF` we observe that:-

`[BF]/ [AB] = sin 25^o => BF = AB*sin 25^o = 0.423*l_1`

`therefore` using 2., `DE=0.423*l_1` `->` 3.

`[AF]/[AB] = cos 25^o => AF = AB*cos25^o = 0.906*l_1` `->` 4.

From `triangle BCD` we find:-

`[CD]/[BC] = sin40^o => CD = BC*sin40^o = 0.643*l_2` `->` 5.

`[BD]/[BC] = cos40^0 => BD = BC*cos40^o = 0.766*l_2`

`therefore` form 2., `FE=0.766*l_2` `->` 6.

Now,
`tan theta = [CE]/[AE] = [CD+DE]/[AF+FE]`

Substituting values of `DE, AF, CD, FE` from 3., 4., 5., 6.

`tan theta = [0.643*l_2 + 0.423*l_1]/[0.766*l_2 + 0.906*l_1]`

Dividing the numerator and deniminator by `l_2`

`=> tan theta = [0.643*cancel(l_2/l_2) + 0.423*l_1/l_2]/[0.766*cancel(l_2/l_2) + 0.906*l_1/l_2]`

Substituting value of `l_1/l_2` from 1.

`=> tan theta = [0.643 + 0.423*2/3]/[0.766+0.906*2/3] = 0.675`

`therefore theta = tan^-1 0.675 approx 34^o`

`therefore color(red)[angleCAE = 34^o]` which is the angle of elevation of `C` from `A`.

Answer 2

The angle of elevation is: `hatA = 34°`

Explanation:

I am referring to the same image as posted by Veerpaksh S.
Thank you.

The required angle of elevation of `C` from `A` can be seen as
`hat A` in `Delta CAE`, a right-angled triangle.

The opposite side `color(blue)(CE=CD +DE)`

The adjacent side `color(red)(AE = AF +FE)`

You do not need the actual lengths of any of the lines, just a ratio is good enough. `AB : BC = 2:3`

In `Delta CDB: " " (CD)/(BC) = sin 40°, :. color(blue)(CD = 3sin40°)`

In `Delta BFA: " "(BF)/(AB) = sin25°, :. color(blue)(BF = 2sin25°)`

`color(blue)(CE = 3sin40°+2sin25°)`

In `Delta CDB: " " (BD)/(BC) = cos 40°, :. color(red)(CD=3cos40°)`

In `Delta BFA: " "(AF)/(AB) = cos25°, :. color(red)(BF=2cos25°)`

`color(red)(AE = 3cos40°+2cos25°)`

In `Delta CAE:`

`tan A = (color(blue)(CE))/(color(red)(AE))= color(blue)((3sin40°+2sin25°))/(color(red)((3cos40°+2cos25°))`

`tan A = 0.6747`

`hatA = 34°`