Question

How to do the following? I NEED HELP !

Answer 1

Below

Explanation:

a. The `triangle TOB` is a right-angled triangle

`tan30=h/(OB)`

`OB=h/tan30`

`OB=h/(1/sqrt3)`

`OB=hsqrt3`

b. Similarly, `triangle AOT` is a right angled triangle

`tan45=h/(OA)`

`1=h/(OA)`

`h=OA`

Since you have already find the side OB and OA, then you can make use the right angled `triangle AOB` and the pythagoras theorem

Using pythagoras theorem,

`(OA)^2+(AB)^2=OB^2`

`h^2+(100)^2=(hsqrt3)^2`

`h^2+10000=3h^2`

`2h^2=10000`

`h^2=5000`

`h=+-sqrt(2500times2)`

`h=+-50sqrt2`

Now, since h is a height then `h>0`
so `h=50sqrt2`m only

c. "bearing of B from the base of the tower" is in other words "find the `angle TOB`

From a, we found that `OB=hsqrt3`
So subbing in `h=50sqrt2`, we get `OB=sqrt3times50sqrt2 = 50sqrt6`

We can find `angleOTB`.

`tan angleOTB = (50sqrt6)/(50sqrt2)`

`tan angle OTB=sqrt3`

`angle OTB=60'`

So `angle TOB = 180-60-30=90`

Therefore, the bearing of B from the base of the tower is 90

Answer 2

`(OT)/(OA)=tan45^@=1`

`=>h=OA`

Again

`(OB)/(OT)=cot30^@=sqrt3`

`=>OB=sqrt3h`

In `Delta OAB`, `angleBAO=90^@`

So by Pythagoras theorem

`OA^2+AB^2=OB^2`

`=>h^2+100^2=(sqrt3h)^2`

`=>h=50sqrt2`
Again
`tanangle AOB=(AB)/(AO)=100/(50sqrt2)=sqrt2`

`=>angle AOB=tan^-1(sqrt2)~~54.7^@`

So bearing of B from base of the tower `180^@-54.7^@=125.3^@`