Need help with a trig question?? (1 Viewer)

[fatima96]

Q 17 please?

 

[cheepy5]

Let the 3rd vertice be C (the top one)

You find the length of BC:

BC / sine50 = 40/sine(180-50-65)

BC = 33.81

use SOHCAHTOA to find x

sine65= X/BC

X= BC x sine65

X=30.6

 

[fatima96]

i feel so silly i uploaded the wrong qn, can u please do this one, thanks and sorryy

 

[RealiseNothing]

Draw a perpendicular line from A and B to the ground. This is the height of the balloons, let it be "h".

Let the distance from "O" to the spot where B meets the ground be some distance "k".

So the distance from the "O" to the spot where A meets the ground is "k-1000".

Use the two triangles you have formed to find a value for "k" by solving the "tan" of the known angles simultaneously.

Substitute this value into one of the original "tan" equations.

?????

Profit.

 

[Timske]

Draw a perpendicular line from point B to the line OX and lets call that line 'h' and the point 'C'
AB = 1000m
angle AOX = 14'15'
angle BOX = 10'30'
angle AOB = angle AOX - angle BOX = 3'45'
angle OAB = 104'15'

Find BO using the formula; a/sinA = b/sinB

1000/sin3'45' = BO/sin104'15'
:. (1000sin104'15')/(sin3'45') = BO
:. BO = 14, 819m

Find BC using the same formula

14819/sin90 = h/sin10'30'
:. 14819sin10'30'/sin90 = h = 2700.54m (2.d.p)

 

[fatima96]

The answers say 686m, could they be wrong?? :S Ive been trying to get the answer for a few hrs now. feelingg hopeless

 

[Timske]

oo nope im wrong lol

 

[fatima96]

appreciate ur help though, but i dont get @Realisenothing's method.

 

[Timske]

i worked it out ill post solutions asap

 

[Timske]

Draw a perpendicular line from point B to the line OX and lets call that line 'h' and the point 'C'
AB = 1000m
angle AOX = 14'15'
angle BOX = 10'30'
angle AOB = angle AOX - angle BOX = 3'45'
angle ABO = angle BOX ( AB is perpendicular to OX ) they are alternate angles.
:. angle OAB = 165'45' ((180 - (3'45' + 10'30'))

Find BO using the formula; a/sinA = b/sinB
BO/sin165'45' = 1000/sin3'45'
BO = 1000sin165'45'/sin3'45'
= 3763.63 m (2dp)

Do the same for 'h'
3763.63/sin90 = h/sin10'30'
h = 3763.63sin10'30'/sin90 = 686 m (1dp)

TADAH!

 

[fatima96]

Thanks a lot

 

[fatima96]

Draw a perpendicular line from point B to the line OX and lets call that line 'h' and the point 'C'
AB = 1000m
angle AOX = 14'15'
angle BOX = 10'30'
angle AOB = angle AOX - angle BOX = 3'45'
angle ABO = angle BOX ( AB is perpendicular to OX ) they are alternate angles.
:. angle OAB = 165'45' ((180 - (3'45' + 10'30'))

Find BO using the formula; a/sinA = b/sinB
BO/sin165'45' = 1000/sin3'45'
BO = 1000sin165'45'/sin3'45'
= 3763.63 m (2dp)

Do the same for 'h'
3763.63/sin90 = h/sin10'30'
h = 3763.63sin10'30'/sin90 = 686 m (1dp)

TADAH!

Thankyouu soo much!! Repped you!
I actually get it now!