Trigonometry!!

[S-M-A-R-T-I-E-S]

HI!

I'm having problem with these questions and i was wondering if anyone could help me please?

Q1. A ship sails 16km due north then 5km due east. Find the bearing and distance of the ship from its start position.

I drew the triangle and worked out the angle which was 17 degrees (rounded down)So i managed to find the bearing.

tan = opp divide by adj
tan x = 5 divide by 16
x = tan -1 (5 divide by 16)
x = 17


however i'm stuck on working out the distance, so far i managed:

sin = opp divide hyp
sin17 = 5 divide by x

Can someone please help! Thanks in advance!!

Sorry but i'm stuck on one more question!

Q2. An owl sits on a branch 9.5cm from the ground vertically over the base of a tree. If he sees a mouse on the ground 6.2 m from the tree at what angle to the vertical must he swoop to catch his supper.

I drew out the triangle and converted 6.2 m to 620 cm.
Then i did:
tan = opp divide by adj
tan x = 620 divide by 9.5
x = tan -1 (620 divide by 9.5)
x = 89.12

I think this is wrong though because i'm not sure if you were suppose to convert the units??

Any help will be greatly appreciated!! THANKS!!

 

[TchrQbic]

HI!

I'm having problem with these questions and i was wondering if anyone could help me please?

Q1. A ship sails 16km due north then 5km due east. Find the bearing and distance of the ship from its start position.

I drew the triangle and worked out the angle which was 17 degrees (rounded down)So i managed to find the bearing.

tan = opp divide by adj
tan x = 5 divide by 16
x = tan -1 (5 divide by 16)
x = 17


however i'm stuck on working out the distance, so far i managed:

sin = opp divide hyp
sin17 = 5 divide by x

Can someone please help! Thanks in advance!!
The distance can be found using the Pythagorean theorem, since the triangle here is a right triangle.
Sorry but i'm stuck on one more question!

Q2. An owl sits on a branch 9.5cm from the ground vertically over the base of a tree. If he sees a mouse on the ground 6.2 m from the tree at what angle to the vertical must he swoop to catch his supper.

I drew out the triangle and converted 6.2 m to 620 cm.
Then i did:
tan = opp divide by adj
tan x = 620 divide by 9.5
x = tan -1 (620 divide by 9.5)
x = 89.12

I think this is wrong though because i'm not sure if you were suppose to convert the units??
Both units should be converted similarly or left alone -- changing just one won't work. The tangent of your angle = opposite/adjacent, which I believe you have inverted.
Any help will be greatly appreciated!! THANKS!!

 

[arthur ohlsten]

16 km due north
5km due east

bearing angle from north : tan @= opposite/adjacent
tan@=5/16
@=tan^-1 5/16
@=4.9 degrees

distance is the hypoteneuse of the right triangle
distance= square root [ 16^2+5^2]
distance= sq.rt. 281
distance= 16.76 km

you do it another way to check on me!
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we have a right triangle
side one= 9.5 centimeters
side two = 6.2 meters
side two= 6,200 centimeters

tan@= 6,200/9.5
@=tan^-1 6,200/9.5
@=89.9 degrees from the vertical

are you sure side one is 9.5 cms? This is only about 4 inchs from the ground
Arthur

 

[S-M-A-R-T-I-E-S]

yep i'm positive its 9.5cm because the question is in the textbbok!

Thanks for all ur help everyone!!!!!!!!!

 

[Gene]

Arthur erred.
tan^-1(5/16) = 16.8°
6.2m = 620 cm but the method is right.
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Gene