Question

1) A bullet is found embedded in the wall of a room 2.9 m above the floor. The bullet entered the wall going upward at an angle of 39.6°. How far from the wall was the bullet fired if the gun was held 1.1 m above the floor?

2) Let sin θ = 2/7. Find the exact value of cos θ.
for question 2, I did
cos (θ) =sin⁡(90degrees −θ)
cos (θ) = sin (90 - 2/7)
cos (θ) = 0.999987566

Answer 1

1. I don't see how the 1.1 m has anything to do with the problem
I simply see:
cot 39.6° = x/2.9
x = 2.9 cot 39/6° or 2.9/tan39.6° = ...

2. first of all, when you did sin(90° - 2/7)
you really found sin(90° - sin θ), which is incorrect

given sin θ = 2/7 which is y/r , let's make y = 2, and r = 7
we know x^2 + y^2 = r^2
x^2 + 4 = 49
x = ± √45 = ± 3√5

so cos θ = 3√5/7 or -3√5/7 , you didn't say where θ was.

Answer 2

for question 1, I got 3.5055
for question 2, I entered -3√5/7 in my calculator and I got -2.5355

Answer 3

is my answer correct?

Answer 4

I also got 3.505 for the first question,
but....
Whenever it asks for exact values, they expect either exact fractions or
exact square roots,
that is why the correct exact value is ± 3√5/7

besides all that (3√5)/7 = appr .95831.... not 2.5355

These kind of questions can be checked on a calculator
press the following sequence: (my calculator is a scientific SHARP)
make sure it is set to DEG
press: 2 ÷7 =
you should get .2857...
press: 2ndF sin =
you should get 16.6015... °
now press: cos =
and you will get .95831..
now do:
- 3 x √5 ÷ 7 =
you should get 0 , showing that cosθ is indeed 3√5/7 exactly!!!

Answer 5

ok thank you