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2 years ago

PLEASE HELP ME WITH THIS!!!!

Mathematics

1 answer:

posledela2 years ago

4 0

Answer:

y ≈ 24.6°

Step-by-step explanation:

sin(y) = 5/12

y = 24.62431...

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I forget how to subtract, so can you help?

vodka [1.7K]

Yes subtracting is taking away from another. 5-3 is two because you take 3 away from 5.

7 0

3 years ago

The captain of a ship at point A and sailing toward point B observes a lighthouse at L and finds angle LAC to be 36*30'. After s

wel

Answer:

The distance between B and lighthouse is 3.8688 km

Step-by-step explanation:

Given:

The angle made from ship to lighthouse is 36.5 degrees

and that of point B is 73 degrees.

To Find:

Distance Between Point B and Lighthouse

Solution:

<em>Consider a triangle LAB(Refer the attachment )</em>

And Point C is on the line AB as A i.e. ship is sailing to B

So C is at 5 km from A.

Now In triangle LAC,

Using Trigonometry Functions as ,

tan(36.5)=LC/AC

LC=tan(36.5)*AC

=0.7399*5

=3.6998 km

Now In triangle LBC,

As,

Sin(73)=LC/LB

LB=LC/(Sin(73))

=3.6998/0.9563

=3.8688 km

LB=3.8688 km

6 0

3 years ago

Simplify 3ab + 4ab + 5

adelina 88 [10]

7ab+5 is the simplest it can be

4 0

3 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,

$\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}$

The answer is none of the choices

7 0

1 year ago

Answer for brainliest!!!! YOU CAN DO IT!!!

laiz [17]

what is the question pls.

can you pls tell me

3 0

3 years ago

Read 2 more answers