# I have three questions on my homework in which I am stuck on. Thanks

for the help!
#1
find all solutions of the equation in the interval [0,2pie). In cases
where a calculator is necessary, round the answers to two decimal
places.
sin(sinx)=pie/6
#2
A 100 ft vertical antenna is on the roof of a building. From a point on
the ground, the angles of elevation to the top and the bottom of the
antenna are 51 degrees and 37 degrees, respectively. Find the height of
the building.
#3
the length of the three sides of a triangle are denoted a, b, and c;
the angles opposite these sides are A,B, and C, respectively. In each
exercise, use the given information to find the required quantities.
c=4, a=2, B=90 degrees
find sin^2(A) + cos^2(B)

## Sure, I can help you with your homework questions!

#1: To find all the solutions of the equation sin(sin(x)) = pi/6 in the interval [0, 2pi), you can start by finding the solutions for sin(x) = pi/6. Take the inverse sine of both sides to get x = arcsin(pi/6). Since the interval is [0, 2pi), you need to find all the values of x within this interval that satisfy the equation.

To solve the equation sin(x) = pi/6, you can use a calculator to find the inverse sine of pi/6, which is approximately 0.551. So x = 0.551 is one solution.

However, sin(x) has a periodicity of 2pi, which means it repeats every 2pi radians. So to find all solutions in the interval [0, 2pi), you can add or subtract multiples of 2pi from the initial solution.

In this case, you can add 2pi to the initial solution: x = 0.551 + 2pi. This will give you another solution. Keep adding 2pi until you are within the interval [0, 2pi).

So, the solutions for the equation sin(sin(x)) = pi/6 in the interval [0, 2pi) are 0.551, 0.551 + 2pi, 0.551 + 4pi, and so on, until you reach the value within the interval.

Remember to round the answers to two decimal places, so your final solutions will be something like 0.55, 6.69, 12.83, and so on.

#2: To find the height of the building, you can use the concept of trigonometry and the angles of elevation.

Since the antenna is vertical, the angle of elevation of the top of the antenna will be equal to the angle of depression (or negative angle of elevation) of the bottom of the antenna.

Let's denote the height of the building as h.

Using the tangent function, you can set up a right triangle with the opposite side (height of the building) and the adjacent side (distance from the point on the ground to the bottom of the antenna).

For the top angle of elevation, you have tan(51 degrees) = h / x, where x is the distance from the point on the ground to the bottom of the antenna.

For the bottom angle of elevation, you have tan(37 degrees) = (h + 100 ft) / x.

Since the point on the ground is the same for both angles, x will be the same in both equations. You can solve one equation for x and substitute it into the other equation to solve for h.

Using trigonometric identities, you can rewrite the equations as follows:

h = x * tan(51 degrees) (*)
h + 100 ft = x * tan(37 degrees) (**)

From equation (*), you can solve for x:

x = h / tan(51 degrees)

Now substitute this value of x into equation (**):

h + 100 ft = (h / tan(51 degrees)) * tan(37 degrees)

Simplify the equation and solve for h:

h + 100 ft = h * (tan(37 degrees) / tan(51 degrees))
h * (1 - (tan(37 degrees) / tan(51 degrees))) = 100 ft
h = 100 ft / (1 - (tan(37 degrees) / tan(51 degrees)))

Use a calculator to find the values of tangent for 37 degrees and 51 degrees, then substitute them into the equation to solve for h. Round the answer to the appropriate units (in this case, feet).

#3: To find sin^2(A) + cos^2(B), you can use the trigonometric identities.

sin^2(A) + cos^2(B) = 1

According to the given information, B = 90 degrees, which means cos(B) = 0.

Substitute cos(B) = 0 into the equation:

sin^2(A) + 0 = 1

sin^2(A) = 1

Therefore, sin^2(A) + cos^2(B) = 1.

And that's the final answer for the third question.