Questions on Algebra: Trigonometry answered by real tutors!

We assume that all measurements relate to horizontal ground surface.


If so, then we write the proportion for the Tony's house height 'T'

     = ,


which gives  the Tony's house height  T =  = 26 feet.



Next, we write the proportion for the Dermont's house height 'D'

     = ,


which gives  the Dermont's house height  D =  = 20 feet.


ANSWER.  Tony's hous is taller.

NEVER pack more than one problem per post.

The period for  2sin(2x)  is   = .


The period for   cos(3x)  is  .


The period for  y = 2sin(2x) + cos(3x)  is the Least Common Multiple of    and  , which is  .


ANSWER.  The period for  y = 2sin(2x) + cos(3x)  is  .

450°  is  360°  plus  90°.


Therefore, cos(450°) is the same as cos(90°), which is 0 (zero)>


ANSWER.  cos(450°) = cos(90°) = 0 (zero).

 = -sin(x).



ANSWER.  Option (C).

This problem is to find the sum of the infinite geometric progression 

having the first term  1 =   and the common ratio  .


The general formula is  S = ,  where 'a'  is the first term of a GP,
and  'r'  is the common ratio.


For this problem, the sum is  S =  =  = .


ANSWER.  S = .

I will use the letter t to represent .


The Pythagorean identify is sin^2(t) + cos^2(t) = 1.

You can write it in the form

    cos^2(t) = 1 - sin^2(t).


You are given that  1 - sin^2(t) = 0.5.

Hence,

    cos^2(t) = 0.5.


Take the square root of both sides of this equation to get:

    cos(t) = +/- .


If  cos(t) = + ,  then angle  't'  is either   45°  or -45°,  which geometrically is the same as 315°.

If  cos(t) = - ,  then angle  't'  is either  135°  or  225°.



Thus, in the interval  [0°,360°)  the given equation   = 0.5  has 4 (four) solutions
      that are  45°,  135°,  225°  and  315°.

If a trigonometric function is given in the form

    f(t) = a + b*sin(c(t-d)),


where  c > 0, then

    - the midline ("the axis") is y = a;

    - the amplitude is  |b|;

    - the period is  ;

    - the shift is  'd'  units to the right, if  d >= 0, and/or  |d|  units to the left, if  d < 0,

          comparing to the parent function f(t) = a + b*sin(ct).


Therefore, in this case

    (a)  the amplitude is 3 units.

    (b)  the period is   =  = 0.5.

    (c)  the maximum is  2 + 3 = 5;  the minimum is 2 - 3 = -1.

    (d)  To plot the function, go to website https://www.desmos.com/calculator/
         and use free of charge plotting tool there.
     
         Print the formula  and get the plot immediately.

 =   means, by the definition


     =   and   <=  <= .


For this standard definition, see, for example, this Wikipedia article

https://en.wikipedia.org/wiki/Inverse_trigonometric_functions



In turn,   is cosecant 

     = .



So, the problem wants you find angle    such that

     =   and   <=  <= .



In other words, you want to find angle    such that

     =  =   and   <=  <= .



There is only one such angle, and this angle is the standard    of angles, 
which students learn in Trigonometry course.  Its measure is 

     =   in radians,  or   = 45 degrees.



ANSWER.  The angle    satisfying the problem's condition is    radians, or  45 degrees.
         Such an angle is  .

The starting equation is in degrees

    sin(2x) + cos(60°) = 0.


As you know, cos(60°) = -1/2,  so, you equation becomes

    sin(2x) + 1/2 = 0,  or  sin(2x) = -1/2.


You should know the table angles.  If sin(2x) = -1/2, it means
that angle 2x is in 3rd or 4th quadrant

    2x = 210°  or  2x = 330°.


It implies that  x = 210°/2 = 105°  or  x = 330°/3 = 165°.


ANSWER.  The given equation has two solutions,  x = 105°  and/or  x = 165°.

Let F be the midpoint of the edge AB at the base of the pyramid.


(a)  Then triangle EOF is a right-angled triangle (recall that point O is the foot
     of the altitude EO, and is, at the same time, the center of the square ABCD).


     Triangle EOF has the legs  EO = 8/2 = 4 cm  and  OF = 5 cm.  
     Hence, the length of the hypotenuse EF is


            EF =  =  =  cm.


     Then tangent of angle EAB is   =  = .

     Hence, angle EAB is   = arctan() = 58.007183 degrees, approximately.

     Thus, angle EAB is about 58 degrees.



(b)  Angle    between a slant edge and the plane of the base is the angle EAO of triangle EAO. 
     This triangle is a right-angled triangle.
     Its leg EO is 5 cm long; its leg OA is  cm long.

     Therefore,   = .


     Hence,  =  = arctan(0.883883476) = 41.4729343 degrees.


     Thus, angle   is about  41.47 degrees.



(c)  angle    between a slant face and the plane on the base is the angle EFO of triangle EFO.

     Triangle EFO is a right-angled triangle.
     Its leg  EO is 5 cm long.  Its leg OF is 4 cm long.

     Therefore,   =  = tan(1.25)}}}.


     Hence,   =  = arctan(1.25) = 51.3401917 degrees.

     Thus, angle  is about 51.34 degrees.



ANSWER.  Angle EAB is about 58 degrees.

         Angle    is about 41.47 degrees.
   
         Angle    is about 51.34 degrees.

This is a problem I solved many years ago, maybe 10 years ago. I am using it to
inform you tutors about some history of this site. 

I picked a random date, October 16, 2020. At present the posts that you see on
the problem page were posted 4.5 years ago on October 16, 2020 are these:

1167408 (2020-10-16 05:27:24)
1167440 (2020-10-16 11:36:06)
1167453 (2020-10-16 13:36:10)
1167480 (2020-10-16 21:39:15)

Notice that the earliest one of those posted on that date at 5:27 AM was
numbered 1167408, and the last one posted on that date at 9:39 PM was numbered
1167480.

That means that on that date between 5:27 AM and 9:39PM there were 73 problems
posted. Can you believe?  I can because I remember.  Most of the 73 problems
posted on that date were solved by the many tutors who were around back then.
The posted problems that are still there now are the ones that the tutors back
then either didn't get to or weren't as interesting as the ones the tutors chose
to solve -- before they scrolled off.

I've been here since this site first opened way back in the early 2000's.  This
site was really hopping back then. I have seen many tutors come and go.

Now compare the fact that there were 73+ problems posted on October 16, 2020 to
the fact that 0 have been posted in the last few days!  Only the ads are keeping
this site going.  Perhaps some students are using this site as a source for
problems.  I don't know, but I fear this site is dying and is almost dead.    

Since I've been here from the start, I'm just informing you about this site in
case you haven't figured out what's going on.  
 
Cheers, Edwin

I agree with greenestamps that the way this problem was typed, it made no sense.
Here's how it was typed:
-------------
"Figure 12.32 shows, a, cross-sectional of a sheetis a series of ark of
radius 10 cm each acts of stained 120° at eating time if there are 14 such ax in
one sheet how wide would a sheet be if flatted out"
-------------
But, believe it or not, I googled and found this very problem online. Here's the
way it was stated online:

Figure 12.36 shows a cross-section of a sheet of corrugated iron.



The sheet is a series of arcs of radius 10 cm., each arc subtending 120o at its centre.  
If there are 14 such arcs in one sheet, how wide would the sheet be if flattened out?

I only drew two of the 14 arcs above.  So I'll find the length of one arc and
multiply it by 14. 

The length of an arc is  of a circumference. So the length
of one arc is



Multiplying by 14:

, approximately 293.2 cm.

Edwin