civonamzuk

how to find the angles of a 5 sided polygon if one angle is 100 Degrees

Like this...

"how do you find a measure of an angle if they only give you the measure of one side(?)"

Answer:

Right triangle:   A = a*b/2

Equilateral triangle:     A = 

Isosceles triangle:        $${\mathtt{A}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{ah}}$$

For all the other triangles, you basically have to find a height and multiply it by one of the sides of a triangle.

In this case, the height(h) is multiplied by the side (b) and their product divided by 2.

A = b*h/2

Triangle ABC is given and angle B = 38 degrees Angle C is the right angle CA is X and CB is 4.5 I have no clue how to find x ?

B = 38^o,      CB = 4.5,     CA = X

tan(B) = X/CB

tan38^o = X/4.5

X = tan38^o * 4.5

X = 3.52

19 is what percentage of 58?

Make a proportion:     58/100 = 19/x

                                x = (100*19)/58

                                x = 32.76

The answer is:  19 is ≈32.76%  of 58!

I'm too tired; I'll work on this one tomorrow; I'm going to 

@heureka:/  This time your numbers are correct! I came up with the same answer --with a little help from you. Thanks for showing me how to find a side of the equilateral triangle when only the area is known!

I guess..., I can go back to (see above image) 

If sin(8)+0.54337 what's half the answer?

(sin(8) + 0.54337)/2 = (0.13917310096 + 0.54337)/2 = 0.34127155048

Because...,that we can have fun with maths -- free  of charge!

Thanks, Admin !

Thanks, CPhill! It took me too long...about half an hour. 

Write a ten-digit number so that the first digit tells how many zeros there are in the number, the second how many ones, the third how many twos, and so forth.       

         0 1 2 3 4 5 6 7 8 9

         6 2 1 0 0 0 1 0 0 0