I need some help with these questions
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Ooh this is a good one!
Firstly, let's find RT so that we can solve the top triangle.
We could actually use 3 methods to find this side: 1) trig to find RT
2) trig to find ST, then Pythagorean Theorem to determine RT
3) use the special 30-60-90 right triangle rule ... hmm, I'm going with the last because it's easiest ;)
For 30-60-90 triangles, the sides are always as follows: smallest side (opposite the 30) is s, the middle side is sSR3, and the hypotenuse is 2×s
Thus, since we have the side opposite 60, it is = sSR3, so:
So now, the hypotenuse (RT) = 2s = 2(2) = 4
Secondly, we are looking at the other type of special right triangle, the 45-45-90:
The side opposite the 45s are s (remember side lengths are equal whose opposite angles are equal); and the hypotenuse is always
we only need QR, which is s, which is also = RT
So the answer is:
QR = x = 4
x = 4 IS THE FINAL ANSWER!!!
[ i hope by my explanations you understand how better to do special right triangles in the future] :-D
Firstly, let's find RT so that we can solve the top triangle.
We could actually use 3 methods to find this side: 1) trig to find RT
2) trig to find ST, then Pythagorean Theorem to determine RT
3) use the special 30-60-90 right triangle rule ... hmm, I'm going with the last because it's easiest ;)
For 30-60-90 triangles, the sides are always as follows: smallest side (opposite the 30) is s, the middle side is sSR3, and the hypotenuse is 2×s
Thus, since we have the side opposite 60, it is = sSR3, so:
So now, the hypotenuse (RT) = 2s = 2(2) = 4
Secondly, we are looking at the other type of special right triangle, the 45-45-90:
The side opposite the 45s are s (remember side lengths are equal whose opposite angles are equal); and the hypotenuse is always
we only need QR, which is s, which is also = RT
So the answer is:
QR = x = 4
x = 4 IS THE FINAL ANSWER!!!
[ i hope by my explanations you understand how better to do special right triangles in the future] :-D
Check the picture below.
so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -2 &,& -4~) % (c,d) &&(~ 3 &,& 8~) \end{array}~~~ % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[3-(-2)]^2+[8-(-4)]^2}\implies d=\sqrt{(3+2)^2+(8+4)^2} \\\\\\ d=\sqrt{25+144}\implies d=\sqrt{169}\implies d=13\qquad\qquad \qquad \stackrel{radius}{\frac{13}{2}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-2%20%26%2C%26%20-4~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%203%20%26%2C%26%208~%29%0A%5Cend%7Barray%7D~~~%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B3-%28-2%29%5D%5E2%2B%5B8-%28-4%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%283%2B2%29%5E2%2B%288%2B4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B25%2B144%7D%5Cimplies%20d%3D%5Csqrt%7B169%7D%5Cimplies%20d%3D13%5Cqquad%5Cqquad%20%5Cqquad%20%20%5Cstackrel%7Bradius%7D%7B%5Cfrac%7B13%7D%7B2%7D%7D)
so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.