Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
Multiply 3x9 and it gives u 27
And -3 plus -9 equal -12
From the reference of the 18 degree angle, 'h' is the opposite side and 100 is the adjacent side.
The trig ratio which uses both the opposite and adjacent sides is the tangent.
tan(18) = opp/adj = h/100
Trig equation:
tan(18) = h/100
Answer: yes it is
Step-by-step explanation:
Answer:
If they are same side interior or same side exterior they will always add up to 180 so if you have m<7 = 131 then you subtract 131-180 and then you have the angle for one and you do it so on and so forth. If you have A corresponding angle, alternate exterior,alternate interior or a vertical angle they will be the same number at the angle given.
