According to Euclidean Geometry, two tangents to a circle are symmetrical to each other and the axis of symmetry passes through the center of the circle and, hence, each tangent is perpendicular to a respective radius. We represent the statement in the diagram included below.

Then, we calculate the angle of the radius with respect to the axis of symmetry by knowing the fact that sum of internal angles within triangle equals 180º. That is to say:

$\theta = 180^{\circ}-90^{\circ}-30^{\circ}$

$\theta = 60^{\circ}$

And the angle between these two radii is twice the result.

$\theta' = 2\cdot \theta$

$\theta' = 120^{\circ}$

The angle between these radii must be 120º.

5 0

3 years ago

I don’t know how to use the triangle area to calculate the area of the inscribed polygons

Trava [24]

First of all, let's recall the area of a triangle, knowing its base (b) and height (h):

$A = \cfrac{bh}{2}$

The exercise is showing you that, if you inscribe a polynomial with more and more side, the area of the polynomial will approximate the area of the circle better and better (you can see youself that the polygon is "filling" the circle more and more as the number of sides increase).

Now, the second column tells you the area of each of the triangles the polygon is split into. So, we can see that the first polygon is split into 3 triangles, each of them having base 1.73 and height 0.5.

So, the area of each triangle is

$A = \cfrac{1.73 \cdot 0.5}{2} = 0.4325$

There are three of these triangles, so the area of the whole polygon is

$A = 0.4325 \cdot 3 = 1.2975$

In the second case, you have six triangles, each with base 1 and height 0.87. So, the whole area is

$A = 6 \cdot \cfrac{1\cdot 0.87}{2} = 0.87 \cdot 3 = 2.61$

Finally, in the last case you have 8 triangles, each with base 0.77 and height 0.92. So, the whole area is

$A = 8 \cdot \cfrac{0.77\cdot 0.92}{2} = 0.77\cdot 0.92 \cdot 4 = 2.8336$

3 0

3 years ago