It is a triangle inside a circle..
Notice both sides of the triangle (Left and right) are radii of the circle so they are equal in length.
The angle 120 is between both sides (4cm each)
If we want the area, there is 2 ways to find it..
We can really just constract a right angle triangle from the 120 degrees and use trig to find the rest.
Other method (I prefer) Is to find the opposite side of 120 degrees, and then apply Heron's formula.
First of all let's label the sides
Let the side opposite to angle 120 be c,
Let the side on the left be b,
Let the side on the right be a.
Using the law of cosines, we know that:
c^2=a^2+b^2-2ab*cos(x) , Let the angle 120=x
So just applying the formula we get,
c^2=16+16-32*cos(120)
cos(120)=-0.5
32-32(-0.5)=48
c^2=48
c=sqrt(48) Positive or negative but since we are finding a side it must be positive.
Now let's apply heron's formula
which is:
Area=sqrt((s(s-a)(s-b)(s-c)) where s=(a+b+c)/2
we know that,
a=4
b=4
c=sqrt(48)
adding these together we get:
4+4+sqrt(48) =14.928203 approx.=15
so s=15
just subsituiting,
sqrt((15(15-4)(15-4)(15-sqrt(48))=121.038469 approx=121
So the area is 121 cm^2
I am pretty sure there is a faster solution.
Maybe using inscribed theorem etc..
Thanx, Guest...... you should check to see if you think your answer is 'reasonable'....121 cm^2 seems wwwwaaaayyyy high doesn't it? (even a square of 4cm sides has but 16 cm^2 area)
Area of isoceles triangle = 1/2 s2 sin (theta) = 1/2 42 sin 120 = 6.93 cm^2
+2863 Notice that if you divided that isosceles triangle into two congruent parts, they form 30-60-90 triangles.
You can easily solve from there.
+2863 This is why there is a controversy to whether someone should learn calculus at a young age.
In my opinion, it is not good to learn calculus. Because the extra knowledge my blur your vision on seeing simple solutions.
+2863 Guest's solution doesn't use calculus, it uses triginometry.
But you get what I mean.
A simple knowledge of 30 60 90 well help!
+321 Using the method that CalculatorUser said, you split the triangle into two 30-60-90 triangles. We get that the height of the isosceles triangle is simply just 2, and the base is 4sqrt3. Using the formula for triangle area, we get that the area of the triangle 4sqrt3.
(That calculus method looks hard!)
+37146 Drop a line from the apex of the triangle to form 2 triangles then use law of sines:
Sin30 / height = sin 90 / 4
4 sin 30 = height = 2
sin 60 / base = sin 90 / 4
4 sin 60 = base = 2 sqrt 3
area of each triangle = 1/2 base * height
1/2 (2) * 2 sqrt 3 = 2 sqrt 3 for each triangle (remember, there are two) Area = 4 sqrt 3
