Hello folks, this is Precious posting for our blog for today. At the start of our class Mr. K give us some problems to answer. Well as we saw these problems; we noticed that these questions was on our exercise 2 assignment. Mr. K just gave us other information on how to solve these questions.

Mr. K gave us 4 important questions that we need to study and to review.

The questions are:

1. How many radians are there between the minute and hour hands of clock at 4:00?

Solution:

= 4π / 6

= 2π / 3

Half of the circle is π. In the diagram above, the half of the clock is 6. Divide the π into 6 equal parts. We need to find the radian between the minute and hour of the 4:00. Therefore;

In 4:00 = 4π / 6 or 2π / 3

2. A bicycle wheel has a diameter of 0.4 meters. A point on the outer rim travelled 2.6 meters. How many radians has the wheel turned?

given: 0.4 diameter

0.2 radius

2.6 meters travelled

Solution:

= x / 2π = L / 2πr

= x / 2π = 2.6 / 2π(0.2)

= x / 2π = 2.6 / 0.4π

= x = 2π(2.6) / 0.4π

= x = 5.2π / 0.4π

= x = 13

which is 13 is in quadrant 1.

What we do here is we make the equation proportional. So we have x / 2π = L / 2πr (which is the circumference of the circle).

Mr. K also discuss on how to find the quadrants of a given number. For example:

1. 17 = it is located in Quadrant 4.

2. 02 = it is located in Quadrant 2.

3. 17π / 12 = it is located in Quadrant 3.

How to find these quadrants?

o Half of the circle is π which is equal to 3.14 or approximately 3. π is equivalent to 180 degrees. The whole circle which is 360 degrees or 2π is equivalent to approximately 6.

If you want to find 11:

180 degrees is equivalent to 3

360 degrees is equivalent to 6

back again; 180 degrees is equivalent to 9

360 degrees is equivalent to 12 so move down a little bit therefore 11 is in quadrant 4.

3. In a circle of radius 6 cm, a sector has a central angle of 30 degrees. What is the area of this sector in square centimeter?

Solution:

One of our classmates came up to a great answer. He makes the solution proportional. He put “A” as an area over πr^2 which is the area of the circle = 30 / 360.

= 30 / 360 = A / πr^2

= 30 / 360 = A / π(6)^2

= 30 / 360 = A / 36π

= A = 36π(30) / 360

= A = 1080π / 360

= A = 3π cm^2

We used πr^2 because we are looking for the “area” of the sector.

4. Is point (1 / square root of 5 , 2 / square root of 5) on the unit circle?

Solution:

By simpy looking at the illustration the hypotenuse of the triangle must be 1.

We can used the equation of the unit circle.

= x^2 + y^2 = 1

=(1 / square root of 5)^2 + (2 / square root of 5)^2 = 1

=(1 / 5) + (4 / 5) = 1

=(5 / 5) = 1

Therefore, (1 / square root of 5 , 2 / square root of 5) is on the unit circle.

Mr. K also told us to find the sin(tetta), cos (tetta), tan (tetta) of the illustration in question #4.

We all know the magic word SOHCAHTOA.

The blog that Luis post, he explain the meaning of SOHCAHTOA and what the used of it.

Let’s review again what we had in our class about how to find sin (tetta), cos (tetta) and tan (tetta) of the question number #4.

sin (tetta) = opposite / hypotenuse

= (2 / square root of 5) / 1

= 2 square root of 5

cos (tetta) = adjacent / hypotenuse

= (1 square root of 5) / 1

= 1 square root 5

tan (tetta) = opposite / adjacent

= ( 2 square root of 5) / ( 1 square root of 5)

= 2

How about let’s try this one.

Find the sin (tetta), cos (tetta) and tan ( tetta) of:

***P(tetta) is on a ray passing through the point (6,8)

In this diagram we can find the hypotenuse of this triangle by using pythagorian theorem.

= c^2 = a^2 + b^2

= c^2 = (6)^2 + (8)^2

= c^2 = 36 +84

= c^2 = 100

= c = 10

therefore the hypotenuse is 10.

sin (tetta) = opposite / hypotenuse

= 8 / 10

= 4 / 5

cos (tetta) = adjacent / hypotenuse

= 6 / 10

= 3 / 5

tan (tetta) = opposite / adjacent

= 8 / 6

= 4 / 3

I hope you learn something new this day, like I am.

And tomorrow make sure to asked Mr. K for the story of the guy who died because of triangles.

The next scribe is Jordan.

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