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Mathematics 4 Online
OpenStudy (anonymous):

The ratio of the base of an isosceles triangle to its altitude is 43.85:4. Find the measure of one of the congruent angles (in radians) of the triangle. Round off your answer to two decimal places.

OpenStudy (anonymous):

can someone help me please? my answe is 31.05 but got a wrong answer

OpenStudy (irishboy123):

tan ø = 4 / (43.85/2) ie tan is height over half-base draw it and see. it's a small angle so i'd use tan ø = ø and leave it there.

OpenStudy (anonymous):

i got an anwer of 10.33 degrees and in radians 0.180 rad

OpenStudy (anonymous):

@IrishBoy123 am I on the right path?

OpenStudy (irishboy123):

yep!! well done.

OpenStudy (anonymous):

@IrishBoy123 mind if I ask 2 more questions?

OpenStudy (irishboy123):

sure

OpenStudy (anonymous):

I really need to ace this one I only got 120/150 by now, 3 questions are in the way...

OpenStudy (anonymous):

Determine the area of the segment of a circle if the length of the chord is 15 inches and located 4.14 inches from the center of the circle. Round off your answer to two decimal places. (Thank you so much for the help)

OpenStudy (michele_laino):

|dw:1425120039830:dw| is that right?

OpenStudy (anonymous):

Yep!!! exactly

OpenStudy (michele_laino):

ok! Now please keep in mind that the area A of the triangle AOB is given by the subsequent formula: \[A = \frac{{AB \times OH}}{2}\] so, what is A?

OpenStudy (michele_laino):

hint: \[AB = 15,\quad OH = 4.14\]

OpenStudy (michele_laino):

hint: what is: \[A = \frac{{15 \times 4.14}}{2} = ...?\]

OpenStudy (anonymous):

31.05 - exact anser I had when I submitted it and got an incoorrect answer @Michele_Laino

OpenStudy (michele_laino):

please note that that's the area of the triangle, not the area of the circular segment

OpenStudy (michele_laino):

now, you have to find the area S of the circular sector

OpenStudy (michele_laino):

that area S is given by the subsequent formula: \[S = \frac{{ arc(AB)\times r}}{2}\] where r is the radius of our circumference, and: arc(AB) is the length of the arc AB

OpenStudy (michele_laino):

plase note that I can compute the area S using the subequent formula: \[S = \frac{{{r^2}}}{2}\sin \alpha \]

OpenStudy (michele_laino):

|dw:1425122429952:dw|

OpenStudy (anonymous):

@Michele_Laino ahh ok so I derived a formula for finding the area of the segmen = 1/2 r^2(Theta - sin theta) (is it correct?

OpenStudy (michele_laino):

ok!

OpenStudy (michele_laino):

hint: \[{r^2} = {7.5^2} + {4.15^2} = ...?\]

OpenStudy (michele_laino):

please wait, I'm gong to check your result...

OpenStudy (anonymous):

I got 0.58

OpenStudy (michele_laino):

I got this: \[\sin \theta = \frac{{2S}}{{{r^2}}} = \frac{{2 \cdot 31.05}}{{73.38}} = \frac{{62.1}}{{73.38}} = 0.8462\]

OpenStudy (michele_laino):

so: \[\theta = 1.009\;radians\]

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