Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (anonymous):

I have a math question if anybody can help?? :) thnx!!

OpenStudy (anonymous):

yeah what is your question?

OpenStudy (anonymous):

In the figure below, the length of GE is 97 square root of three units and the length of BG is 149 units. What is the length of AC? 52 square root of three 97 52 square root of two 194

OpenStudy (anonymous):

OpenStudy (anonymous):

there is the picture

OpenStudy (anonymous):

okay So AG=97, and AB=54, so AC=54 sqrt(2). Justification for my answer above: The ratio between the the two legs of a 30-60-90 triangle are 1/sqrt(3), so giving the length of AG. The given length of the base gives us the size of AB, and the known ratio of a 45-45-90 triangle give us the dimension of AC. OOPS!! Bad arithmetic!!! So AG=97, and AB=52, so AC=52 sqrt(2)

OpenStudy (anonymous):

wait hold on.

OpenStudy (anonymous):

okay :) i'll wait

OpenStudy (anonymous):

okay thanks for your help :)

OpenStudy (anonymous):

what do you think genius12

OpenStudy (anonymous):

so....

OpenStudy (anonymous):

We will be using trig ratios here. But before we do that, to find BC, we must first know atleast of the side of triangle ABC and angle of that triangle is already given. Once we find on the sides of the triangle, we can use trig to solve for the others using the angle of 45 degrees given already. So what we can do is find the length of BA. We can do that by using triangle AGE to find the length of side AG, and then subtract that value from 149 to find BA. Then we can solve for side BC. So to find AG, we already have a side and angle given, so here we can use tangent. Since AG is adjacent to the angle, 97sqrt(3) is opposite, and tan(theta) = Opposite/Adjacent, we can use tangent to find side AG.\[\tan(60)=Opposite/Adjacent=\frac{97\sqrt3 }{ AG }\rightarrow AG=\frac{ 97\sqrt3 }{ \tan(60) }=97\]So AG = 97, now we subtract this from 149 to find BA.\[BA=149-AG=149-97=52\]Now we know that side BA = 52. Now, let's move on to triangle ABC. For this triangle we now know one side BA, and an angle of 45 degrees, which we can use to find side AC (the Hypotenuse of the right triangle) once again using trig. Here we need to find the hypotenuse of the triangle, which side AC, and we know the side adjacent to the angle which is 42 along with an of 45 degrees. This time, we use cosine. Because cos(theta) = Adjacent / Hypotenuse, we can use it fond AC.\[\cos(\theta)=\frac{ Adjacent }{ Hypotenuse } \rightarrow \cos(45)=\frac{ 52 }{ AC } \rightarrow AC =\frac{ 52 }{ \cos(45) }=\frac{ 52 }{ \frac{ \sqrt2 }{ 2 } }=\frac{ 104 }{ \sqrt2 }\]Therefore, side AC = 104 / Sqrt(2) @space4cadet101

OpenStudy (anonymous):

wow!!! thank you sooooo much :) this is very helpful (: I will give you the medal :D thank you again! have a great day :) :)

OpenStudy (anonymous):

Change 'BC' to 'AC' in the last sentence of the first paragraph. It was a typo. Btw, fan me? =D

OpenStudy (anonymous):

I fanned you!! thank you sooo much :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!