Sam Z. answered • 06/13/19
Math/Science Tutor
In order to get the imfo of a triangle; 2 sides and an angle are needed;
or 2 angles and a side; or 3 sides.
If this is a rt triangle a^2+b^2=c^2; so 37^2+37^2=2738^.5; c=52.326
root=(r*cisθ)^(1/n)
r(cosθ+i*sinθ)=r(cisθ)
Another way to write it is: (r cis(θ+k*360))^(1/n)
=r^(1/n)cis((θ+k*360)/n)
=2^(1/2)cis((37+37*360)/n)
=1.414cis(13,357/2)
=1.414cis6678.5
=1.414(cos6678.5+2*sin6678.5)
=1.414*(-.948+2*-.317)
=-2.238
Sam Z.
No; acute means 3 angles; each less than 90deg. You mentioned "root"; so I figured with the formula above. -2.238 is "root". With only 2 sides given; not much can be done. Need more imfo.06/13/19
Mark H.
my mistake--i said "sides", but meant angles. "root" appears in the problem statements as "37 root 2". I assume this means 37 times the square root of 2. Regardless, the problem asks us to find the RANGE of possible values for x, which I assume to be the third side. Can anyone explain why my answer is not correct? (Assuming of course that I'm reading the problem correctly!!)06/14/19
Sam Z.
Since this triangle is "acute" and 2 angles are 37 degrees ea; angle "c"=116degrees. Since there is no length; there's nothing more I can say.06/14/19
Mark H.
What?? The problem statement states the LENGTH of 2 sides. Please tell me why my solution is not correct.....06/15/19
Sam Z.
Let's say each angle is 60 deg. Then each side is 37". The area is c^2-a^2=b^2. 37^2-18.5^2=b^2 so b=33.19 so 33.19*18.5=614.06sq/in. As for "root" (formula above) would be zk=(__cis37)^1/2. =__^1/2cis((37+k*360)/2) k=0,1,2 zo=2cis37/2deg z1= " " 55.5 " z2= " " 93 "06/15/19
Mark H.
This simply does not make sense. The problem gives 2 sides, and asks for the range of values for the 3rd. the 2 sides are given as 37 and 37*sqrt(2) From the problem statement: "what would the range of values be for x?" 1. What is your answer to that question? 2. If "x" is not the third side, then what is it.?06/16/19
Maddy G.
I am just as confused as you both. All I am given is a triangle with the values of 37, 37 root 2, and x as the third value. I am asked to construct a range of values that will cause for a valid, acute triangle. If you may know the answer, please help!06/20/19
Maddy G.
Also, trig is not required for this question. Simple inequalities.06/20/19
Mark H.
Maddy; Do you think my previous answer (below) is correct? The two cases are the limits for x if one of the angles is allowed to go to 90 degrees. To keep it as an acute triangle, x must be greater than 37 and less than 37 root 3.06/21/19
Mark H.
I added a picture to my previous answer below06/21/19
Mark H.
I'm not understanding this. The problem tells us that the triangle is acute, meaning that all 3 sides are less than 90 degrees. There are a range of possibilities for the 3rd side. (the problem requests the range for x) Also, what is the second method doing---e.g. what is 2.238?06/13/19