<span>Let A be the center of a circle and two angles at the adjacent center AOB and BOC. Knowing the measure of the angle AOB = 120 and the measure BOC = 150, find the measures of the angles of the ABC triangle.
</span>solution
Given the above information;
AC=AB, therefore ABC is an isosceles triangle.
therefore, BAO=ABO=(180-120)/2=30
OAC=OCA=(180-90)/2=45
OBC=BCO=(180-150)/2=15
THUS;
BAC=BAO+OAC=45+30=75
ABC=OBA+CBO=15+30=45
ACB=ACO+BCO=15+45=60
The answer to this question is
. Hope this helps.
Step-by-step explanation:
roots as exponents are fractions
if there is no root number next to the radical sign its assumed that its 1/2
3√2 is 2^(1/3)
√3 is 3^(1/2)
4√3 is 3^(1/4)
5√2 is 2^(1/5)
Answer:
Acute.
Step-by-step explanation:
Using the Triangle Inequality Theorem you know that the sum of the lengths of two sides of a triangle are greater than the length of the third side.
Using this we can narrow it down that this DOES have a solution, since 10+12>15.
Now, we can determine if this triangle is a right triangle using the Pythagorean Theorem. C is the longest length.
If c^2= a^2+b^2, then it is a right triangle.
If c^2< a^2+b^2 then it is an acute triangle.
If c^2>a^2+b^2 then it is an obtuse triangle.
We can now substitute. (A and B are interchangeable, but C is the longest length.
A=10
B=12
C=15
A^2=100
B^2=144
C^2=255
We can now figure out that this triangle is acute because A^2+ B^2 (244) < C^2 (255).
Hope this helps!
This problem is a projectile problem. The initial velocity is 50m/s at an optimum angle of 45. The maximum height or the height of the y component is at it's highest is asked.
Informtion needed:
Acceleration due to gravity = 9.81m/s^2
Formula:
Hmax = ((V^2)(sin(angle))^2)/2g
Hmax = ((50^2)(sin45)^2)/(2*9.81) = 63.71 or letter B 63.5 m is the closest answer.
