If i read this correctly then the answer would be
57-12= 45
45/9= 5
The answer is 5
Answer:
Step-by-step explanation:
c^2=a^2+b^2-2abcosC
cosC= (a^2+b^2-c^2)/(2ab)
C=arccos[(a^2+b^2-c^2)/(2ab)]
C=arccos(90^2+55^2-50^2)/(2(90)55)
C=arccos(8625)/(9900)
C=29.4 degrees
Answer:
Draw a perpendicular line from point A to line segment BC. Name the intersection of said line at BC “E.” You now have a right angled triangle AED.
Now, you know AD = 6 m. Next, given that the trapezoid is a normal one, you know that the midpoints of AB and DC coincide. Therefore, you can find the length of DE like so, DE = (20–14)/2 = 3 m.
Next, we will use the cosign trigonometric function. We know, cos() = adjacent / hypotenuse. Hence, cosx = 3/6 = 1/2. Looking it up on a trigonometric table we know, cos(60 degrees) = 1/2. Therefore, x = 60 degrees.
Alternatively, you could simply use the Theorem for normal trapezoids that states that the base angles will be 60 degrees. Hope this helps!
T + S + R = 180
R = 180 - (T + S)
T + S = 68 + 76 = 144
So R = 180 - 144 = 36
<R = 36
