Answer:
5.98
Step-by-step explanation:
A!! Yes, because adjacent sides aren’t perpendicular
Assume 0 < <em>x</em>/2 < <em>π</em>/2. Then
tan²(<em>x</em>/2) + 1 = sec²(<em>x</em>/2) ===> sec(<em>x</em>/2) = √(1 - tan²(<em>x</em>/2))
===> cos(<em>x</em>/2) = 1/√(1 - tan²(<em>x</em>/2))
===> cos(<em>x</em>/2) = 1/√(1 - <em>t</em> ²)
We also know that
sin²(<em>x</em>/2) + cos²(<em>x</em>/2) = 1 ===> sin(<em>x</em>/2) = √(1 - cos²(<em>x</em>/2))
Recall the double angle identities:
cos(<em>x</em>) = 2 cos²(<em>x</em>/2) - 1
sin(<em>x</em>) = 2 sin(<em>x</em>/2) cos(<em>x</em>/2)
Then
cos(<em>x</em>) = 2/(1 - <em>t</em> ²) - 1 = (1 + <em>t</em> ²)/(1 - <em>t</em> ²)
sin(<em>x</em>) = 2 √(1 - 1/(1 - <em>t</em> ²)) / √(1 - <em>t</em> ²) = 2<em>t</em>/(1 - <em>t</em> ²)
Answer:
10·b = 60
Step-by-step explanation:
The given system of equation is presented as follows;
5·a + 5·b = 25...(1)
-5·a + 5·b = 35...(2)
Given that the coefficient of a in equation (1) is equal in magnitude but opposite in sign to the coefficient of 'a' in equation (2), to eliminate the variable 'a' when using the elimination method, we add both equations as follows;
5·a + 5·b + (-5·a + 5·b) = 25 + 35 = 60
5·a - 5·a + 5·b + 5·b = 60
5·a - 5·a = 0
5·b + 5·b = 10·b
∴ 5·a - 5·a + 5·b + 5·b = 60 = 0 + 10·b = 10·b
∴ 10·b = 60
