(a} In the diagram, O is the centre of the circle ABCDE, = I\(\overline{BC}\)I = |\(\overline{CD}\)| and < BCD = 108°. Find < CDE.
(b) Given that tan x = \(\sqrt{3}\), 0\(^o\) \(\geq\) x \(\geq\) 90\(^o\), evaluate
(a) To find < CDE, we need to first find < BOC.
Since O is the center of the circle, < BOC is twice the angle < BAC. Therefore,
< BAC = 1/2 * < BOC
= 1/2 * (360° - < BCD - < ABC)
= 1/2 * (360° - 108° - 36°)
= 1/2 * 216°
= 108°
Similarly, we can find that < BDC = 1/2 * (360° - < BCD - < CDE)
= 1/2 * (360° - 108° - < CDE)
= 126° - 1/2 * < CDE
Since < BDC and < BCD are equal, we have:
126° - 1/2 * < CDE = 108°
1/2 * < CDE = 18°
< CDE = 36°
Therefore, < CDE is 36°.
(b) We are given that tan x = √3, and we need to find the value of:
(cos x)^2 - sin x
---------------------
(sin x)^2 + cos x
Using the identity (cos x)^2 + (sin x)^2 = 1, we can write:
(cos x)^2 = 1 - (sin x)^2
Substituting this into the expression, we get:
[(1 - (sin x)^2) - sin x] / [(sin x)^2 + cos x]
Simplifying, we get:
(1 - 2(sin x)^2 - sin x) / [(sin x)^2 + cos x]
Substituting tan x = √3, we have:
sin x / cos x = √3
sin x = √3 cos x
Substituting this into the expression, we get:
(1 - 2(3cos^2 x) - √3 cos x) / (3cos^2 x + cos x)
Multiplying both the numerator and denominator by -1, we get:
(2(3cos^2 x) + √3 cos x - 1) / (3cos^2 x + cos x)
Simplifying, we get:
(6cos^2 x + √3 cos x - 1) / (3cos^2 x + cos x)
Substituting √3 for sin x / cos x, we have:
(6/4 + √3/4 - 1) / (3/4 + 1/4)
= (9/4 + √3/4) / 1
= 9 + √3
Therefore, the value of the expression is
