CHALL
ii) cos 315º
1. Find the value of
i) sin 210
°
2. Find the value of
i) sin(-210°)
v) sec(-135)
3. Find the value of
ii) cos(-30°)
vi) cosec(-120°)
-TC
-Зл
i) sin
Answers
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Answer:
The given polynomial is 2x3 - 15x2 + 37x - 30.
Since the roots of the polynomial are in AP, so let the roots be a - d, a, a + d.
Now, using the relation between zeroes and the coefficients of the given cubic polynomial, we have:
Sum of roots = a - d + a + a + d = - (-15)/ 2 = 15/ 2
So, 3a = 15/ 2
a = 5/ 2 = 2.5
Now, product of roots = (a - d) (a) (a + d) = -(-30)/ 2 = 15
2.5(a2 - d2) = 15
(2.5)2 - d2 = 6
d2 = 0.25
d = 0.5
Thus, the zeroes of the given polynomial are 2, 2.5, and 3.
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