Grade 12th passTrigonometry
Suppose √9 − 8 cos 40◦ = a + b sec 40◦, where a and b are rational numbers. Then |a + b| equals?
Suppose √9 − 8 cos 40◦ = a + b sec 40◦, where a and b are rational numbers. Then |a + b| equals?
3 Answers
Vikas TU
we can write --->
√9 − 8 cos 40◦ = a + b sec 40◦
as
√9 − 8 cos 40◦ = a + b /cos 40◦
Let cos40’ = t
then
√9 − 8 t = a + b/t
squaring both sides,
and arranging after we get,
8t^3 + (a^2 -9)t^2 + 2abt + b^2 = 0
cos40 will have some constant numeric values,
comparing then coefficient of t^2 = R.H.S = 0
a^2 – 9 =0
a = 3 or a = -3
coeff. of of t = 0
2ab= 0
ab= 0
3b = 0 or -3b =0
b = 0
|a + b| = |3 + 0| or |-3 +0| = 3
is the answer.
√9 − 8 cos 40◦ = a + b sec 40◦
as
√9 − 8 cos 40◦ = a + b /cos 40◦
Let cos40’ = t
then
√9 − 8 t = a + b/t
squaring both sides,
and arranging after we get,
8t^3 + (a^2 -9)t^2 + 2abt + b^2 = 0
cos40 will have some constant numeric values,
comparing then coefficient of t^2 = R.H.S = 0
a^2 – 9 =0
a = 3 or a = -3
coeff. of of t = 0
2ab= 0
ab= 0
3b = 0 or -3b =0
b = 0
|a + b| = |3 + 0| or |-3 +0| = 3
is the answer.
GauRav Ginni
Vikas yadav , according to answer key, right answer is 2 not 3 .
i’m also confused!
any other method you have
Unmani Shinde
√(9 − 8 cos 40) = a + b sec 40= (acos40+b)/cos40
Squaring both sides, we get,
(9-8cos40)(cos240) = a2cos240 + b2 + 2abcos40
Re-arranging,
8cos340 + (a2-9)cos240 + 2abcos40 + b2 = 0 (Equation i)
Now, cos(120) = cos(3*40) = 4cos340 – 3cos 40 = -sin30 = -1/2
Rearranging,
8cos340 – 6cos40 +1 = 0 (Equation ii)
Comparing the coefficients of equations i and ii, we get,
a2-9 = 0, 2ab = -6, and b2 = 1
From above equations, (a,b) = (3,-1) or (-3,1)
Thus a+b = 3-1 or -3+1 = +2 or -2
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