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Trigonometry Problem

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ashwinkandel:
I just came to one problem of trigonometry. Solved it and looked the marking scheme , but my answer didn't matched and i did not understand the process given in marking scheme. the question paper has been attached below and the question no. is 12. It would be great help if someone can clarify me very soon(with in few hours) as because tomorrow i have got my test of mathematics.

EMO123:
excuse me there is no Q no. 12 in this paper  >:(

ashwinkandel:
I am sorry, it was question no. 11.  :)

Vin:
a)
3 (2 sin x - cos x) = 2 (sin x - 3 cos x)
6 sin x - 3 cos x = 2 sin x - 6 cos x
6 sin x - 2 sin x = -6 cox s + 3 cos x
4 sin x = -3 cos x

sin x      -3
----  =   ---
cos x      4

sin x/ cos x = tan x.

tan x = -3/4

b) for this you have multiple methods.
I do it this way.

x = tan-1 (3/4)
(neglecting the sign)

x = 36.9

tan x is negative, so it lies in the 2nd and the 4th quadrant . (check the attachment )

x1 = 180 - 36.9
= 143.1

x2  = 360 - 36.9
= 323.1

Given domain is 0 to 360 so your two values are 143.1 and 323.1

Wait, now I typed all of it, is it Q.11 or Q1. ii) ?
-.-

ashwinkandel:
It's question no. 11  ;D

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