Find value of y when x = 45?

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Find value of y when x = 45?

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Answer 1 · 2018-05-10T18:22:35

$\Rightarrow y = 3 \sqrt{2} + 4$ $=> y=3sqrt2+4$

Explanation:

$cos 90 = 0$ $cos 90=0$ so from the table, $b = 4$ $b=4$
$cos 0 = 1$ $cos 0=1$ , putting this into the equation gives
$a \times 1 + 4 = 10$ $a xx 1+4=10$
$a = 10 - 4, a = 6$ $a=10-4, a=6$

So $y = 6 cos x + 4$ $y=6cos x +4$ is the equation.
Put $x = 45$ $x=45$ into the equation

$\Rightarrow y = 6 \times \frac{\sqrt{2}}{2} + 4$ $=> y=6 xx sqrt2/2+4$

$\Rightarrow y = 3 \sqrt{2} + 4$ $=> y=3sqrt2+4$

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Find value of y when x = 45?

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