Please help me, I'm not really good with trigo.. A pilot flew for 210 miles at 21degree and then 147 miles at 162 degree. How far was he from starting point?
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start by making a sketch of the problem ... looks to me like a task involving the cosine law.
Thanks for the diagram sir.. ∠ABC = 21 + (180 - 162) = 39° Is that correct sir?
Originally Posted by Adikz Thanks for the diagram sir.. ∠ABC = 21 + (180 - 162) = 39° Is that correct sir? if you mean the acute angle at the top of the position triangle, yes.
Using the law of cosines: This will be my new formula right? b = √(210² + 147² - 2 * 210 * 147 * cos 39°)
formula is a^2+b^2 - c^2 / 2ab = cos 39 deg Solve for c given a and b
Originally Posted by Adikz Using the law of cosines: This will be my new formula right? b = √(210² + 147² - 2 * 210 * 147 * cos 39°) correct
Thanks..
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