bookishkelly
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How do I answer this Trig function question?
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OK. This problem is designed to show a very simple practical application of trig functions (indeed this trype of problem is why trig functions were originally invented over 2000 years ago).
What is the measure of each angle in the triangle ABC?
The cosine of angle ACB = the ratio of what 2 sides of the triangle?
The sine of angle BAC = the ratio of what 2 of the triangle?
How do those help answer this problem?
bookishkelly
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What is the measure of each angle in the triangle ABC?= m<A is 25, m<b is 90, m,C is 65
The cosine of angle ACB = the ratio of what 2 sides of the triangle? cosine of angle ACB is 25/hypotenuse
The sine of angle BAC = the ratio of what 2 of the triangle? - sine of angle BAC is opposite leg/hypotenuse
How do those help answer this problem? - Unfortunately, I still don't know how to solve the problem. The sides don't have any values so I can't plug any number in.
The cosine of angle ACB = the ratio of what 2 sides of the triangle? cosine of angle ACB is 25/hypotenuse
The sine of angle BAC = the ratio of what 2 of the triangle? - sine of angle BAC is opposite leg/hypotenuse
How do those help answer this problem? - Unfortunately, I still don't know how to solve the problem. The sides don't have any values so I can't plug any number in.
D
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Are you allowed to use calculator?
bookishkelly
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Yes I can use a calculator
But there are no values on the triangle other than 25 and 65 degrees, so what would be the point of using it? Just like I said in the previous comment, I tried using sine and cosine to work out the problems but it doesn't work without any numbers.
But there are no values on the triangle other than 25 and 65 degrees, so what would be the point of using it? Just like I said in the previous comment, I tried using sine and cosine to work out the problems but it doesn't work without any numbers.
Well it did not help that I did not indicate the functions most useful for this problem. Not the most helpful clue. Sorry about that.
\(\displaystyle x = \text {length of side AC} \implies\)
\(\displaystyle sin(65^o) = \dfrac{25}{x} \implies x * sin(65^o) = 25 \implies x = \dfrac{25}{sin(65^o)} = WHAT?\)
The sine is a function. You ask it a number, and it responds with a number. To use the trigonometric functions, you need tables or a good calculator.
\(\displaystyle sin(65^o) \approx 0.9063 \implies x \approx \dfrac{25}{0.9063} \approx 27.58.\)
How do you check your work? By using the cosine and the Pythagorean Theorem.
\(\displaystyle y = \text {length of side BC} \implies\)
\(\displaystyle cos(65^o) = \dfrac{y}{x} \implies y = x * cos(65^o) \approx 11.66.\)
\(\displaystyle \sqrt{11.66^2 + 25^2} = \sqrt{135.9556 + 625} = \sqrt{760.9556} \approx 27.58.\)
They match.
Again, I apologize for not providing the best hint.
a) the question is what is the length of side ac
sin65=ab/ac; you know ab=25 feet, solve for ac
b) the question is what is the length of side ad
sin62=ad/ab; you know ab=25 feet, solve for ad
redraw the picture for yourself with sides labeled,
and apply sin=opposite/hypotenuse.
[rule 1- always draw a picture, do not spare the pencils and paper.]
it will come to you.
Maybe in a few more years.
bookishkelly
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Thank you to both!!
I was foolishly forgetting that the calculator determines the functions for you.
I was foolishly forgetting that the calculator determines the functions for you.
I disagree.
Learning math and geometry/trig is a series of small steps of learning.
Each step concludes with an 'aha' moment, when what seemed to be complex becomes simple.