Find one solution for each equation. Assume all angles involved a...

Question

Find one solution for each equation. Assume all angles involved are acute angles. See Example 3.sec(3β + 10°) = csc(β + 8°)

Accepted Answer

Welcome back. I am so glad you're here. We're asked to solve the following trigonometric equation. Consider all the angles as acute angles. Our equation is the C can of five alpha plus 14 degrees is equal to the K. He can't of alpha plus 40 degrees. Our answer choices are answer choice. A 36 degrees. Answer choice B six degrees. Answer choice, C seven degrees and answer choice D 13 degrees. All right. C can't and cos can't are co functions. So this statement is true if the angle measures so C can't five A five alpha plus 14 degrees added with Koi cans angle alpha plus 40 degrees, add up to 90 degrees. So we can set these two angles added up equal to 90 degrees. And solve for alpha. We have five alpha plus 14 degrees plus alpha plus 40 degrees equals 90 degrees. Now, we can combine like terms five alpha plus alpha is six alphas and then we have 14 degrees plus 40 degrees is a positive 54 degrees. Those together equal 90 degrees. Solving for alpha will subtract 54 degrees from both sides. Now six alphas are equal to 90 degrees minus 54 degrees is 36 degrees, dividing both sides by six and alpha equals 36 degrees divided by six is six degrees. And there's our solution, we look at our answer choices and this matches with answer choice. B well done. We'll catch you on the next one.

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Find one solution for each equation. Assume all angles involved are acute angles. See Example 3.sec(3β + 10°) = csc(β + 8°)

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