This problem asks for the correct setup to find the opposite side using trigonometry. We are given the adjacent side PQ = 12 and the angle at P = 22°, and need to find the opposite side QR. Since we have the adjacent side and need the opposite side, we use the tangent ratio: tan(angle) = opposite/adjacent. The correct setup is: tan(22°) = QR/12, which can be rearranged to find QR. This is justified because tangent relates the opposite side to the adjacent side. A common mistake is using sine or cosine, which would require the hypotenuse. When setting up trigonometric equations, identify which sides you have and need relative to the given angle.

This problem requires finding the side opposite to a given angle in a right triangle. We have hypotenuse BC = 30, angle C = 22°, and angle A is the right angle. Since AB is opposite to angle C and BC is the hypotenuse, we use sine: sin(C) = opposite/hypotenuse = AB/BC. Rearranging: AB = BC × sin(C) = 30sin(22°). The answer is 30sin(22°) because sine relates the opposite side to the hypotenuse. A common error would be using cosine (30cos(22°)), which would give the adjacent side AC instead. When solving with trigonometry, always identify which side you need relative to the given angle: opposite uses sine, adjacent uses cosine.

Since angles A and B are complementary in a right triangle, $$\cos(B) = \sin(A) = \frac{3}{5}$$. To find $$\sin(B)$$, we use $$\sin(B) = \cos(A)$$. Since $$\sin(A) = \frac{3}{5}$$, we can find $$\cos(A)$$ using the Pythagorean identity: $$\cos(A) = \sqrt{1 - \sin^2(A)} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$. Therefore, $$\sin(B) = \frac{4}{5}$$, and $$\cos(B) + \sin(B) = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}$$. Choice B gives only $$\cos(A)$$, choice C gives only $$\sin(A)$$, and choice D incorrectly adds the squares.

When you encounter trigonometric expressions involving complementary angles like $$90° - \alpha$$, remember that complementary angle identities are key: $$\sin(90° - \alpha) = \cos(\alpha)$$ and $$\cos(90° - \alpha) = \sin(\alpha)$$.

Since you're given $$\cos(\alpha) = \frac{5}{13}$$ with $$\alpha$$ acute, you first need to find $$\sin(\alpha)$$. Using the Pythagorean identity $$\sin^2(\alpha) + \cos^2(\alpha) = 1$$:

$$\sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}$$

Since $$\alpha$$ is acute, $$\sin(\alpha) = \frac{12}{13}$$.

Now you can evaluate the expression:

$$\sin(90° - \alpha) + \cos(90° - \alpha) = \cos(\alpha) + \sin(\alpha) = \frac{5}{13} + \frac{12}{13} = \frac{17}{13}$$

Looking at the wrong answers: Choice A ($$\frac{18}{13}$$) likely comes from incorrectly calculating $$\sin(\alpha)$$ or making an arithmetic error. Choice B ($$\frac{12}{13}$$) represents just $$\sin(\alpha)$$ alone, missing the cosine term entirely. Choice D ($$1$$) might result from incorrectly thinking that $$\sin(90° - \alpha) + \cos(90° - \alpha)$$ equals the Pythagorean identity, but that's $$\sin^2(\alpha) + \cos^2(\alpha) = 1$$, not the sum of the functions themselves.

The correct answer is C.

Study tip: Always memorize complementary angle identities and practice finding missing trigonometric values using the Pythagorean identity—these appear frequently together on geometry exams.

This problem involves finding the hypotenuse of a right triangle using trigonometry. We are given the adjacent side DE = 9 and the angle at D = 35°, and need to find the hypotenuse DF. Since we have the adjacent side and need the hypotenuse, we use the cosine ratio: cos(angle) = adjacent/hypotenuse. Setting up the equation: cos(35°) = 9/DF, which rearranges to DF = 9/cos(35°). This is justified because cosine relates the adjacent side to the hypotenuse in a right triangle. A common mistake is using sine instead of cosine, which would give 9/sin(35°). Before computing, identify which sides you have relative to the given angle to select the correct trigonometric ratio.

This problem requires finding the hypotenuse when given a leg and an acute angle. We have a right angle at S, leg RS = 10, and angle R = 28°. Since we know an angle and a leg, we use trigonometry. To find hypotenuse RT, we note that RS is adjacent to angle R, so we use cosine: cos(28°) = adjacent/hypotenuse = 10/RT. Solving for RT: RT = 10/cos(28°). This correctly isolates the hypotenuse by dividing the adjacent side by cosine. A common error would be multiplying by cosine (10·cos(28°)), which would give a smaller value. When finding the hypotenuse using trigonometry, you divide the known side by the appropriate trig function.

This problem asks us to find the hypotenuse of a right triangle using the Pythagorean theorem. We're given the two legs: AB = 9 and BC = 12, with angle B being the right angle. Since we have both legs and need the hypotenuse, we apply the Pythagorean theorem: c² = a² + b². Setting up the equation: AC² = AB² + BC² = 9² + 12² = 81 + 144 = 225, so AC = √225 = 15. The answer is 15 because we take the positive square root when finding a length. A common error would be to subtract instead of add (√63 = √(144-81)), which happens when students confuse finding a leg versus finding the hypotenuse. Remember: when finding the hypotenuse, always add the squares of the legs.

This problem requires finding an angle when given two sides of a right triangle. We have leg GH = 4 (adjacent to angle G) and hypotenuse GI = 9. Since we have the adjacent side and hypotenuse, we use cosine: cos(G) = adjacent/hypotenuse = 4/9. To find angle G, we use the inverse cosine: G = cos⁻¹(4/9). This gives the angle measure in degrees. A common error is using regular cosine cos(4/9) instead of inverse cosine, or choosing the wrong trig function. When finding angles from sides, use inverse trig functions with the correct ratio.

Solving right triangles involves using the Pythagorean theorem or trigonometry to find unknown sides or angles. In this problem, we are given the hypotenuse GI = 15 and one leg GH = 9. Since we have the hypotenuse and one leg, and need the other leg HI, the Pythagorean theorem applies. The equation is HI² = 15² - 9². This setup correctly finds HI, matching the method in choice A. A common misconception is assuming a special triangle like 30-60-90 without angle information, as in choice D. To transfer this strategy, always choose whether to use the Pythagorean theorem or trig ratios based on the given information before computing.

Solving right triangles involves using the Pythagorean theorem or trigonometry to find unknown sides or angles. In this problem, we are given the hypotenuse $ \overline{JL} = 10 $ and the acute angle at $ J = 35^\circ $. Since we have the hypotenuse and an angle, and need the adjacent leg $ \overline{JK} $, trigonometric ratios apply. The setup is $ \cos(35^\circ) = \frac{\text{JK}}{10} $. This gives $ \text{JK} = 10 \cos(35^\circ) $, matching choice B. A common misconception is using sine for the adjacent side, as in choice A, confusing opposite and adjacent. To transfer this strategy, always choose whether to use the Pythagorean theorem or trig ratios based on the given information before computing.