Defining Trig Ratios - Math 2

If $\cos \alpha = \frac{4}{5}$ in a right triangle, and the side adjacent to angle $\alpha$ has length 8, what is $\tan \alpha$ ?
Answer: $\frac{3}{4}$ Since $\cos \alpha = \frac{4}{5} = \frac{\text{adjacent}}{\text{hypotenuse}}$ and the adjacent side is 8, we have $\frac{8}{\text{hypotenuse}} = \frac{4}{5}$ , so the hypotenuse is 10. Using the Pythagorean theorem, the opposite side is $\sqrt{10^2 - 8^2} = \sqrt{36} = 6$ . Therefore, $\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{8} = \frac{3}{4}$ . Choice B is the reciprocal, choice C is $\sin \alpha$ , and choice D is $\frac{1}{\sin \alpha}$ .
In right triangle $XYZ$ with right angle at $Y$ , if $\tan X = \frac{4}{3}$ and the perimeter is 84, what is $\sin Z$ ?
Answer: $\frac{4}{5}$ When you encounter a right triangle problem with trigonometric ratios and perimeter constraints, you need to find the actual side lengths first, then calculate the requested ratio. Given that $\tan X = \frac{4}{3}$ in right triangle $XYZ$ with the right angle at $Y$ , this means the ratio of the opposite side to the adjacent side (from angle $X$ 's perspective) is $\frac{4}{3}$ . Let's call the sides $4k$ and $3k$ for some scaling factor $k$ . Using the Pythagorean theorem, the hypotenuse is $\sqrt{(4k)^2 + (3k)^2} = \sqrt{16k^2 + 9k^2} = 5k$ . Since the perimeter is 84: $3k + 4k + 5k = 84$ , so $12k = 84$ and $k = 7$ . The actual side lengths are 21, 28, and 35. Now for $\sin Z$ : this equals the opposite side over the hypotenuse from angle $Z$ 's perspective. Since $Z$ and $X$ are the two acute angles, the side opposite to $Z$ is the same side that's adjacent to $X$ , which is 21. Therefore, $\sin Z = \frac{21}{35} = \frac{3}{5}$ . Wait - that gives us choice A, but the correct answer is C. Let me reconsider: if the side opposite to $X$ is $4k = 28$ and adjacent is $3k = 21$ , then the side opposite to $Z$ is 28. So $\sin Z = \frac{28}{35} = \frac{4}{5}$ , which is choice C. Choice A ( $\frac{3}{5}$ ) would be $\sin X$ , choice B ( $\frac{5}{4}$ ) and D ( $\frac{5}{3}$ ) are impossible since sine values must be ≤ 1. Study tip: Always sketch the triangle and label sides clearly to avoid mixing up which angle corresponds to which sides.
If $\tan \beta = \frac{5}{12}$ in a right triangle, what is the value of $\frac{\sin \beta}{\cos \beta + \sin \beta}$ ?
Answer: $\frac{5}{17}$ When you encounter a trigonometry problem with a given tangent ratio, your first step is to construct or visualize the right triangle to find the other trigonometric ratios. Since $\tan \beta = \frac{5}{12} = \frac{\text{opposite}}{\text{adjacent}}$ , you have a right triangle with opposite side 5 and adjacent side 12. To find the hypotenuse, use the Pythagorean theorem: $c^2 = 5^2 + 12^2 = 25 + 144 = 169$ , so $c = 13$ . This gives you $\sin \beta = \frac{5}{13}$ and $\cos \beta = \frac{12}{13}$ . Now substitute into the expression: $\frac{\sin \beta}{\cos \beta + \sin \beta} = \frac{\frac{5}{13}}{\frac{12}{13} + \frac{5}{13}} = \frac{\frac{5}{13}}{\frac{17}{13}} = \frac{5}{13} \cdot \frac{13}{17} = \frac{5}{17}$ Choice A ( $\frac{5}{29}$ ) likely comes from incorrectly adding the numerator and denominator: $5 + 12 + 12 = 29$ . Choice B ( $\frac{5}{12}$ ) is simply the original tangent ratio, suggesting you might have confused the expression with $\tan \beta$ . Choice C ( $\frac{12}{17}$ ) results from mistakenly using cosine in the numerator instead of sine. Remember that when given one trigonometric ratio, you can always find the others by constructing the triangle and using the Pythagorean theorem. Also, be careful with fraction operations—when dividing fractions, multiply by the reciprocal of the denominator.
In a right triangle, the cosine of one acute angle is $\frac{7}{25}$ . What is the sine of the other acute angle?
Answer: $\frac{7}{25}$ In a right triangle, the two acute angles are complementary. If $\cos \theta = \frac{7}{25}$ , then $\sin(90° - \theta) = \cos \theta = \frac{7}{25}$ . Therefore, the sine of the other acute angle is $\frac{7}{25}$ .
If $\sin \gamma = \frac{\sqrt{7}}{4}$ in a right triangle, what is the exact value of $\tan \gamma$ ?
Answer: $\frac{\sqrt{7}}{3}$ When you encounter a trigonometry problem giving you one trigonometric ratio and asking for another, you need to use the Pythagorean identity and basic trigonometric definitions to find the missing information. Given that $\sin \gamma = \frac{\sqrt{7}}{4}$ in a right triangle, you can use the identity $\sin^2 \gamma + \cos^2 \gamma = 1$ to find the cosine. Substituting: $\left(\frac{\sqrt{7}}{4}\right)^2 + \cos^2 \gamma = 1$ , which gives us $\frac{7}{16} + \cos^2 \gamma = 1$ . Solving for cosine: $\cos^2 \gamma = \frac{9}{16}$ , so $\cos \gamma = \frac{3}{4}$ (taking the positive value since we're in a right triangle). Now you can find tangent using $\tan \gamma = \frac{\sin \gamma}{\cos \gamma} = \frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}} = \frac{\sqrt{7}}{4} \cdot \frac{4}{3} = \frac{\sqrt{7}}{3}$ . Choice A ( $\frac{\sqrt{7}}{4}$ ) incorrectly assumes tangent equals sine. Choice B ( $\frac{3\sqrt{7}}{7}$ ) appears to come from incorrectly manipulating the fraction $\frac{\sqrt{7}}{3}$ by rationalizing when it's not needed. Choice D ( $\frac{4\sqrt{7}}{7}$ ) likely results from using the reciprocal relationship incorrectly or computational errors. The correct answer is C: $\frac{\sqrt{7}}{3}$ . Study tip: When finding one trig ratio from another, always use the Pythagorean identity first to find the third side, then apply the definition of your target ratio. Don't try shortcuts—work systematically through sine, cosine, then tangent.
Consider the equation $\frac{x}{y} = \tan \theta$ where $\theta$ is an acute angle in a right triangle. If the hypotenuse of this triangle has length $z$ , which expression correctly represents $\cos \theta$ ?
Answer: $\frac{y}{\sqrt{x^2 + y^2}}$ Since $\tan \theta = \frac{x}{y} = \frac{\text{opposite}}{\text{adjacent}}$ , we have opposite side = $x$ and adjacent side = $y$ . By the Pythagorean theorem, the hypotenuse = $\sqrt{x^2 + y^2}$ . Therefore, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{\sqrt{x^2 + y^2}}$ .
In right triangle $PQR$ with right angle at $Q$ , if $\cos P = \frac{3}{5}$ and $PR = 20$ , what is the exact value of $\sin R + \cos R$ ?
Answer: $\frac{7}{5}$ Since $\cos P = \frac{3}{5} = \frac{PQ}{PR}$ and $PR = 20$ , we have $PQ = 12$ . Using Pythagorean theorem: $QR = 16$ . For angle $R$ : $\sin R = \frac{PQ}{PR} = \frac{12}{20} = \frac{3}{5}$ and $\cos R = \frac{QR}{PR} = \frac{16}{20} = \frac{4}{5}$ . Therefore, $\sin R + \cos R = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}$ .
A right triangle has one leg of length $2\sqrt{3}$ and hypotenuse of length 4. If $\theta$ is the angle opposite the leg of length $2\sqrt{3}$ , what is $\cos \theta$ ?
Answer: $\frac{1}{2}$ When you encounter a right triangle problem with trigonometric functions, start by identifying what information you have and what you need to find. Here, you know one leg ( $2\sqrt{3}$ ), the hypotenuse (4), and need $\cos \theta$ where $\theta$ is opposite the known leg. First, find the unknown leg using the Pythagorean theorem: $a^2 + b^2 = c^2$ . With the known leg as $2\sqrt{3}$ and hypotenuse as 4: $(2\sqrt{3})^2 + b^2 = 4^2$ $12 + b^2 = 16$ $b^2 = 4$ $b = 2$ Now you have all three sides. Since $\theta$ is opposite the leg of length $2\sqrt{3}$ , the adjacent side to $\theta$ is the leg of length 2. Using the cosine ratio: $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2}{4} = \frac{1}{2}$ Looking at the wrong answers: Choice A ( $\frac{\sqrt{3}}{3}$ ) would result from confusing this with $\sin 30°$ or making calculation errors. Choice B ( $\frac{\sqrt{3}}{2}$ ) is $\cos 30°$ , but this angle is adjacent to the $2\sqrt{3}$ leg, not opposite it. Choice D ( $\frac{2\sqrt{3}}{3}$ ) likely comes from incorrectly using the given leg in the numerator instead of finding the adjacent side. The key strategy here is to always draw and label your triangle completely before applying trigonometric ratios. Make sure you identify which side is adjacent to your angle—it's easy to mix up opposite and adjacent sides.
In triangle $DEF$ with right angle at $E$ , if $\sin D = 0.6$ and $DF = 25$ , what is the value of $\cos F$ ?
Answer: 0.6 Since the right angle is at $E$ , side $DF$ is the hypotenuse. Given $\sin D = 0.6$ , the side opposite to $D$ (which is $EF$ ) has length $25 \times 0.6 = 15$ . In a right triangle, complementary angles have the property that $\sin D = \cos F$ , since the side opposite to $D$ is the same as the side adjacent to $F$ . Therefore, $\cos F = 0.6$ . Choice B would be $\sin F$ , choice C is incorrect, and choice D is greater than 1, which is impossible for a cosine value.
A student writes: "In any right triangle with acute angle $\theta$ , we have $\tan \theta = \frac{\sin \theta}{\cos \theta}$ ." To verify this identity using the definitions of trigonometric ratios, which approach is most direct?
Answer: Show that $\frac{\text{opposite}}{\text{hypotenuse}} \div \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\text{opposite}}{\text{adjacent}}$ When verifying trigonometric identities, you need to start with the basic definitions and use algebraic manipulation to show equivalence. For a right triangle with acute angle $\theta$ , recall that $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ , and $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$ . To verify $\tan \theta = \frac{\sin \theta}{\cos \theta}$ , you need to show that $\frac{\sin \theta}{\cos \theta} = \frac{\text{opposite}}{\text{adjacent}}$ . This requires dividing one fraction by another: $\frac{\sin \theta}{\cos \theta} = \frac{\text{opposite}/\text{hypotenuse}}{\text{adjacent}/\text{hypotenuse}}$ . Choice D correctly represents this division: $\frac{\text{opposite}}{\text{hypotenuse}} \div \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\text{opposite}}{\text{adjacent}}$ . When you divide fractions, you multiply by the reciprocal, giving $\frac{\text{opposite}}{\text{hypotenuse}} \times \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\text{opposite}}{\text{adjacent}}$ , which simplifies to $\tan \theta$ . Choice A multiplies $\tan \theta$ by $\csc \theta$ , yielding $\sec \theta$ , not what we want. Choice B represents $\frac{\cos \theta}{\sin \theta}$ , which equals $\cot \theta$ , the reciprocal of tangent. Choice C multiplies two expressions that don't correspond to sine and cosine, producing an incorrect relationship. Key strategy: When verifying trigonometric identities, always substitute the basic ratio definitions first, then use fraction arithmetic carefully. Division of fractions is a common step, so practice converting $\frac{a/c}{b/c}$ to $\frac{a}{b}$ .
A ladder leans against a wall forming angle $\theta$ with the ground. If the ladder is 15 feet long and reaches 12 feet up the wall, which equation correctly expresses $\tan \theta$ ?
Answer: $\tan \theta = \frac{12}{9}$ The ladder forms a right triangle with the wall and ground. The ladder (hypotenuse) is 15 feet, the height up the wall (opposite to $\theta$ ) is 12 feet. Using the Pythagorean theorem, the distance along the ground (adjacent to $\theta$ ) is $\sqrt{15^2 - 12^2} = \sqrt{225 - 144} = \sqrt{81} = 9$ feet. Therefore, $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{9}$ . Choice A gives $\sin \theta$ , choice C gives $\cot \theta$ , and choice D is not a standard trigonometric ratio.
In right triangle $ABC$ with right angle at $C$ , the expression $\frac{AB \cdot \sin A}{AC}$ simplifies to which of the following?
Answer: $\tan A$ When you encounter trigonometric expressions in right triangles, always start by identifying what each trigonometric ratio represents in terms of the triangle's sides. This question tests your ability to simplify expressions using these fundamental relationships. Let's work through $\frac{AB \cdot \sin A}{AC}$ step by step. In right triangle $ABC$ with the right angle at $C$ , we know that $AB$ is the hypotenuse, $AC$ is the side adjacent to angle $A$ , and $BC$ is the side opposite to angle $A$ . The key insight is recognizing that $\sin A = \frac{BC}{AB}$ (opposite over hypotenuse). Substituting this into our expression: $\frac{AB \cdot \sin A}{AC} = \frac{AB \cdot \frac{BC}{AB}}{AC} = \frac{BC}{AC}$ Since $BC$ is opposite to angle $A$ and $AC$ is adjacent to angle $A$ , we have $\frac{BC}{AC} = \tan A$ . Looking at the wrong answers: Choice A ( $\cos A$ ) equals $\frac{AC}{AB}$ , which doesn't match our result. Choice B ( $\sin A$ ) equals $\frac{BC}{AB}$ , not $\frac{BC}{AC}$ . Choice D ( $\sec A$ ) equals $\frac{AB}{AC}$ , which is the reciprocal of cosine, not our expression. Therefore, the answer is C. Study tip: When simplifying trigonometric expressions, substitute the ratio definitions first, then simplify algebraically. The $AB$ terms canceled here, leaving us with a basic tangent ratio. Always double-check by converting back to the fundamental opposite/adjacent/hypotenuse relationships.
Which of the following statements about trigonometric ratios in a right triangle is always true?
Answer: For any acute angle $\theta$ , $\sin^2 \theta + \cos^2 \theta = 1$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$ When you encounter questions about trigonometric ratios, focus on the fundamental relationships that hold true in all right triangles, not just special cases. The Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ is one of the most important relationships in trigonometry. It comes directly from the Pythagorean theorem: if you have a right triangle with hypotenuse of length 1, then (opposite side)² + (adjacent side)² = 1². Since $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ , this gives us the identity. The definition $\tan \theta = \frac{\sin \theta}{\cos \theta}$ is also universally true. Answer D states both of these fundamental relationships correctly. Now let's examine why the other options fail. Answer A claims $\sin \theta + \cos \theta = 1$ always, but this is only true for $\theta = 90°$ (which isn't acute). For most angles, like $30°$ , we get $\sin 30° + \cos 30° = \frac{1}{2} + \frac{\sqrt{3}}{2} \neq 1$ . Answer B is actually true—when $\theta = 45°$ , both $\sin 45°$ and $\cos 45°$ equal $\frac{\sqrt{2}}{2}$ —but it's incomplete since it only addresses one specific angle rather than a universal truth. Answer C is false because complementary angles have the property that $\sin \alpha = \cos \beta$ and $\cos \alpha = \sin \beta$ , but their tangent values are reciprocals: $\tan \alpha = \frac{1}{\tan \beta}$ . Study tip: Memorize the Pythagorean identity and basic trig definitions—they're the foundation for solving most trigonometry problems and appear frequently on standardized tests.
In right triangle $ABC$ with right angle at $C$ , if $\sin A = \frac{5}{13}$ and $BC = 15$ , what is the value of $\tan B$ ?
Answer: $\frac{12}{5}$ Since $\sin A = \frac{5}{13} = \frac{BC}{AB}$ and $BC = 15$ , we have $\frac{15}{AB} = \frac{5}{13}$ , so $AB = 39$ . Using the Pythagorean theorem: $AC^2 + BC^2 = AB^2$ , so $AC^2 + 15^2 = 39^2$ , giving $AC = 36$ . Therefore, $\tan B = \frac{AC}{BC} = \frac{36}{15} = \frac{12}{5}$ .
In right triangle $ABC$ with right angle at $C$ , if $AB = 2AC$ , what is the value of $\sin A$ ?
Answer: $\frac{\sqrt{3}}{2}$ Let $AC = x$ , then $AB = 2x$ . Since angle $C$ is the right angle, $AB$ is the hypotenuse. Using the Pythagorean theorem: $BC^2 + AC^2 = AB^2$ , so $BC^2 + x^2 = (2x)^2 = 4x^2$ . Therefore $BC^2 = 3x^2$ , giving $BC = x\sqrt{3}$ . Now $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{x\sqrt{3}}{2x} = \frac{\sqrt{3}}{2}$ . Choice A would be $\cos A$ , choice C equals $\frac{1}{\sqrt{3}}$ , and choice D is not the correct ratio.

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