Algebra 3 Quiz: Right Triangle Trigonometry
12 questions · exam conditions
0:00
Right Triangle TrigonometryQuestion 1 of 12

From the top of a 42-meter cliff, a lifeguard observes a kayak at an angle of depression of 23∘23^\circ. The kayak then paddles straight toward the cliff until the angle of depression is 46∘46^\circ. Which expression represents the distance, in meters, the kayak traveled?

42(tan⁡46∘−tan⁡23∘)42(\tan 46^\circ - \tan 23^\circ)
42(cot⁡23∘−cot⁡46∘)42(\cot 23^\circ - \cot 46^\circ)
42(tan⁡23∘−tan⁡46∘)42(\tan 23^\circ - \tan 46^\circ)
42(cot⁡46∘−cot⁡23∘)42(\cot 46^\circ - \cot 23^\circ)
← Back to quizzes

Algebra 3 Quiz

Algebra 3 Quiz: Right Triangle Trigonometry

Practice Right Triangle Trigonometry in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Right Triangle Trigonometry, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

From the top of a 42-meter cliff, a lifeguard observes a kayak at an angle of depression of 23∘23^\circ. The kayak then paddles straight toward the cliff until the angle of depression is 46∘46^\circ. Which expression represents the distance, in meters, the kayak traveled?

  1. 42(tan⁡46∘−tan⁡23∘)42(\tan 46^\circ - \tan 23^\circ)
  2. 42(cot⁡23∘−cot⁡46∘)42(\cot 23^\circ - \cot 46^\circ) (correct answer)
  3. 42(tan⁡23∘−tan⁡46∘)42(\tan 23^\circ - \tan 46^\circ)
  4. 42(cot⁡46∘−cot⁡23∘)42(\cot 46^\circ - \cot 23^\circ)
Explanation: When you see an angle of depression problem, remember that the angle of depression from the cliff equals the angle of elevation from the kayak back to the top. That means each observation creates a right triangle with vertical height 42 meters and horizontal distance unknown. Since tan⁡(angle)=verticalhorizontal\tan(\text{angle}) = \frac{\text{vertical}}{\text{horizontal}}, the horizontal distance is 42tan⁡(angle)=42cot⁡(angle)\frac{42}{\tan(\text{angle})} = 42\cot(\text{angle}). At 23∘23^\circ, the kayak is 42cot⁡23∘42\cot 23^\circ meters from the cliff base. After moving straight closer, the angle increases to 46∘46^\circ, so the new distance is 42cot⁡46∘42\cot 46^\circ. The distance traveled is the original distance minus the final distance: 42cot⁡23∘−42cot⁡46∘=42(cot⁡23∘−cot⁡46∘)42\cot 23^\circ - 42\cot 46^\circ = 42(\cot 23^\circ - \cot 46^\circ). The choice 42(cot⁡46∘−cot⁡23∘)42(\cot 46^\circ - \cot 23^\circ) reverses the subtraction, giving a negative distance. The choices with tan⁡\tan instead of cot⁡\cot confuse the ratio: tan⁡\tan gives 42d\frac{42}{d}, so multiplying by 42 would not produce a distance. Specifically, 42(tan⁡46∘−tan⁡23∘)42(\tan 46^\circ - \tan 23^\circ) and 42(tan⁡23∘−tan⁡46∘)42(\tan 23^\circ - \tan 46^\circ) both use the wrong trigonometric function; the latter also has the wrong order. Your strategy: draw the right triangle, label the known vertical side and the unknown horizontal side, then decide which trig function relates them. When the unknown is the horizontal distance and the known side is vertical, use cotangent. Also, as the kayak moves closer, the angle of depression increases, so the distance traveled must be the larger initial distance minus the smaller final distance.

Question 2

A tree casts a shadow of length xx feet when the sun's angle of elevation is 41∘41^\circ. Later, the shadow is 8 feet shorter and the sun's angle of elevation is 58∘58^\circ. Which equation correctly relates the height hh of the tree to these two measurements?

  1. h=xtan⁡41∘=(x−8)tan⁡58∘h = x\tan 41^\circ = (x-8)\tan 58^\circ (correct answer)
  2. h=xcot⁡41∘=(x−8)cot⁡58∘h = x\cot 41^\circ = (x-8)\cot 58^\circ
  3. h=xtan⁡41∘=(x+8)tan⁡58∘h = x\tan 41^\circ = (x+8)\tan 58^\circ
  4. h=xtan⁡41∘=x−8tan⁡58∘h = \frac{x}{\tan 41^\circ} = \frac{x-8}{\tan 58^\circ}
Explanation: When you see a shadow-and-angle-of-elevation problem, think of a right triangle: the tree is the vertical side, the shadow is the horizontal side, and the angle of elevation is at the tip of the shadow. The key relationship is tan⁡(angle)=heightshadow length\tan(\text{angle}) = \frac{\text{height}}{\text{shadow length}}, so solving for height gives h=shadow⋅tan⁡(angle)h = \text{shadow} \cdot \tan(\text{angle}). Initially the shadow is xx, so h=xtan⁡41∘h = x\tan 41^\circ. Later the shadow is 8 feet shorter, so its length is x−8x-8, and with angle 58∘58^\circ, h=(x−8)tan⁡58∘h = (x-8)\tan 58^\circ. Therefore the correct equation is h=xtan⁡41∘=(x−8)tan⁡58∘h = x\tan 41^\circ = (x-8)\tan 58^\circ. The choice h=xcot⁡41∘=(x−8)cot⁡58∘h = x\cot 41^\circ = (x-8)\cot 58^\circ uses cotangent, which would relate the shadow to the height incorrectly; cot⁡θ=shadowheight\cot\theta = \frac{\text{shadow}}{\text{height}}, not the height itself. The choice h=xtan⁡41∘=(x+8)tan⁡58∘h = x\tan 41^\circ = (x+8)\tan 58^\circ uses x+8x+8, but the shadow got shorter, not longer, so it must be x−8x-8. Finally, h=xtan⁡41∘=x−8tan⁡58∘h = \frac{x}{\tan 41^\circ} = \frac{x-8}{\tan 58^\circ} divides by tangent instead of multiplying, which reverses the relationship and cannot give the tree's height. Your takeaway: always identify the shadow as the adjacent side, then use h=shadow⋅tan⁡θh = \text{shadow}\cdot\tan\theta. Watch for the words "shorter" or "longer" to get the correct expression for the new shadow length.

Question 3

To measure the distance across a river, a surveyor stands at point AA on one bank and locates a tree at point BB directly across the river. She walks 80 meters along the bank to point CC, where ∠ACB=64∘\angle ACB = 64^\circ. Which expression gives the straight-line distance from CC to the tree?

  1. 80tan⁡64∘80\tan 64^\circ
  2. 80cos⁡64∘\frac{80}{\cos 64^\circ} (correct answer)
  3. 80sin⁡64∘80\sin 64^\circ
  4. 80cos⁡64∘80\cos 64^\circ
Explanation: This is a right-triangle trigonometry setup. The moment you read "directly across," you know ABAB is perpendicular to the bank, so the right angle is at AA. Therefore triangle ABCABC has hypotenuse BCBC, which is exactly the distance from CC to the tree, and AC=80AC = 80 is the leg adjacent to the 64∘64^\circ angle at CC. Since cos⁡64∘=adjacenthypotenuse=80BC,\cos 64^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{80}{BC}, you solve to get BC=80cos⁡64∘.BC = \frac{80}{\cos 64^\circ}. That is why the correct expression is 80cos⁡64∘\frac{80}{\cos 64^\circ}. Now for the traps: 80tan⁡64∘80\tan 64^\circ uses tangent as opposite over adjacent, so it computes ABAB, the width of the river, not the distance from CC to the tree. 80sin⁡64∘80\sin 64^\circ misapplies sine, which would involve the opposite side ABAB over the hypotenuse BCBC; multiplying the known adjacent side by sine doesn't give either side. 80cos⁡64∘80\cos 64^\circ reverses the cosine relationship — cosine is adjacent over hypotenuse, so multiplying the known adjacent side by cosine gives no meaningful side here; you need to divide, not multiply. Strategy: on any right-triangle word problem, sketch the triangle and label the right angle first. Then ask: which side do I know, which side do I want, and what trig ratio connects them? Here the key insight is that the line to the tree across the river is the hypotenuse, not a leg.

Question 4

A road sign says the road has a 35% grade, meaning it rises 35 feet for every 100 feet of horizontal run. Which expression gives the angle of inclination θ\theta of the road?

  1. θ=cos⁡−1(0.35)\theta = \cos^{-1}(0.35)
  2. θ=sin⁡−1(0.35)\theta = \sin^{-1}(0.35)
  3. θ=tan⁡−1(0.35)\theta = \tan^{-1}(0.35) (correct answer)
  4. θ=tan⁡−1(100/35)\theta = \tan^{-1}(100/35)
Explanation: Whenever you see a percent grade, translate it as slope: rise over horizontal run. A 35% grade means 35 feet up for every 100 feet forward. In the right triangle formed by the road, the angle of inclination θ\theta has opposite side 35 and adjacent side 100, so tan⁡θ=oppositeadjacent=35100=0.35.\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{35}{100} = 0.35. To undo the tangent and solve for θ\theta, take the inverse tangent: θ=tan⁡−1(0.35).\theta = \tan^{-1}(0.35). That is why the choice with tan⁡−1(0.35)\tan^{-1}(0.35) is correct. The other expressions mix up which trig ratio corresponds to the given numbers. θ=sin⁡−1(0.35)\theta = \sin^{-1}(0.35) would treat 0.35 as opposite over hypotenuse, but 100 is the horizontal run, not the road's length. θ=cos⁡−1(0.35)\theta = \cos^{-1}(0.35) would treat 0.35 as adjacent over hypotenuse, again using the wrong sides. And θ=tan⁡−1(100/35)\theta = \tan^{-1}(100/35) uses the reciprocal slope, run over rise, giving about 70.7∘70.7^\circ — far too steep for a 35% grade. The correct angle is about 19.3∘19.3^\circ. Study tip: when you see "grade" or "slope," think tangent. Write the ratio riserun\frac{\text{rise}}{\text{run}}, then apply the corresponding inverse trig function.

Question 5

An isosceles triangle has two equal sides of length 10 and a vertex angle of 40∘40^\circ. Which expression gives the length of its base?

  1. 20tan⁡20∘20\tan 20^\circ
  2. 20cos⁡20∘20\cos 20^\circ
  3. 10sin⁡40∘10\sin 40^\circ
  4. 20sin⁡20∘20\sin 20^\circ (correct answer)
Explanation: Whenever you see an isosceles triangle with a vertex angle, the key move is to drop an altitude from the vertex to the base. That altitude does two things: it bisects the 40∘40^\circ vertex angle into two 20∘20^\circ angles, and it splits the base into two equal halves. In each resulting right triangle, the hypotenuse is the equal side of length 10, and the side opposite the 20∘20^\circ angle is half the base. Using sine: half-base10=sin⁡20∘\frac{\text{half-base}}{10}=\sin 20^\circ so half-base is 10sin⁡20∘10\sin 20^\circ, and the full base is 2(10sin⁡20∘)=20sin⁡20∘.2(10\sin 20^\circ)=20\sin 20^\circ. Now look at the traps. 20tan⁡20∘20\tan 20^\circ would be correct only if 10 were the adjacent leg, but 10 is the hypotenuse. 20cos⁡20∘20\cos 20^\circ actually gives twice the altitude, not the base — cosine finds the adjacent side, which is the height. And 10sin⁡40∘10\sin 40^\circ uses the original vertex angle directly, but in the original triangle the sides of length 10 are not a hypotenuse, so applying sine that way skips the essential step of splitting the triangle. For any isosceles base question, split the triangle first. Then identify the hypotenuse, opposite, and adjacent sides in one of the right halves. If you need the side opposite an angle, use sine; if you need the adjacent side, use cosine.

Question 6

A guy wire is attached to the top of a vertical pole and anchored 12 feet from the base of the pole. The wire makes a 55∘55^\circ angle with the pole, not with the ground. Which expression gives the height of the pole?

  1. 12tan⁡55∘12\tan 55^\circ
  2. 12cot⁡55∘12\cot 55^\circ (correct answer)
  3. 12sin⁡55∘12\sin 55^\circ
  4. 12sin⁡55∘\frac{12}{\sin 55^\circ}
Explanation: Whenever you see a right-triangle word problem like this, first ask: "Which side is known, and which side do I need?" Drawing the pole, ground, and wire helps you see that the pole height is the vertical leg, and the 12-foot anchor distance is the horizontal leg. The wire makes a 55∘55^\circ angle with the pole, so the angle at the top of the pole is between the wire and the vertical height. In that triangle, the height is adjacent to the 55∘55^\circ angle, while the 12-foot ground distance is opposite it. Therefore tan⁡55∘=12h\tan 55^\circ = \frac{12}{h}, so solving for height gives h=12tan⁡55∘=12cot⁡55∘h = \frac{12}{\tan 55^\circ} = 12\cot 55^\circ. That is why 12cot⁡55∘12\cot 55^\circ is correct. The choice 12tan⁡55∘12\tan 55^\circ would be correct only if the wire made 55∘55^\circ with the ground, not with the pole — a classic misreading trap. 12sin⁡55∘12\sin 55^\circ incorrectly treats 12 as the hypotenuse or the wrong side relationship. And 12sin⁡55∘\frac{12}{\sin 55^\circ} actually finds the length of the wire, not the pole height, because it treats 12 as the side opposite the angle and solves for the hypotenuse. Your key strategy: always read whether the given angle is with the ground or with the pole, then label the triangle sides relative to that angle. Once you know opposite, adjacent, and hypotenuse, pick the trig ratio that connects the known side to the unknown side you need.

Question 7

From a point on level ground, a surveyor measures the angle of elevation to the top of a vertical cliff and obtains ́40°. After walking 100 feet directly toward the base of the cliff, she measures the angle of elevation again and obtains ́55°. If both measurements are made from ground level, approximately how tall is the cliff?

  1. 58.9 ft
  2. 169.8 ft
  3. 203.4 ft (correct answer)
  4. 142.8 ft
Explanation: Whenever you see two angle-of-elevation measurements from different ground locations, picture two right triangles that share the same height hh. Let the initial distance from the cliff be xx. After walking 100 feet closer, the distance is x−100x-100. So tan⁡40∘=hx,tan⁡55∘=hx−100.\tan 40^\circ=\frac{h}{x},\qquad \tan 55^\circ=\frac{h}{x-100}. Rewrite these as x=hcot⁡40∘x=h\cot 40^\circ and x−100=hcot⁡55∘x-100=h\cot 55^\circ. Subtracting gives 100=h(cot⁡40∘−cot⁡55∘)⇒h=100cot⁡40∘−cot⁡55∘≈203.4.100=h(\cot 40^\circ-\cot 55^\circ) \quad\Rightarrow\quad h=\frac{100}{\cot 40^\circ-\cot 55^\circ}\approx 203.4. So the cliff is about 203.4 ft tall. The other choices come from common setup errors. The 58.9 ft value is 100(tan⁡55∘−tan⁡40∘)100(\tan 55^\circ-\tan 40^\circ), which subtracts tangent ratios directly, but the 100 feet is a horizontal distance, not a vertical difference. The 142.8 ft value is 100tan⁡55∘100\tan 55^\circ, which incorrectly treats the 100-foot walk as the current horizontal distance from the cliff. The 169.8 ft value is 100/(tan⁡55∘−tan⁡40∘)100/(\tan 55^\circ-\tan 40^\circ), which uses tangent difference in the denominator instead of cotangent difference, reversing the roles of height and horizontal distance. A reliable strategy: in two-angle height problems, express both horizontal distances in terms of hh using cotangent, or set up xtan⁡40∘=(x−100)tan⁡55∘x\tan 40^\circ=(x-100)\tan 55^\circ. The 100-foot difference should become the difference of two distance expressions, never a direct tangent subtraction.

Question 8

A 25-foot ladder is leaning against a vertical wall. The base of the ladder is 7 feet from the wall. Which expression gives the measure of the acute angle between the ladder and the wall?

  1. arcsin⁡(725)\arcsin\left(\frac{7}{25}\right) (correct answer)
  2. arccos⁡(725)\arccos\left(\frac{7}{25}\right)
  3. arctan⁡(725)\arctan\left(\frac{7}{25}\right)
  4. arcsin⁡(2425)\arcsin\left(\frac{24}{25}\right)
Explanation: When you see a ladder leaning against a wall, picture a right triangle: the ladder is the hypotenuse, the wall is one leg, and the ground is the other leg. The key is to identify which angle the question asks for, because that determines which sides are opposite and adjacent. Here, the acute angle is between the ladder and the wall. That angle is at the top of the triangle, where the ladder touches the wall. The side opposite that angle is the distance from the wall to the ladder's base, which is 7 feet. The hypotenuse is the ladder itself, 25 feet. Since sine is opposite over hypotenuse, you get sin⁡(θ)=725\sin(\theta)=\frac{7}{25}, so θ=arcsin⁡(725)\theta=\arcsin\left(\frac{7}{25}\right). Now look at the traps. arccos⁡(725)\arccos\left(\frac{7}{25}\right) would be correct for the angle between the ladder and the ground, because then the 7-foot side is adjacent to the angle. arctan⁡(725)\arctan\left(\frac{7}{25}\right) incorrectly uses the hypotenuse as the adjacent side; tangent is opposite over adjacent, and the adjacent side here is the wall height, 24 feet. arcsin⁡(2425)\arcsin\left(\frac{24}{25}\right) uses the wall height, 24, as the opposite side, which gives the angle with the ground, not the wall. A useful habit: always label the sides relative to the exact angle named in the problem. For "angle between ladder and wall," the opposite side is the ground distance, not the wall height.

Question 9

In a right triangle, an acute angle θ has cos θ = ́3/7. The side adjacent to θ is 9 centimeters long. What is the length of the side opposite θ?

  1. 21
  2. 12√10
  3. 18
  4. 6√10 (correct answer)
Explanation: When you see a right-triangle trig question, translate the ratio into sides right away: cos⁡θ=adjacenthypotenuse\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}. Here cos⁡θ=37\cos\theta=\frac{3}{7}, so think of a ratio triangle with adjacent side 33, hypotenuse 77, and opposite side 72−32=40=210\sqrt{7^2-3^2}=\sqrt{40}=2\sqrt{10}. The given adjacent side is 99, which is 33 times the 33 in that ratio triangle, so the hypotenuse is 7⋅3=217\cdot 3=21. The opposite side is therefore (210)⋅3=610(2\sqrt{10})\cdot 3=6\sqrt{10}. You can also verify with the Pythagorean theorem: 212−92=360=610\sqrt{21^2-9^2}=\sqrt{360}=6\sqrt{10}. The choice 2121 is the hypotenuse, not the opposite side; it comes from computing 93/7\frac{9}{3/7} and stopping. The choice 121012\sqrt{10} is exactly double the correct opposite length, which would result from using a scale factor of 66 or treating the adjacent side as 1818 instead of 99. The choice 1818 is twice the given adjacent side; it cannot be the opposite side because 92+182≠2129^2+18^2\neq 21^2 — the hypotenuse would be 959\sqrt5, not 2121. Study tip: after finding a side from a trig ratio, check it against the triangle. The hypotenuse is the longest side, and the Pythagorean theorem is a quick way to confirm your opposite-side answer.

Question 10

An airplane descends along a straight path that makes a 6∘6^\circ angle with the horizontal. During the descent it loses 1200 feet of altitude. Which expression gives the horizontal distance the airplane covers during this descent?

  1. 1200tan⁡6∘1200\tan 6^\circ
  2. 1200tan⁡6∘\frac{1200}{\tan 6^\circ} (correct answer)
  3. 1200sin⁡6∘1200\sin 6^\circ
  4. 1200sin⁡6∘\frac{1200}{\sin 6^\circ}
Explanation: Whenever you see an angle "with the horizontal" in a descent or ramp problem, sketch a right triangle: the horizontal distance is one leg, the change in altitude is the other leg, and the flight path is the hypotenuse. In this question, the 12001200 feet lost is the vertical drop, so it is the side opposite the 6∘6^\circ angle. The horizontal distance you want is the side adjacent to that angle. Since tangent is opposite over adjacent, you get tan⁡6∘=1200horizontal distance.\tan 6^\circ=\frac{1200}{\text{horizontal distance}}. Solving for the horizontal distance: horizontal distance=1200tan⁡6∘.\text{horizontal distance}=\frac{1200}{\tan 6^\circ}. So the correct choice is the 1200tan⁡6∘\frac{1200}{\tan 6^\circ} expression. The other expressions come from switching the roles of the sides. The expression 1200tan⁡6∘1200\tan 6^\circ treats the 12001200 as if it were the adjacent side and solves for the opposite side, which would be wrong. The expressions 1200sin⁡6∘1200\sin 6^\circ and 1200sin⁡6∘\frac{1200}{\sin 6^\circ} involve the hypotenuse, not the horizontal leg. In fact, 1200sin⁡6∘\frac{1200}{\sin 6^\circ} gives the length of the actual flight path, because sine is opposite over hypotenuse. And 1200sin⁡6∘1200\sin 6^\circ would be the altitude lost if the flight path itself were 12001200 feet long. For quick success on these problems, ask yourself: "Which side is the 12001200? Which side do I need?" Then pick sine, cosine, or tangent based on those two sides. Drawing the triangle is the step that prevents these side-role errors.

Question 11

If θ\theta is an acute angle in a right triangle and sin⁡θ=513\sin\theta = \frac{5}{13}, what is tan⁡(90∘−θ)\tan(90^\circ - \theta)?

  1. 513\frac{5}{13}
  2. 512\frac{5}{12}
  3. 1312\frac{13}{12}
  4. 125\frac{12}{5} (correct answer)
Explanation: Whenever you see a question involving complementary angles in a right triangle, recall the cofunction identities: tan⁡(90∘−θ)=cot⁡θ\tan(90^\circ - \theta) = \cot \theta. Start by drawing the triangle. Since sin⁡θ=513\sin\theta = \frac{5}{13}, the side opposite θ\theta is 5 and the hypotenuse is 13. Use the Pythagorean theorem to find the adjacent side: 132−52=169−25=144=12\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12. Therefore, cot⁡θ=adjacentopposite=125\cot \theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{12}{5}, so tan⁡(90∘−θ)=125\tan(90^\circ - \theta) = \frac{12}{5}. Now look at the distractors. The choice 513\frac{5}{13} is just sin⁡θ\sin\theta itself, and 512\frac{5}{12} is tan⁡θ\tan\theta (opposite over adjacent) — both ignore the complement. The choice 1312\frac{13}{12} is hypotenuse over adjacent, or sec⁡θ\sec\theta, which might come from mixing up reciprocal trig ratios. Only 125\frac{12}{5} uses the correct complementary relationship. For future problems: when you see (90∘−θ)(90^\circ - \theta), immediately rewrite it using cofunctions. Also remember SOH-CAH-TOA and that cot⁡θ\cot\theta is the reciprocal of tan⁡θ\tan\theta, not the reciprocal of sin⁡θ\sin\theta.

Question 12

From the top of a 30-meter building, a person sees a car at an angle of depression of 25∘25^\circ. The car drives 20 meters straight toward the building. Which expression gives the tangent of the new angle of depression?

  1. 3030cot⁡25∘−20\frac{30}{30\cot 25^\circ - 20} (correct answer)
  2. 3030tan⁡25∘−20\frac{30}{30\tan 25^\circ - 20}
  3. 3030cot⁡25∘+20\frac{30}{30\cot 25^\circ + 20}
  4. 30cot⁡25∘−2030\frac{30\cot 25^\circ - 20}{30}
Explanation: Whenever you see an angle of depression problem, draw the right triangle and remember that the angle of depression from the top equals the angle of elevation from the car. Here, the building height is fixed at 30 meters, and the car's original horizontal distance is unknown. Let that original distance be xx. At the car's first position, tan⁡25∘=30x\tan 25^\circ = \frac{30}{x}, so x=30cot⁡25∘x = 30\cot 25^\circ. After the car drives 20 meters closer, the new horizontal distance is 30cot⁡25∘−2030\cot 25^\circ - 20. Therefore, the tangent of the new angle of depression is verticalhorizontal=3030cot⁡25∘−20.\frac{\text{vertical}}{\text{horizontal}}=\frac{30}{30\cot 25^\circ - 20}. That matches the expression with the cotangent in the denominator and the minus 20. Now examine the traps. The choice 3030tan⁡25∘−20\frac{30}{30\tan 25^\circ - 20} uses tan⁡25∘\tan 25^\circ instead of cot⁡25∘\cot 25^\circ, which would incorrectly treat 30tan⁡25∘30\tan 25^\circ as the original horizontal distance. The choice 3030cot⁡25∘+20\frac{30}{30\cot 25^\circ + 20} adds 20, which would mean the car drove away from the building rather than toward it. The choice 30cot⁡25∘−2030\frac{30\cot 25^\circ - 20}{30} is the reciprocal relationship, giving horizontal over vertical, which is the cotangent of the new angle, not the tangent. Study tip: when working with angles of depression, always identify where the angle actually sits in the triangle. The 30-meter height is opposite the angle at the car's position, and the horizontal distance is adjacent. Set up tan⁡=heighthorizontal distance\tan = \frac{\text{height}}{\text{horizontal distance}}, then adjust the distance by how far the car moves.