5.3 Solving Right Triangles & Angle of Elevation/Depression

Solving a Right Triangle

To solve a right triangle means to find every missing side and every missing angle. Two facts organize the work: the acute angles always sum to 90° (since the three angles total 180° and one is the right angle), and each missing part points to a specific tool.

Multi-step example

A right triangle has one leg of 6 and a hypotenuse of 10. First, the missing leg is √(10² − 6²) = √64 = 8. Next, the angle opposite the leg of 6 is sin⁻¹(6/10) = sin⁻¹(0.6) ≈ 36.87°. The remaining acute angle is 90° − 36.87° ≈ 53.13°. The triangle is now fully solved: sides 6, 8, 10 and angles 36.87°, 53.13°, 90°.

Angle of Elevation and Angle of Depression

Both angles are measured from a horizontal line of sight, never from the vertical.

• Angle of elevation: the angle you look up from horizontal to see a higher object.
• Angle of depression: the angle you look down from horizontal to see a lower object.

Because the two horizontal lines are parallel, an angle of elevation and the matching angle of depression are equal alternate interior angles. That equality is the key to depression problems: the angle of depression from a tower to a boat equals the angle of elevation from the boat back up to the tower.

Elevation word problem

An observer standing 50 feet from the base of a building measures the angle of elevation to the top as 62°. How tall is the building? The height is opposite the 62° angle and 50 is adjacent, so tangent applies: tan 62° = h / 50, giving h = 50 · tan 62° ≈ 50(1.8807) ≈ 94 feet. If the problem gave the observer's eye height, you would add it to this result.

Depression word problem

From the top of a 120-foot lighthouse, the angle of depression to a boat is 8°. How far is the boat from the base? The angle of depression (8°) equals the angle of elevation from the boat, and the 120-foot height is opposite that angle while the horizontal distance d is adjacent. So tan 8° = 120 / d, which rearranges to d = 120 / tan 8° ≈ 120 / 0.1405 ≈ 854 feet.

Solving From an Angle and a Side

Not every solve starts with two sides. Suppose a right triangle has a 34° angle and the leg adjacent to it is 18. The opposite leg comes from tangent: opp = 18 · tan 34° ≈ 18(0.6745) ≈ 12.1. The hypotenuse comes from cosine: hyp = 18 / cos 34° ≈ 18 / 0.8290 ≈ 21.7. The other acute angle is 90° − 34° = 56°. All six parts are now known from a single angle and a single side.

Two-Position Word Problems

Harder Regents items use two observation points. From a point on level ground the angle of elevation to the top of a tower is 40°; moving 30 feet closer, the angle becomes 55°. Each position defines a right triangle that shares the same tower height h. From the near point, tan 55° = h / d; from the far point, tan 40° = h / (d + 30). Solving the system gives the height. On multiple-choice versions you can often test the given answer choices against tan 55° = h / d rather than solving the full system by hand.

Draw and Label First

The most reliable Regents strategy is to sketch the situation before computing:

1. Draw the right triangle and mark the right angle.
2. Mark the given angle of elevation or depression at the correct vertex, measured from the horizontal.
3. Label each side as opposite, adjacent, or hypotenuse relative to that angle.
4. Choose sin, cos, or tan based on which two sides are involved, then solve.

For example, "a kite flies on a 60-foot string held taut at a 50° angle of elevation; how high is the kite?" Draw the right triangle, place the 50° angle at the ground observer, mark the 60-foot string as the hypotenuse, and label the kite's height as the side opposite the 50° angle. Opposite and hypotenuse mean sine, so height = 60 · sin 50° ≈ 46 feet. Labeling first converts a wordy paragraph into a single one-line equation and removes almost all room for a setup mistake.

Common Traps

• Measuring a depression angle from the vertical instead of the horizontal; it is always from the horizontal line of sight.
• Forgetting to add the observer's height when the problem gives an eye level or instrument height above the ground.
• Placing the unknown as the hypotenuse when it is actually a leg, which swaps sine and tangent.
• Rounding partway through a multi-step solve; carry full precision and round once at the end.