Heights and Distances: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 1)

Let the pole’s height be h = 35 m. Let its shadow length be s = 35 m too. Let theta be the angle of elevation of the sun. The pole stands vertical and the shadow lies horizontal. So tan(theta) = opposite/adjacent = height/shadow = h/s. Substituting gives tan(theta) = 35/35 = 1. We know tan(45°) = 1. So theta = 45 degrees. Check: equal height and shadow always give a 45-degree angle.

Use the same split equilateral triangle as for tan(60°): base = 1, height = √3. The 30-degree angle sits opposite the base, the shorter side. So tan(30°) = perpendicular/base = 1/√3. This is the reciprocal of tan(60°), since the two acute angles swap their opposite and adjacent sides. Check: tan(30°) x tan(60°) = (1/√3) x √3 = 1.

Take an equilateral triangle with each side = 2, and split it down the middle into two right triangles. Each half has base = 1. By the Pythagorean theorem, its height = sqrt(2² – 1²) = sqrt(3). The 60-degree angle sits opposite that height. So tan(60°) = perpendicular/base = √3/1 = √3, approximately 1.732.

A right triangle with a 45-degree angle is isosceles. So its perpendicular and base sides are equal in length. Tangent is perpendicular divided by base. Equal sides give a ratio of exactly 1. So tan(45°) = 1.

In a right triangle, label the sides relative to angle theta. The side opposite theta is called the perpendicular. The side next to theta, other than the hypotenuse, is called the base. Tangent is defined as tan(theta) = perpendicular / base = opposite / adjacent. Unlike sine and cosine, this ratio never involves the hypotenuse.

Picture an observer looking at an object below their eye level, like a boat seen from a cliff. Draw a horizontal line out from the observer’s eye. The angle of depression is the angle between that horizontal line and the line of sight. It is always measured downward, since the object sits below the horizontal. Check: by alternate angles, this equals the angle of elevation measured from the object back up to the observer.

Picture an observer looking at an object above their eye level. Draw a horizontal line out from the observer’s eye. The angle of elevation is the angle between that horizontal line and the line of sight. It is always measured upward, since the object sits above the horizontal. Looking up at a kite, for example, involves an angle of elevation.

Let the river breadth be x and the tree’s height be h. At the bank, the angle of elevation of the tree is 60 degrees. So tan(60°) = h/x, giving h = x√3. After retiring 78 m, the new distance from the tree is x + 78. The angle of elevation there is 30 degrees. So tan(30°) = h/(x+78), giving h = (x+78)/√3. Both expressions equal h, so x√3 = (x+78)/√3. Multiply both sides by √3: 3x = x + 78. So 2x = 78, giving x = 39 m. Check: substituting x = 39 back in, both x√3 and (x+78)/√3 equal 39√3.

Let the river breadth be x and the tree’s height be h. At the bank, the angle of elevation of the tree is 60 degrees. So tan(60°) = h/x, giving h = x√3. After retiring 52 m, the new distance from the tree is x + 52. The angle of elevation there is 30 degrees. So tan(30°) = h/(x+52), giving h = (x+52)/√3. Both expressions equal h, so x√3 = (x+52)/√3. Multiply both sides by √3: 3x = x + 52. So 2x = 52, giving x = 26 m. Check: substituting x = 26 back in, both x√3 and (x+52)/√3 equal 26√3.

Let the river breadth be x and the tree’s height be h. At the bank, the angle of elevation of the tree is 60 degrees. So tan(60°) = h/x, giving h = x√3. After retiring 44 m, the new distance from the tree is x + 44. The angle of elevation there is 30 degrees. So tan(30°) = h/(x+44), giving h = (x+44)/√3. Both expressions equal h, so x√3 = (x+44)/√3. Multiply both sides by √3: 3x = x + 44. So 2x = 44, giving x = 22 m. Check: substituting x = 22 back in, both x√3 and (x+44)/√3 equal 22√3.

Let the river breadth be x and the tree’s height be h. At the bank, the angle of elevation of the tree is 60 degrees. So tan(60°) = h/x, giving h = x√3. After retiring 102 m, the new distance from the tree is x + 102. The angle of elevation there is 30 degrees. So tan(30°) = h/(x+102), giving h = (x+102)/√3. Both expressions equal h, so x√3 = (x+102)/√3. Multiply both sides by √3: 3x = x + 102. So 2x = 102, giving x = 51 m. Check: substituting x = 51 back in, both x√3 and (x+102)/√3 equal 51√3.

Let the river breadth be x and the tree’s height be h. At the bank, the angle of elevation of the tree is 60 degrees. So tan(60°) = h/x, giving h = x√3. After retiring 70 m, the new distance from the tree is x + 70. The angle of elevation there is 30 degrees. So tan(30°) = h/(x+70), giving h = (x+70)/√3. Both expressions equal h, so x√3 = (x+70)/√3. Multiply both sides by √3: 3x = x + 70. So 2x = 70, giving x = 35 m. Check: substituting x = 35 back in, both x√3 and (x+70)/√3 equal 35√3.

Let the river breadth be x and the tree’s height be h. At the bank, the angle of elevation of the tree is 60 degrees. So tan(60°) = h/x, giving h = x√3. After retiring 26 m, the new distance from the tree is x + 26. The angle of elevation there is 30 degrees. So tan(30°) = h/(x+26), giving h = (x+26)/√3. Both expressions equal h, so x√3 = (x+26)/√3. Multiply both sides by √3: 3x = x + 26. So 2x = 26, giving x = 13 m. Check: substituting x = 13 back in, both x√3 and (x+26)/√3 equal 13√3.

Let the river breadth be x and the tree’s height be h. At the bank, the angle of elevation of the tree is 60 degrees. So tan(60°) = h/x, giving h = x√3. After retiring 16 m, the new distance from the tree is x + 16. The angle of elevation there is 30 degrees. So tan(30°) = h/(x+16), giving h = (x+16)/√3. Both expressions equal h, so x√3 = (x+16)/√3. Multiply both sides by √3: 3x = x + 16. So 2x = 16, giving x = 8 m. Check: substituting x = 8 back in, both x√3 and (x+16)/√3 equal 8√3.

Let the river breadth be x and the tree’s height be h. At the bank, the angle of elevation of the tree is 60 degrees. So tan(60°) = h/x, giving h = x√3. After retiring 72 m, the new distance from the tree is x + 72. The angle of elevation there is 30 degrees. So tan(30°) = h/(x+72), giving h = (x+72)/√3. Both expressions equal h, so x√3 = (x+72)/√3. Multiply both sides by √3: 3x = x + 72. So 2x = 72, giving x = 36 m. Check: substituting x = 36 back in, both x√3 and (x+72)/√3 equal 36√3.

Let the river breadth be x and the tree’s height be h. At the bank, the angle of elevation of the tree is 60 degrees. So tan(60°) = h/x, giving h = x√3. After retiring 10 m, the new distance from the tree is x + 10. The angle of elevation there is 30 degrees. So tan(30°) = h/(x+10), giving h = (x+10)/√3. Both expressions equal h, so x√3 = (x+10)/√3. Multiply both sides by √3: 3x = x + 10. So 2x = 10, giving x = 5 m. Check: substituting x = 5 back in, both x√3 and (x+10)/√3 equal 5√3.

Let the river breadth be x and the tree’s height be h. At the bank, the angle of elevation of the tree is 60 degrees. So tan(60°) = h/x, giving h = x√3. After retiring 36 m, the new distance from the tree is x + 36. The angle of elevation there is 30 degrees. So tan(30°) = h/(x+36), giving h = (x+36)/√3. Both expressions equal h, so x√3 = (x+36)/√3. Multiply both sides by √3: 3x = x + 36. So 2x = 36, giving x = 18 m. Check: substituting x = 18 back in, both x√3 and (x+36)/√3 equal 18√3.

Let the tower height be h. Let A be the first point, B the point after walking 26 m closer, and F the tower’s foot. At A the angle of elevation is 30 degrees. So tan(30°) = h/AF, giving AF = h√3. At B the angle of elevation is 60 degrees. So tan(60°) = h/BF, giving BF = h/√3. The walked distance is AF minus BF, so AF – BF = 26 m. That means h√3 – h/√3 = 26. Combine the left side over √3: (3h – h)/√3 = 2h/√3 = 26. So h = 26√3/2 = 13√3 m. Check: with h = 13√3, AF = 3 x 13 = 39 m and BF = 13 m, so AF – BF = 26 m, matching the 26 m walked.

Let the tower height be h. Let A be the first point, B the point after walking 30 m closer, and F the tower’s foot. At A the angle of elevation is 30 degrees. So tan(30°) = h/AF, giving AF = h√3. At B the angle of elevation is 60 degrees. So tan(60°) = h/BF, giving BF = h/√3. The walked distance is AF minus BF, so AF – BF = 30 m. That means h√3 – h/√3 = 30. Combine the left side over √3: (3h – h)/√3 = 2h/√3 = 30. So h = 30√3/2 = 15√3 m. Check: with h = 15√3, AF = 3 x 15 = 45 m and BF = 15 m, so AF – BF = 30 m, matching the 30 m walked.

Let the tower height be h. Let A be the first point, B the point after walking 34 m closer, and F the tower’s foot. At A the angle of elevation is 30 degrees. So tan(30°) = h/AF, giving AF = h√3. At B the angle of elevation is 60 degrees. So tan(60°) = h/BF, giving BF = h/√3. The walked distance is AF minus BF, so AF – BF = 34 m. That means h√3 – h/√3 = 34. Combine the left side over √3: (3h – h)/√3 = 2h/√3 = 34. So h = 34√3/2 = 17√3 m. Check: with h = 17√3, AF = 3 x 17 = 51 m and BF = 17 m, so AF – BF = 34 m, matching the 34 m walked.

Let the tower height be h. Let A be the first point, B the point after walking 54 m closer, and F the tower’s foot. At A the angle of elevation is 30 degrees. So tan(30°) = h/AF, giving AF = h√3. At B the angle of elevation is 60 degrees. So tan(60°) = h/BF, giving BF = h/√3. The walked distance is AF minus BF, so AF – BF = 54 m. That means h√3 – h/√3 = 54. Combine the left side over √3: (3h – h)/√3 = 2h/√3 = 54. So h = 54√3/2 = 27√3 m. Check: with h = 27√3, AF = 3 x 27 = 81 m and BF = 27 m, so AF – BF = 54 m, matching the 54 m walked.

Let the tower height be h. Let A be the first point, B the point after walking 68 m closer, and F the tower’s foot. At A the angle of elevation is 30 degrees. So tan(30°) = h/AF, giving AF = h√3. At B the angle of elevation is 60 degrees. So tan(60°) = h/BF, giving BF = h/√3. The walked distance is AF minus BF, so AF – BF = 68 m. That means h√3 – h/√3 = 68. Combine the left side over √3: (3h – h)/√3 = 2h/√3 = 68. So h = 68√3/2 = 34√3 m. Check: with h = 34√3, AF = 3 x 34 = 102 m and BF = 34 m, so AF – BF = 68 m, matching the 68 m walked.

Let the tower height be h. Let A be the first point, B the point after walking 4 m closer, and F the tower’s foot. At A the angle of elevation is 30 degrees. So tan(30°) = h/AF, giving AF = h√3. At B the angle of elevation is 60 degrees. So tan(60°) = h/BF, giving BF = h/√3. The walked distance is AF minus BF, so AF – BF = 4 m. That means h√3 – h/√3 = 4. Combine the left side over √3: (3h – h)/√3 = 2h/√3 = 4. So h = 4√3/2 = 2√3 m. Check: with h = 2√3, AF = 3 x 2 = 6 m and BF = 2 m, so AF – BF = 4 m, matching the 4 m walked.

Let the tower height be h. Let A be the first point, B the point after walking 38 m closer, and F the tower’s foot. At A the angle of elevation is 30 degrees. So tan(30°) = h/AF, giving AF = h√3. At B the angle of elevation is 60 degrees. So tan(60°) = h/BF, giving BF = h/√3. The walked distance is AF minus BF, so AF – BF = 38 m. That means h√3 – h/√3 = 38. Combine the left side over √3: (3h – h)/√3 = 2h/√3 = 38. So h = 38√3/2 = 19√3 m. Check: with h = 19√3, AF = 3 x 19 = 57 m and BF = 19 m, so AF – BF = 38 m, matching the 38 m walked.

Let the tower height be h. Let A be the first point, B the point after walking 66 m closer, and F the tower’s foot. At A the angle of elevation is 30 degrees. So tan(30°) = h/AF, giving AF = h√3. At B the angle of elevation is 60 degrees. So tan(60°) = h/BF, giving BF = h/√3. The walked distance is AF minus BF, so AF – BF = 66 m. That means h√3 – h/√3 = 66. Combine the left side over √3: (3h – h)/√3 = 2h/√3 = 66. So h = 66√3/2 = 33√3 m. Check: with h = 33√3, AF = 3 x 33 = 99 m and BF = 33 m, so AF – BF = 66 m, matching the 66 m walked.