Heights and Distances: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 2)

Let the tower height be h. Let A be the first point, B the point after walking 62 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 = 62 m. That means h√3 – h/√3 = 62. Combine the left side over √3: (3h – h)/√3 = 2h/√3 = 62. So h = 62√3/2 = 31√3 m. Check: with h = 31√3, AF = 3 x 31 = 93 m and BF = 31 m, so AF – BF = 62 m, matching the 62 m walked.

Let the tower height be h. Let A be the first point, B the point after walking 8 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 = 8 m. That means h√3 – h/√3 = 8. Combine the left side over √3: (3h – h)/√3 = 2h/√3 = 8. So h = 8√3/2 = 4√3 m. Check: with h = 4√3, AF = 3 x 4 = 12 m and BF = 4 m, so AF – BF = 8 m, matching the 8 m walked.

Let the tower’s height be h. Let the horizontal distance from its base be d = 25√3 m. Let theta = 30 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d. So h = d x tan(30°) = d x (1/√3). Substituting gives h = 25√3 x (1/√3) = 25 m. Check: tan(30°) = 1/√3 is less than 1, so the height should be less than the distance, and it is.

Let the tower’s height be h. Let the horizontal distance from its base be d = 11√3 m. Let theta = 30 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d. So h = d x tan(30°) = d x (1/√3). Substituting gives h = 11√3 x (1/√3) = 11 m. Check: tan(30°) = 1/√3 is less than 1, so the height should be less than the distance, and it is.

Let the tower’s height be h. Let the horizontal distance from its base be d = 16√3 m. Let theta = 30 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d. So h = d x tan(30°) = d x (1/√3). Substituting gives h = 16√3 x (1/√3) = 16 m. Check: tan(30°) = 1/√3 is less than 1, so the height should be less than the distance, and it is.

Let the tower’s height be h. Let the horizontal distance from its base be d = 6√3 m. Let theta = 30 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d. So h = d x tan(30°) = d x (1/√3). Substituting gives h = 6√3 x (1/√3) = 6 m. Check: tan(30°) = 1/√3 is less than 1, so the height should be less than the distance, and it is.

Let the tower’s height be h. Let the horizontal distance from its base be d = 12√3 m. Let theta = 30 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d. So h = d x tan(30°) = d x (1/√3). Substituting gives h = 12√3 x (1/√3) = 12 m. Check: tan(30°) = 1/√3 is less than 1, so the height should be less than the distance, and it is.

Let the tower’s height be h. Let the horizontal distance from its base be d = 17√3 m. Let theta = 30 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d. So h = d x tan(30°) = d x (1/√3). Substituting gives h = 17√3 x (1/√3) = 17 m. Check: tan(30°) = 1/√3 is less than 1, so the height should be less than the distance, and it is.

Let the tower’s height be h. Let the horizontal distance from its base be d = 3√3 m. Let theta = 30 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d. So h = d x tan(30°) = d x (1/√3). Substituting gives h = 3√3 x (1/√3) = 3 m. Check: tan(30°) = 1/√3 is less than 1, so the height should be less than the distance, and it is.

Let the tower’s height be h. Let the horizontal distance from its base be d = 19√3 m. Let theta = 30 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d. So h = d x tan(30°) = d x (1/√3). Substituting gives h = 19√3 x (1/√3) = 19 m. Check: tan(30°) = 1/√3 is less than 1, so the height should be less than the distance, and it is.

Let the tower’s height be h. Let the horizontal distance from its base be d = 16 m. Let theta = 60 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d, giving h = d x tan(theta). Substituting gives h = 16 x tan(60°) = 16 x √3 = 16√3 m. Check: tan(60°) = √3 is more than 1, so the height should exceed the distance, and it does.

Let the tower’s height be h. Let the horizontal distance from its base be d = 17 m. Let theta = 45 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d, giving h = d x tan(theta). Substituting gives h = 17 x tan(45°) = 17 x 1 = 17 m. Check: at 45 degrees the height always equals the distance, and it does here.

Let the tower’s height be h. Let the horizontal distance from its base be d = 35 m. Let theta = 45 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d, giving h = d x tan(theta). Substituting gives h = 35 x tan(45°) = 35 x 1 = 35 m. Check: at 45 degrees the height always equals the distance, and it does here.

Let the tower’s height be h. Let the horizontal distance from its base be d = 20 m. Let theta = 60 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d, giving h = d x tan(theta). Substituting gives h = 20 x tan(60°) = 20 x √3 = 20√3 m. Check: tan(60°) = √3 is more than 1, so the height should exceed the distance, and it does.

Let the tower’s height be h. Let the horizontal distance from its base be d = 11 m. Let theta = 45 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d, giving h = d x tan(theta). Substituting gives h = 11 x tan(45°) = 11 x 1 = 11 m. Check: at 45 degrees the height always equals the distance, and it does here.

Let the tower’s height be h. Let the horizontal distance from its base be d = 7 m. Let theta = 60 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d, giving h = d x tan(theta). Substituting gives h = 7 x tan(60°) = 7 x √3 = 7√3 m. Check: tan(60°) = √3 is more than 1, so the height should exceed the distance, and it does.

Let the tower’s height be h. Let the horizontal distance from its base be d = 43 m. Let theta = 60 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d, giving h = d x tan(theta). Substituting gives h = 43 x tan(60°) = 43 x √3 = 43√3 m. Check: tan(60°) = √3 is more than 1, so the height should exceed the distance, and it does.

Let the tower’s height be h. Let the horizontal distance from its base be d = 14 m. Let theta = 45 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d, giving h = d x tan(theta). Substituting gives h = 14 x tan(45°) = 14 x 1 = 14 m. Check: at 45 degrees the height always equals the distance, and it does here.

Let the tower’s height be h. Let the horizontal distance from its base be d = 48 m. Let theta = 45 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d, giving h = d x tan(theta). Substituting gives h = 48 x tan(45°) = 48 x 1 = 48 m. Check: at 45 degrees the height always equals the distance, and it does here.

Let the tower’s height be h. Let the horizontal distance from its base be d = 4 m. Let theta = 60 degrees be the angle of elevation of the top. The tower stands vertical. So tan(theta) = opposite/adjacent = height/distance = h/d, giving h = d x tan(theta). Substituting gives h = 4 x tan(60°) = 4 x √3 = 4√3 m. Check: tan(60°) = √3 is more than 1, so the height should exceed the distance, and it does.

Let the ladder length be L = 20 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 30 degrees. Let y be the height where the ladder touches the wall. That height y lies opposite theta. So sin(theta) = opposite/hypotenuse = y/L, giving y = L x sin(theta). Substituting gives y = 20 x sin(30°) = 20 x (1/2) = 10 m. Check: 10 m is less than the ladder length 20 m, as it must be.

Let the ladder length be L = 92 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 30 degrees. Let y be the height where the ladder touches the wall. That height y lies opposite theta. So sin(theta) = opposite/hypotenuse = y/L, giving y = L x sin(theta). Substituting gives y = 92 x sin(30°) = 92 x (1/2) = 46 m. Check: 46 m is less than the ladder length 92 m, as it must be.

Let the ladder length be L = 44 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 30 degrees. Let y be the height where the ladder touches the wall. That height y lies opposite theta. So sin(theta) = opposite/hypotenuse = y/L, giving y = L x sin(theta). Substituting gives y = 44 x sin(30°) = 44 x (1/2) = 22 m. Check: 22 m is less than the ladder length 44 m, as it must be.

Let the ladder length be L = 84 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 30 degrees. Let y be the height where the ladder touches the wall. That height y lies opposite theta. So sin(theta) = opposite/hypotenuse = y/L, giving y = L x sin(theta). Substituting gives y = 84 x sin(30°) = 84 x (1/2) = 42 m. Check: 42 m is less than the ladder length 84 m, as it must be.

Let the ladder length be L = 64 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 30 degrees. Let y be the height where the ladder touches the wall. That height y lies opposite theta. So sin(theta) = opposite/hypotenuse = y/L, giving y = L x sin(theta). Substituting gives y = 64 x sin(30°) = 64 x (1/2) = 32 m. Check: 32 m is less than the ladder length 64 m, as it must be.