Heights and Distances: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 3)

Let the ladder length be L = 12 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 = 12 x sin(30°) = 12 x (1/2) = 6 m. Check: 6 m is less than the ladder length 12 m, as it must be.

Let the ladder length be L = 90 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 = 90 x sin(30°) = 90 x (1/2) = 45 m. Check: 45 m is less than the ladder length 90 m, as it must be.

Let the ladder length be L = 104 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 = 104 x sin(30°) = 104 x (1/2) = 52 m. Check: 52 m is less than the ladder length 104 m, as it must be.

Let the ladder length be L = 60 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 = 60 x sin(30°) = 60 x (1/2) = 30 m. Check: 30 m is less than the ladder length 60 m, as it must be.

Let the ladder length be L = 106 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 = 106 x sin(30°) = 106 x (1/2) = 53 m. Check: 53 m is less than the ladder length 106 m, as it must be.

Let the ladder length be L = 120 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 60 degrees. Let x be the distance of the foot of the ladder from the wall. That distance x lies adjacent to theta. So cos(theta) = adjacent/hypotenuse = x/L, giving x = L x cos(theta). Substituting gives x = 120 x cos(60°) = 120 x (1/2) = 60 m. Check: 60 m is less than the ladder length 120 m, as it must be.

Let the ladder length be L = 18 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 60 degrees. Let x be the distance of the foot of the ladder from the wall. That distance x lies adjacent to theta. So cos(theta) = adjacent/hypotenuse = x/L, giving x = L x cos(theta). Substituting gives x = 18 x cos(60°) = 18 x (1/2) = 9 m. Check: 9 m is less than the ladder length 18 m, as it must be.

Let the ladder length be L = 70 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 60 degrees. Let x be the distance of the foot of the ladder from the wall. That distance x lies adjacent to theta. So cos(theta) = adjacent/hypotenuse = x/L, giving x = L x cos(theta). Substituting gives x = 70 x cos(60°) = 70 x (1/2) = 35 m. Check: 35 m is less than the ladder length 70 m, as it must be.

Let the ladder length be L = 6 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 60 degrees. Let x be the distance of the foot of the ladder from the wall. That distance x lies adjacent to theta. So cos(theta) = adjacent/hypotenuse = x/L, giving x = L x cos(theta). Substituting gives x = 6 x cos(60°) = 6 x (1/2) = 3 m. Check: 3 m is less than the ladder length 6 m, as it must be.

Let the ladder length be L = 24 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 60 degrees. Let x be the distance of the foot of the ladder from the wall. That distance x lies adjacent to theta. So cos(theta) = adjacent/hypotenuse = x/L, giving x = L x cos(theta). Substituting gives x = 24 x cos(60°) = 24 x (1/2) = 12 m. Check: 12 m is less than the ladder length 24 m, as it must be.

Let the ladder length be L = 48 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 60 degrees. Let x be the distance of the foot of the ladder from the wall. That distance x lies adjacent to theta. So cos(theta) = adjacent/hypotenuse = x/L, giving x = L x cos(theta). Substituting gives x = 48 x cos(60°) = 48 x (1/2) = 24 m. Check: 24 m is less than the ladder length 48 m, as it must be.

Let the ladder length be L = 96 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 60 degrees. Let x be the distance of the foot of the ladder from the wall. That distance x lies adjacent to theta. So cos(theta) = adjacent/hypotenuse = x/L, giving x = L x cos(theta). Substituting gives x = 96 x cos(60°) = 96 x (1/2) = 48 m. Check: 48 m is less than the ladder length 96 m, as it must be.

Let the ladder length be L = 80 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 60 degrees. Let x be the distance of the foot of the ladder from the wall. That distance x lies adjacent to theta. So cos(theta) = adjacent/hypotenuse = x/L, giving x = L x cos(theta). Substituting gives x = 80 x cos(60°) = 80 x (1/2) = 40 m. Check: 40 m is less than the ladder length 80 m, as it must be.

Let the ladder length be L = 60 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 60 degrees. Let x be the distance of the foot of the ladder from the wall. That distance x lies adjacent to theta. So cos(theta) = adjacent/hypotenuse = x/L, giving x = L x cos(theta). Substituting gives x = 60 x cos(60°) = 60 x (1/2) = 30 m. Check: 30 m is less than the ladder length 60 m, as it must be.

Let the ladder length be L = 42 m; this is the hypotenuse. Let the angle it makes with the ground be theta = 60 degrees. Let x be the distance of the foot of the ladder from the wall. That distance x lies adjacent to theta. So cos(theta) = adjacent/hypotenuse = x/L, giving x = L x cos(theta). Substituting gives x = 42 x cos(60°) = 42 x (1/2) = 21 m. Check: 21 m is less than the ladder length 42 m, as it must be.

Let the pole’s height be h = 17 m. Let its shadow length be s = 17 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) = 17/17 = 1. We know tan(45°) = 1. So theta = 45 degrees. Check: equal height and shadow always give a 45-degree angle.

Let the pole’s height be h = 22 m. Let its shadow length be s = 22 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) = 22/22 = 1. We know tan(45°) = 1. So theta = 45 degrees. Check: equal height and shadow always give a 45-degree angle.

Let the pole’s height be h = 36 m. Let its shadow length be s = 36 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) = 36/36 = 1. We know tan(45°) = 1. So theta = 45 degrees. Check: equal height and shadow always give a 45-degree angle.

Let the pole’s height be h = 34 m. Let its shadow length be s = 34 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) = 34/34 = 1. We know tan(45°) = 1. So theta = 45 degrees. Check: equal height and shadow always give a 45-degree angle.

Let the pole’s height be h = 31 m. Let its shadow length be s = 31 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) = 31/31 = 1. We know tan(45°) = 1. So theta = 45 degrees. Check: equal height and shadow always give a 45-degree angle.

Let the pole’s height be h = 25 m. Let its shadow length be s = 25√3 m. 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) = 25/25√3 = 1/√3. We know tan(30°) = 1/√3. So theta = 30 degrees. Check: 25/25√3 simplifies to 1/√3, confirming 30 degrees.

Let the pole’s height be h = 16 m. Let its shadow length be s = 16√3 m. 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) = 16/16√3 = 1/√3. We know tan(30°) = 1/√3. So theta = 30 degrees. Check: 16/16√3 simplifies to 1/√3, confirming 30 degrees.

Let the pole’s height be h = 27 m. Let its shadow length be s = 27√3 m. 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) = 27/27√3 = 1/√3. We know tan(30°) = 1/√3. So theta = 30 degrees. Check: 27/27√3 simplifies to 1/√3, confirming 30 degrees.

Let the pole’s height be h = 12 m. Let its shadow length be s = 12√3 m. 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) = 12/12√3 = 1/√3. We know tan(30°) = 1/√3. So theta = 30 degrees. Check: 12/12√3 simplifies to 1/√3, confirming 30 degrees.

Let the pole’s height be h = 21 m. Let its shadow length be s = 21√3 m. 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) = 21/21√3 = 1/√3. We know tan(30°) = 1/√3. So theta = 30 degrees. Check: 21/21√3 simplifies to 1/√3, confirming 30 degrees.

Let the pole’s height be h = 23 m. Let its shadow length be s = 23√3 m. 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) = 23/23√3 = 1/√3. We know tan(30°) = 1/√3. So theta = 30 degrees. Check: 23/23√3 simplifies to 1/√3, confirming 30 degrees.

Let the pole’s height be h = 28√3 m. Let its shadow length be s = 28 m. 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) = 28√3/28 = √3. We know tan(60°) = √3. So theta = 60 degrees. Check: 28√3/28 simplifies to √3, confirming 60 degrees.

Let the pole’s height be h = 11√3 m. Let its shadow length be s = 11 m. 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) = 11√3/11 = √3. We know tan(60°) = √3. So theta = 60 degrees. Check: 11√3/11 simplifies to √3, confirming 60 degrees.

Let the pole’s height be h = 20√3 m. Let its shadow length be s = 20 m. 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) = 20√3/20 = √3. We know tan(60°) = √3. So theta = 60 degrees. Check: 20√3/20 simplifies to √3, confirming 60 degrees.

Let the pole’s height be h = 22√3 m. Let its shadow length be s = 22 m. 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) = 22√3/22 = √3. We know tan(60°) = √3. So theta = 60 degrees. Check: 22√3/22 simplifies to √3, confirming 60 degrees.

Let the pole’s height be h = 7√3 m. Let its shadow length be s = 7 m. 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) = 7√3/7 = √3. We know tan(60°) = √3. So theta = 60 degrees. Check: 7√3/7 simplifies to √3, confirming 60 degrees.

Let the pole’s height be h = 10√3 m. Let its shadow length be s = 10 m. 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) = 10√3/10 = √3. We know tan(60°) = √3. So theta = 60 degrees. Check: 10√3/10 simplifies to √3, confirming 60 degrees.