Some Applications of Trigonometry | Some Applications of Trigonometry | Assessment | English | Grade 10

A ladder 12 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60⁰ with the ground then the height of the wall is ____

• 12√3 m

• 6√3 m

• 6 m

• Cannot be determined

Answer explanation

With reference to the figure, Let AB = Height of wall AC = Length of ladder = 12m ∠ ACB = 60⁰ In ∆ ABC, sin60⁰ = AB/AC √3/2 = AB/12 AB = 12√3/2 = 6√3 m The Height of the wall = 6√3 m So the correct answer is Option 2

Tags

CCSS.HSG.SRT.C.8

The angle of elevation of the top of a tree from a point on the ground, which is 20 m away from the base of the tree, is 30°. Find the height of the tree.

• 20√3/ 3 m

• 20√3 m

• 10 m

• Cannot be determined

Answer explanation

With reference to the figure, Let AB = Height of tree BC = Distance of the point on ground from the base of tree = 20m ∠ ACB = 30⁰ In ∆ ABC, tan 30⁰ = AB/BC 1/√3 = AB/20 AB = 20 /√3 m The Height of the tree = 20 /√3 m = 20√3/3 m So the correct answer is Option 1

Tags

CCSS.HSG.SRT.C.8

A kite is flying at a height of 40 m above the ground. The string attached to the kite is straight and temporarily tied to a point on the ground. The inclination of the string with the ground 45⁰. Find the length of the string, assuming that there is no slack in the string.

• 40√2 m

• 40 m

• 20 m

• Cannot be determined

Answer explanation

With reference to the figure, Let AB = Height of kite from ground = 40m AC = Length of the string ∠ ACB = 45⁰ In ∆ ABC, sin 45⁰ = AB/AC 1/√2 = 40/AC AC = 40 √2 m The Length of the string = 40√2 m So the correct answer is Option 1

Tags

CCSS.HSG.SRT.C.8

The shadow of a pole, when the angle of elevation of sun is 45⁰ is found to be 10 m longer than when it is 60⁰. Find the height of the pole. (Take √3 = 1.73)

• 6.9 m

• 13.7 m

• 7.3 m

• Cannot be determined

Answer explanation

With reference to the figure, Let AB = Height of the pole BC = Length of the shadow when angle of elevation is 60⁰ = x BD = Length of the shadow when angle of elevation is 45⁰ = x + 10 In ∆ ABC, tan 60⁰ = AB/BC √3 = AB/x ------------- (1) In ∆ ABD, tan 45⁰ = AB/BD 1 = AB/(x + 10) ------------- (2) AB = x + 10 On substituting the value of AB into Eq. 1, we get, √3 = (x + 10)/x √3 x = x + 10 √3 x ─ x = 10 1.73 x - x = 10 (Taking √3 = 1.73) 0.73 x = 10 x = 10/0.73 ≈ 13.7 m Height of the pole = 13.7 m So the correct answer is Option 2

Tags

CCSS.HSG.SRT.C.8

A window in a building is 10 m above the ground. From this window the angles of elevation and depression of the top and foot of another house across the street are 30⁰ and 60⁰ respectively. Find the height of the second house and the width of the street.

• Height of second house = 20m Width of street = 10√3 m

• Height of second house = 3.33 m Width of street = 10/√3 m

• Height of second house = 13.33 m Width of street = 10/√3 m

• Cannot be determined

Answer explanation

With reference to the figure, Let AB = Height of the window = 10m BD = Width of street CD = Height of second house Angle of elevation of top (∠CAO) = 30⁰ Angle of depression at bottom (∠OAD) = 60⁰ Since AD is a transversal to the parallel lines AO and BD. Therefore, ∠OAD and ∠ ADB are alternate angles, and so are equal. So ∠ ADB = 60°. In ∆ ABD, tan 60⁰ = AB/BD √3 = 10/ BD or BD = 10/√3 m So width of street = 10/√3 m In ∆ AOC, tan 30⁰ = OC /AO ( as AO = BD = 10/√3 m) 1/√3 = OC/ 10/√3 OC = 10/3 = 3.33m Height of second house = CD = OD + OC = 10 + 3.33 = 13.33 m So the correct answer is Option 3

Tags

CCSS.HSG.SRT.C.8

A ladder on the truck of the fire brigade can be elevated to an angle of 60⁰ maximum. The length of the ladder can be extended to 25 m. If the platform of the truck is 5 m above the ground, then find the maximum height that the ladder can attain above the ground . ( Take √3 = 1.73)

• 26.625 m

• 21.625 m

• 17.5 m

• Cannot be determined

Answer explanation

With reference to the figure, Let ED = Height of the platform = BC AC = Maximum height from the ground that ladder can reach= AB + BC AE = Max. length of ladder = 25m ∠ AEB = 60⁰ In ∆ ABE, sin 60⁰ = AB/AE = AB /25 √3/2 = AB/ 25 AB = 25.√3/2 = 21.625 m AC = AB + BC = 21.625 + 5 = 26.625 m The Maximum Height that ladder can reach is = 26.625 m So the correct answer is Option 1

Tags

CCSS.HSG.SRT.C.8

A man standing on the top of a tower observes a car travelling on a straight road to the base of the tower. He first observes the car at an angle of depression of 30⁰, which is approaching the foot of the tower with a uniform speed. 18 seconds later the angle of depression of the car is found to be 60⁰. Find the time taken by the car to reach the base of the tower from the time the man first saw the car. (Assume that the car is travelling with uniform speed throughout).

• 9 secs

• 27 secs

• 9√3 secs

• Cannot be determined

Answer explanation

With reference to the figure, Let AB = Height of the tower Point D -- first position of car Point C -- second position of car ∠ADB= 30⁰ ∠ACB= 60⁰ Car takes 18 secs to travel from D to C Let speed of car = y units/sec Distance = speed x time CD = 18y units In ∆ ABC, tan 60⁰ = AB/BC √3 BC= AB ------------- (1) In ∆ ABD, tan 30⁰ = AB/BD 1/√3 = √3 BC/ BD ------------- (2) 3 BC = BD 3 BC = BC + 18y (BD = BC + CD = BC + 18y) 2 BC = 18y BC = 9y, i.e., BC/y = 9 So it takes 9 seconds to travel from C to B Total time taken = 18 + 9 = 27 secs So the correct answer is Option 2

Tags

CCSS.HSG.SRT.C.8

If the height of a pole is 6 m and it casts a shadow 2√3 m long on the ground, then the angle of elevation of the sun is ____

• 30⁰

• 60⁰

• 90⁰

• 45⁰

Answer explanation

In ∆ ABC, BC is height of the pole = 6m, AB is length of shadow = 2√3m The elevation of the sun will be ∠ CAB = θ In ∆ ABC, tanθ = BC/AB = 6/2√3 = √3 Since tan60⁰ = √3 Hence, θ = 60⁰ So, the correct answer is Option 2

Tags

CCSS.HSG.SRT.C.8