Some Applications of Trigonometry Class 10 Competency Questions

20 Questions | 97 Marks

A hot air balloon is observed from two points P and Q on the ground, situated on the same side of the balloon and in a straight line with it. The angles of elevation of the balloon from P and Q are 60° and 45° respectively. If the distance between P and Q is 20 m, find the height of the balloon. 5marks

A student standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is 60°. When he moves 20 m away from the bank, he finds the angle of elevation to be 30°. Find the width of the river and the height of the tree. 5marks

A statue 1.6 m tall stands on top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. 5marks(a) Draw a labelled diagram representing the situation.(b) Find the height of the pedestal.(c) Find the height of the statue.

A vertical tower is built on the bank of a river. From a point on the opposite bank directly opposite to the tower, the angle of elevation of its top is 60°. From another point 20 m away from this point and in line with the foot of the tower, the angle of elevation of the top is 30°. 5marks(a) Calculate the width of the river.(b) Calculate the height of the tower.

Assertion: The angle of depression of a car from the top of a 100 m tall tower is 30°. The distance of the car from the tower is

100 \sqrt{3}

m.Reason: If the angle of depression is

θ

, then

\tan θ = \frac{height of tower}{distance from tower}

. 3marks

A man standing on the top of a 75 m high lighthouse observes two ships approaching the lighthouse from different directions. The angles of depression of the two ships from the top of the lighthouse are 30° and 45°. If the two ships are on the opposite sides of the lighthouse, find the distance between the two ships. 5marks

A soldier is on the top of a cliff 80 m high. He observes a boat at an angle of depression of 30°. He moves the boat towards the cliff. After 2 minutes, the angle of depression is 60°. Find the speed of the boat, assuming it is sailing at a uniform speed. 5marks

From the top of a building 60 m high, the angles of depression of the top and bottom of a vertical lamp post are observed to be 30° and 60° respectively. 5marks(a) Find the horizontal distance between the building and the lamp post.(b) Find the height of the lamp post.

A painter needs to reach a certain height on a building to paint it. The building is 12 m high. She uses a ladder that reaches the top of the building when placed at an angle of 60° to the ground. 3marks(a) What is the length of the ladder?(b) How far is the foot of the ladder from the base of the building?

A spherical balloon of radius r subtends an angle of 60° at the eye of an observer. If the angle of elevation of its centre is 45° from the same point, then prove that the height of the centre of the balloon is

\sqrt{2}

times its radius. 5marks

A kite is flying at a height of 75 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 45°. Find the length of the string, assuming that there is no slack in the string. 3marks

A bird is sitting on the top of a tree 20 m high. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies off vertically upwards. After some time, the angle of elevation of the bird from the same point is 60°. 5marks(a) Find the height of the bird when the angle of elevation is 60°.(b) Find the distance the bird has flown upwards.

A TV tower stands vertically on the edge of a canal. From a point on the opposite edge directly opposite to the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point and in the same line as this point and the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal. 5marks

Assertion: The angle of elevation of the top of a tower from a point on the ground 20 m away from its foot is 45°.Reason: The height of the tower is 20 m. 3marks

Two poles of equal heights are standing opposite each other on either side of a road which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. 5marks(a) Find the height of the poles.(b) Find the distances of the point from the poles.

A cliff is 100 m high. From the top of the cliff, the angle of depression of a boat is 30°. The boat moves further away from the cliff. After 2 minutes, the angle of depression is 15°. Find the speed of the boat. (Assume

\tan 15^{\circ} = 2 - \sqrt{3}

) 5marks

A pole of height 10 m is observed from a point on the ground. The angle of elevation of its top is 30°. The pole breaks due to storm and the broken part bends so that the top touches the ground, making an angle 30° with it. Find the height of the broken part. 5marks

A vertical flagstaff stands on a building. The angles of elevation of the top and bottom of the flagstaff from a point on the ground are 60° and 45° respectively. If the building is 20 m high, find the height of the flagstaff. 5marks

A helicopter is at an altitude of 800 m. It is flying horizontally at a constant speed. From a point on the ground, the angle of elevation of the helicopter is initially 60°. After 1 minute, the angle of elevation reduces to 30°. Find the speed of the helicopter. 5marks

A lighthouse is 75 m high. From its top, the angles of depression of two ships observed at the same time are 30° and 45°. If the ships are on the same side of the lighthouse and one is exactly behind the other, find the distance between the two ships. 5marks

Answers

Let the height of the balloon be h metres and the distance of the balloon from point Q be x metres. The distance of the balloon from point P will be (x + 20) metres.In the right triangle formed by the balloon and point Q:

\tan 45^{\circ} = \frac{h}{x} ⟹ 1 = \frac{h}{x} ⟹ x = h

.In the right triangle formed by the balloon and point P:

\tan 60^{\circ} = \frac{h}{x + 20} ⟹ \sqrt{3} = \frac{h}{x + 20}

.Substituting x = h:

\sqrt{3} = \frac{h}{h + 20} ⟹ \sqrt{3} (h + 20) = h ⟹ \sqrt{3} h + 20 \sqrt{3} = h ⟹ 20 \sqrt{3} = h (1 - \sqrt{3})

h = \frac{20 \sqrt{3}}{\sqrt{3} - 1} = \frac{20 \sqrt{3} (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)} = \frac{20 (3 + \sqrt{3})}{3 - 1} = \frac{20 (3 + \sqrt{3})}{2} = 10 (3 + \sqrt{3})

m.Height of the balloon =

10 (3 + \sqrt{3})

Let the width of the river be x metres and the height of the tree be h metres.From the bank, angle of elevation is 60°:

\tan 60^{\circ} = \frac{h}{x} ⟹ \sqrt{3} = \frac{h}{x} ⟹ h = x \sqrt{3}

.When the student moves 20 m away, the new distance is (x + 20) m. The angle of elevation is 30°.

\tan 30^{\circ} = \frac{h}{x + 20} ⟹ \frac{1}{\sqrt{3}} = \frac{h}{x + 20} ⟹ x + 20 = h \sqrt{3}

.Substitute

h = x \sqrt{3}

x + 20 = (x \sqrt{3}) \sqrt{3} ⟹ x + 20 = 3 x ⟹ 2 x = 20 ⟹ x = 10

m.Width of the river = 10 m.Height of the tree,

h = 10 \sqrt{3}

(a) Let the height of the pedestal be

h_{p}

m and the height of the statue be

h_{s}

m. Let the distance from the point on the ground to the base of the pedestal be d m.Angle of elevation to the top of the pedestal is 45°:

\tan 45^{\circ} = \frac{h_{p}}{d} ⟹ 1 = \frac{h_{p}}{d} ⟹ d = h_{p}

.Angle of elevation to the top of the statue is 60°: The total height is

(h_{p} + h_{s})

\tan 60^{\circ} = \frac{h_{p} + h_{s}}{d} ⟹ \sqrt{3} = \frac{h_{p} + h_{s}}{d}

.Substitute

d = h_{p}

\sqrt{3} = \frac{h_{p} + h_{s}}{h_{p}} ⟹ \sqrt{3} h_{p} = h_{p} + h_{s} ⟹ h_{s} = (\sqrt{3} - 1) h_{p}

.We are given

h_{s} = 1.6

m.(b)

1.6 = (\sqrt{3} - 1) h_{p} ⟹ h_{p} = \frac{1.6}{\sqrt{3} - 1} = \frac{1.6 (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)} = \frac{1.6 (\sqrt{3} + 1)}{3 - 1} = \frac{1.6 (\sqrt{3} + 1)}{2} = 0.8 (\sqrt{3} + 1)

m.Height of the pedestal =

0.8 (\sqrt{3} + 1)

m.(c) The height of the statue is given as 1.6 m. (Note: The question asks for the height of the statue which is already provided in the problem statement. It is likely asking for verification or to check understanding of what 'statue' refers to in contrast to 'pedestal').

(a) Let the width of the river be x metres and the height of the tower be h metres.From the opposite bank, angle of elevation of the top is 60°:

\tan 60^{\circ} = \frac{h}{x} ⟹ \sqrt{3} = \frac{h}{x} ⟹ h = x \sqrt{3}

.From another point 20 m away, angle of elevation is 30°. The distance is (x + 20) m.

\tan 30^{\circ} = \frac{h}{x + 20} ⟹ \frac{1}{\sqrt{3}} = \frac{h}{x + 20} ⟹ x + 20 = h \sqrt{3}

.Substitute

h = x \sqrt{3}

x + 20 = (x \sqrt{3}) \sqrt{3} ⟹ x + 20 = 3 x ⟹ 2 x = 20 ⟹ x = 10

m.Width of the river = 10 m.(b) Height of the tower,

h = 10 \sqrt{3}

Assertion: True. Let the height of the tower be H = 100 m and the distance of the car from the tower be D. The angle of depression is 30°.The angle of elevation from the car to the top of the tower is also 30°.

\tan 30^{\circ} = \frac{H}{D} ⟹ \frac{1}{\sqrt{3}} = \frac{100}{D} ⟹ D = 100 \sqrt{3}

m.Reason: True. For a right-angled triangle,

\tan θ = \frac{opposite side}{adjacent side}

. In this context, the opposite side is the height of the tower and the adjacent side is the distance from the tower.

Let the height of the lighthouse be H = 75 m. Let the distances of the two ships from the base of the lighthouse be

d_{1}

and

d_{2}

.The angle of depression of the first ship is 30°. The angle of elevation from the ship to the top of the lighthouse is also 30°.

\tan 30^{\circ} = \frac{75}{d_{1}} ⟹ \frac{1}{\sqrt{3}} = \frac{75}{d_{1}} ⟹ d_{1} = 75 \sqrt{3}

m.The angle of depression of the second ship is 45°. The angle of elevation from the ship to the top of the lighthouse is also 45°.

\tan 45^{\circ} = \frac{75}{d_{2}} ⟹ 1 = \frac{75}{d_{2}} ⟹ d_{2} = 75

m.Since the ships are on opposite sides, the distance between them is

d_{1} + d_{2} = 75 \sqrt{3} + 75 = 75 (\sqrt{3} + 1)

Let the height of the cliff be 80 m. Let the initial distance of the boat from the cliff be

d_{1}

and the final distance be

d_{2}

.Initial angle of depression is 30°. Angle of elevation from the boat is 30°.

\tan 30^{\circ} = \frac{80}{d_{1}} ⟹ \frac{1}{\sqrt{3}} = \frac{80}{d_{1}} ⟹ d_{1} = 80 \sqrt{3}

m.After 2 minutes, the angle of depression is 60°. Angle of elevation from the boat is 60°.

\tan 60^{\circ} = \frac{80}{d_{2}} ⟹ \sqrt{3} = \frac{80}{d_{2}} ⟹ d_{2} = \frac{80}{\sqrt{3}}

m.The distance travelled by the boat is

d_{1} - d_{2} = 80 \sqrt{3} - \frac{80}{\sqrt{3}} = 80 (\sqrt{3} - \frac{1}{\sqrt{3}}) = 80 (\frac{3 - 1}{\sqrt{3}}) = \frac{160}{\sqrt{3}}

m.Speed of the boat =

\frac{Distance}{Time} = \frac{160 / \sqrt{3} m}{2 min} = \frac{80}{\sqrt{3}} m/min = \frac{80 \sqrt{3}}{3}

m/min.

(a) Let the height of the building be

H_{b} = 60

m and the height of the lamp post be

h_{l}

. Let the horizontal distance between the building and the lamp post be d.From the top of the building, the angle of depression of the bottom of the lamp post is 60°. The angle of elevation from the bottom of the lamp post to the top of the building is also 60°.

\tan 60^{\circ} = \frac{H_{b}}{d} ⟹ \sqrt{3} = \frac{60}{d} ⟹ d = \frac{60}{\sqrt{3}} = 20 \sqrt{3}

m.Horizontal distance =

20 \sqrt{3}

m.(b) From the top of the building, the angle of depression of the top of the lamp post is 30°. Let the height of the lamp post from the ground up to the level of the top of the lamp post be

h_{l}

. The vertical distance from the top of the building to the top of the lamp post is

(H_{b} - h_{l})

\tan 30^{\circ} = \frac{H_{b} - h_{l}}{d} ⟹ \frac{1}{\sqrt{3}} = \frac{60 - h_{l}}{20 \sqrt{3}}

20 \sqrt{3} = \sqrt{3} (60 - h_{l}) ⟹ 20 = 60 - h_{l} ⟹ h_{l} = 60 - 20 = 40

m.Height of the lamp post = 40 m.

(a) Let the height of the building be H = 12 m. Let the length of the ladder be L. The angle the ladder makes with the ground is 60°.

\sin 60^{\circ} = \frac{Height of building}{Length of ladder} ⟹ \frac{\sqrt{3}}{2} = \frac{12}{L}

L = \frac{12 \times 2}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8 \sqrt{3}

m.Length of the ladder =

8 \sqrt{3}

m.(b) Let the distance of the foot of the ladder from the base of the building be d.

\tan 60^{\circ} = \frac{Height of building}{Distance from base} ⟹ \sqrt{3} = \frac{12}{d}

d = \frac{12}{\sqrt{3}} = 4 \sqrt{3}

m.Distance of the foot of the ladder from the base of the building =

4 \sqrt{3}

Let the radius of the balloon be r. Let the height of the centre of the balloon from the observer be H.The angle subtended by the balloon at the eye of the observer is 60°. This means the angle subtended by the diameter of the balloon at the eye is 60°.Consider the triangle formed by the observer and the two extreme points of the diameter of the balloon perpendicular to the line of sight. Let this diameter be 2r. Let the distance of the observer from the center of the balloon be D.

\sin (60^{\circ} / 2) = \frac{r}{D} ⟹ \sin 30^{\circ} = \frac{r}{D} ⟹ \frac{1}{2} = \frac{r}{D} ⟹ D = 2 r

.The angle of elevation of the center of the balloon from the same point is 45°. Let the height of the centre of the balloon from the ground be H.Consider the right-angled triangle formed by the observer, the point directly below the centre of the balloon on the ground, and the centre of the balloon. Let the distance from the observer to the point below the centre be d'.

\tan 45^{\circ} = \frac{H}{d^{'}} ⟹ 1 = \frac{H}{d^{'}} ⟹ d^{'} = H

.Now, consider the observer, the point directly below the centre of the balloon, and the centre of the balloon. The distance from the observer to the center of the balloon is D.The distance of the observer from the point directly below the center is

d^{'}

. The height of the center is H.Using Pythagoras theorem:

D^{2} = (d^{'})^{2} + H^{2}

.Since

d^{'} = H

D^{2} = H^{2} + H^{2} = 2 H^{2}

D = \sqrt{2} H

.We know

D = 2 r

.So,

2 r = \sqrt{2} H ⟹ H = \frac{2 r}{\sqrt{2}} = \sqrt{2} r

.Thus, the height of the centre of the balloon is

\sqrt{2}

times its radius.

Let the height of the kite be h = 60 m. Let the length of the string be L. The inclination of the string with the ground is 60°.

\sin 60^{\circ} = \frac{Height of kite}{Length of string} ⟹ \frac{\sqrt{3}}{2} = \frac{60}{L}

L = \frac{60 \times 2}{\sqrt{3}} = \frac{120}{\sqrt{3}} = 40 \sqrt{3}

m.Length of the string =

40 \sqrt{3}

(a) Let the initial height of the bird be

h_{1} = 20

m. The angle of elevation from the point on the ground is 45°. Let the distance of the point from the tree be d.

\tan 45^{\circ} = \frac{h_{1}}{d} ⟹ 1 = \frac{20}{d} ⟹ d = 20

m.The bird flies off vertically upwards. After some time, the angle of elevation is 60°. Let the new height of the bird be

h_{2}

\tan 60^{\circ} = \frac{h_{2}}{d} ⟹ \sqrt{3} = \frac{h_{2}}{20}

h_{2} = 20 \sqrt{3}

m.Height of the bird when the angle of elevation is 60° =

20 \sqrt{3}

m.(b) The distance the bird has flown upwards =

h_{2} - h_{1} = 20 \sqrt{3} - 20 = 20 (\sqrt{3} - 1)

Let the width of the canal be x metres and the height of the tower be h metres.From the opposite edge, the angle of elevation of the top is 60°:

\tan 60^{\circ} = \frac{h}{x} ⟹ \sqrt{3} = \frac{h}{x} ⟹ h = x \sqrt{3}

.From another point 20 m away, the angle of elevation is 30°. The distance is (x + 20) m.

\tan 30^{\circ} = \frac{h}{x + 20} ⟹ \frac{1}{\sqrt{3}} = \frac{h}{x + 20} ⟹ x + 20 = h \sqrt{3}

.Substitute

h = x \sqrt{3}

x + 20 = (x \sqrt{3}) \sqrt{3} ⟹ x + 20 = 3 x ⟹ 2 x = 20 ⟹ x = 10

m.Height of the tower,

h = 10 \sqrt{3}

Assertion: True. Let the height of the tower be H and the distance from the foot be D = 20 m. The angle of elevation is 45°.

\tan 45^{\circ} = \frac{H}{D} ⟹ 1 = \frac{H}{20} ⟹ H = 20

m.Reason: True. The height calculated from the assertion matches the reason.

Let the height of the poles be h metres. Let the width of the road be 80 m. Let the point on the road be x metres from one pole. Then it is (80 - x) metres from the other pole.From the first pole, angle of elevation is 60°:

\tan 60^{\circ} = \frac{h}{x} ⟹ \sqrt{3} = \frac{h}{x} ⟹ h = x \sqrt{3}

.From the second pole, angle of elevation is 30°:

\tan 30^{\circ} = \frac{h}{80 - x} ⟹ \frac{1}{\sqrt{3}} = \frac{h}{80 - x} ⟹ 80 - x = h \sqrt{3}

.Substitute

h = x \sqrt{3}

80 - x = (x \sqrt{3}) \sqrt{3} ⟹ 80 - x = 3 x ⟹ 4 x = 80 ⟹ x = 20

m.(a) Height of the poles,

h = 20 \sqrt{3}

m.(b) Distances of the point from the poles are x = 20 m and (80 - x) = 60 m.

Let the height of the cliff be H = 100 m. Let the initial distance of the boat from the cliff be

d_{1}

and the final distance be

d_{2}

.The angle of depression of the boat is initially 30°. So, the angle of elevation from the boat is 30°.

\tan 30^{\circ} = \frac{100}{d_{1}} ⟹ \frac{1}{\sqrt{3}} = \frac{100}{d_{1}} ⟹ d_{1} = 100 \sqrt{3}

m.After 2 minutes, the angle of depression is 15°. So, the angle of elevation from the boat is 15°.

\tan 15^{\circ} = \frac{100}{d_{2}} ⟹ (2 - \sqrt{3}) = \frac{100}{d_{2}} ⟹ d_{2} = \frac{100}{2 - \sqrt{3}} = \frac{100 (2 + \sqrt{3})}{(2 - \sqrt{3}) (2 + \sqrt{3})} = \frac{100 (2 + \sqrt{3})}{4 - 3} = 100 (2 + \sqrt{3})

m.The distance travelled by the boat is

d_{2} - d_{1} = 100 (2 + \sqrt{3}) - 100 \sqrt{3} = 200 + 100 \sqrt{3} - 100 \sqrt{3} = 200

m.Speed of the boat =

\frac{Distance}{Time} = \frac{200 m}{2 min} = 100

m/min.

Let the height of the pole be 10 m. Let the broken part be x m. Then the remaining part is (10-x) m.The broken part bends to touch the ground, making an angle of 30° with it.In the right-angled triangle formed by the remaining part, the ground, and the broken part, the remaining part is the adjacent side and the broken part is the hypotenuse.

\cos 30^{\circ} = \frac{10 - x}{hypotenuse}

. This is not helpful.Consider the remaining part as adjacent side and the broken part as hypotenuse.

\cos 30^{\circ} = \frac{remaining part}{broken part} ⟹ \frac{\sqrt{3}}{2} = \frac{10 - x}{x}

\sqrt{3} x = 2 (10 - x) ⟹ \sqrt{3} x = 20 - 2 x ⟹ (\sqrt{3} + 2) x = 20

x = \frac{20}{\sqrt{3} + 2} = \frac{20 (2 - \sqrt{3})}{(2 + \sqrt{3}) (2 - \sqrt{3})} = \frac{20 (2 - \sqrt{3})}{4 - 3} = 20 (2 - \sqrt{3})

m.Height of the broken part =

20 (2 - \sqrt{3})

Let the height of the building be

H_{b} = 20

m. Let the height of the flagstaff be

h_{f}

. Let the distance from the point on the ground to the building be d.The angle of elevation of the bottom of the flagstaff (top of the building) is 45°.

\tan 45^{\circ} = \frac{H_{b}}{d} ⟹ 1 = \frac{20}{d} ⟹ d = 20

m.The angle of elevation of the top of the flagstaff is 60°. The total height is

H_{b} + h_{f}

\tan 60^{\circ} = \frac{H_{b} + h_{f}}{d} ⟹ \sqrt{3} = \frac{20 + h_{f}}{20}

20 \sqrt{3} = 20 + h_{f} ⟹ h_{f} = 20 \sqrt{3} - 20 = 20 (\sqrt{3} - 1)

m.Height of the flagstaff =

20 (\sqrt{3} - 1)

Let the initial altitude of the helicopter be

H = 800

m. Let the initial distance of the helicopter from the point on the ground be

d_{1}

. The angle of elevation is 60°.

\sin 60^{\circ} = \frac{H}{d_{1}} ⟹ \frac{\sqrt{3}}{2} = \frac{800}{d_{1}} ⟹ d_{1} = \frac{1600}{\sqrt{3}}

m.After 1 minute, the angle of elevation reduces to 30°. Let the new distance be

d_{2}

. The altitude remains 800 m.

\sin 30^{\circ} = \frac{H}{d_{2}} ⟹ \frac{1}{2} = \frac{800}{d_{2}} ⟹ d_{2} = 1600

m.The horizontal distance covered by the helicopter in 1 minute is

d_{1} - d_{2}

. However, this is not correct. The angles of elevation are from the same point on the ground. Let the horizontal distances from the point on the ground be

x_{1}

and

x_{2}

.Initial position:

\tan 60^{\circ} = \frac{800}{x_{1}} ⟹ \sqrt{3} = \frac{800}{x_{1}} ⟹ x_{1} = \frac{800}{\sqrt{3}}

m.After 1 minute:

\tan 30^{\circ} = \frac{800}{x_{2}} ⟹ \frac{1}{\sqrt{3}} = \frac{800}{x_{2}} ⟹ x_{2} = 800 \sqrt{3}

m.The distance travelled by the helicopter in 1 minute is

x_{2} - x_{1} = 800 \sqrt{3} - \frac{800}{\sqrt{3}} = 800 (\sqrt{3} - \frac{1}{\sqrt{3}}) = 800 (\frac{3 - 1}{\sqrt{3}}) = \frac{1600}{\sqrt{3}}

m.Speed of the helicopter =

\frac{Distance}{Time} = \frac{1600 / \sqrt{3} m}{1 min} = \frac{1600}{\sqrt{3}} m/min = \frac{1600 \sqrt{3}}{3}

m/min.

Let the height of the lighthouse be H = 75 m. Let the distances of the two ships from the base of the lighthouse be

d_{1}

and

d_{2}

.The angle of depression of the first ship is 30°. The angle of elevation from the ship to the top of the lighthouse is also 30°.

\tan 30^{\circ} = \frac{75}{d_{1}} ⟹ \frac{1}{\sqrt{3}} = \frac{75}{d_{1}} ⟹ d_{1} = 75 \sqrt{3}

m.The angle of depression of the second ship is 45°. The angle of elevation from the ship to the top of the lighthouse is also 45°.

\tan 45^{\circ} = \frac{75}{d_{2}} ⟹ 1 = \frac{75}{d_{2}} ⟹ d_{2} = 75

m.Since the ships are on the same side and one is behind the other, the distance between the two ships is

∣ d_{1} - d_{2} ∣

.Distance between the two ships =

∣ 75 \sqrt{3} - 75 ∣ = 75 (\sqrt{3} - 1)

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