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Consider the diagram is shown above where PR represents the ladder and RQ represent the wall.
Consider the diagram is shown above where PR represents the ladder and RQ represent the wall.
Cos 60° = PQ / PR
⇒ 1 / 2 = 12.4 / PR
⇒ PR = 2 × 12.4 = 24.8 m
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Let us draw a figure below as per given question.
Let A be the position of a person on the bank of a river and OP the tree on the opposite bank and ∠OAP = 60°. When the person retires to the position B, then AB = 40 meter and ∠OBP = 30°
Use the trigonometry formula to solve the given question.
Let us draw a figure below as per given question.
Let A be the position of a person on the bank of a river and OP the tree on the opposite bank and ∠OAP = 60°. When the person retires to the position B, then AB = 40 meter and ∠OBP = 30°
Let us assume OA(Breadth of the river) = x meter and height of tree OP = h meter
In ΔOAP, Use the trigonometry formula
Tan60° = P/B = Perpendicular distance / Base distance
⇒ Tan60° = OP / OA
⇒ OP = OA Tan60°
Put the value of OP and OA, We will get
⇒ h = x√3 ..............(1)
Now in the triangle ΔOBP
Tan30° = OP / OB
⇒ OP = OB Tan30°
⇒ OP = (x + 40)/√3
⇒ h = (x + 40)/√3 ...................(2)
From Equation (1) and (2), We will get
⇒ (x + 40)/√3 = x√3
⇒ (x + 40) = x√3 X √3
⇒ (x + 40) = 3x
⇒ 3x - x = 40
∴ x = 20 m
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Let us draw the figure from the given question.
Let AB = h meter be the height of the tower; C and D are the two points on the ground such that BC = 60 m; ∠ACB = 45° and ∠ADB = 30°
Let us draw the figure from the given question.
Let AB = h meter be the height of the tower; C and D are the two points on the ground such that BC = 60 m; ∠ACB = 45° and ∠ADB = 30°
Now from right triangle ABC,
tan 45° = h/60
⇒ 1 = h/60
∴ h = 60 m;
Again from right triangle ABD;
tan 30° = h/(x + 60)
⇒ 1/√3 = 60/(x + 60)
⇒ x + 60 = 60√3
∴ x = 60(1.73 - 1) = 43.8 meter
Hence, speed of boat = 43.8/5 m/s = 43.8/5 x 18/5 = 31.5 km/hr.
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Let us draw a figure from given question.
Let, AB = 100 m be the height of a tower, P is a point on the ground such that ∠APB = 30°
Let us draw a figure from given question.
Let, AB = 100 m be the height of a tower, P is a point on the ground such that ∠APB = 30°
From right triangle ABP,
BP = 100 cot 30°
⇒ BP = 100 √3
⇒ BP = 100 X 1.73
⇒ BP = 173 meter
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Let us draw a figure below as per given question.
Let AB = 75 m be the height of pole and C is a point on the ground such that BC = 75m
Let us draw a figure below as per given question.
Let AB = 75 m be the height of pole and C is a point on the ground such that BC = 75m
Now, from right triangle ABC.
tan ∝ = AB/BC = 75/75
⇒ tan∝ = 1
∴ ∝ = 45°
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