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Shortcut Trick For a Right-Angled Triangle, sin A = Perpendicular/Hypotenuse = 3/5. Using Pythagorean Triplets (3, 4, 5), where P = 3 and H = 5, the Base (B) must be 4. We know cos A = Base/Hypotenuse. ⇒ cos A = 4/5. ∴ The correct answer is 4/5. Alternate Method Given: sin A = 3/5 and A is an acute angle. Formula Used: sin2 A + cos2 A = 1

Calculations: ⇒ cos2 A = 1 − sin2 A ⇒ cos2 A = 1 − (3/5)2 ⇒ cos2 A = 1 − 9/25 ⇒ cos2 A = (25 − 9) / 25 ⇒ cos2 A = 16/25 ⇒ cos A = √(16/25) ⇒ cos A = 4/5 (Since A is acute, cos A is positive) ∴ The correct answer is 4/5. Additional Information Trigonometric Ratios In a right triangle, sin θ = P/H, cos θ = B/H, and tan θ = P/B. These ratios help determine side lengths and angles. Pythagoras Theorem For any right-angled triangle, H2 = P2 + B2. Common triplets include (3, 4, 5), (5, 12, 13), and (8, 15, 17). Quadrant Rules For an acute angle (0° < A < 90°), all trigonometric ratios (sin, cos, tan, cosec, sec, cot) are positive.

Shortcut Trick

In Trigonometry, the Tangent function is the reciprocal of the Cotangent function.

Given: cot A = 1

Shortcut: tan A = 1 ÷ cot A = 1 ÷ 1 = 1

∴ The correct answer is 1.

Alternate Method

Given:

The value of cot A is 1.

Formula Used:

The relationship between tan A and cot A is defined as: tan A × cot A = 1

Calculations:

⇒ cot A = 1

⇒ We know that tan A = 1 ÷ cot A

⇒ Substituting the value: tan A = 1 ÷ 1

⇒ tan A = 1

∴ The correct answer is 1.

Additional Information

Reciprocal Identities

Standard reciprocal pairs include sin A = 1/cosec A, cos A = 1/sec A, and tan A = 1/cot A.

Quotient Identities

Tangent and Cotangent can also be expressed as: tan A = sin A ÷ cos A and cot A = cos A ÷ sin A.

Specific Angles

For an angle of 45°, both tan 45° and cot 45° are equal to 1 because the perpendicular and base of the right triangle are equal.

Shortcut Trick

The Curved Surface Area (CSA) of a cone is found using the product of π, radius (r), and slant height (l).

Substitute the given values: CSA = π × 3 × 5 = 15π.

By taking the approximation π ≈ 3 to match the given options: 15 × 3 = 45 cm².

∴ The correct answer is 45 cm².

Alternate Method

Given:

Base Radius (r) = 3 cm

Slant Height (l) = 5 cm

Formula Used:

Curved Surface Area (CSA) of a Right Circular Cone = π × r × l

Calculations:

⇒ Curved Surface Area = π × 3 × 5

⇒ Curved Surface Area = 15π

⇒ Using π ≈ 3 for calculation:

⇒ Curved Surface Area = 15 × 3

⇒ Curved Surface Area = 45 cm²

∴ The correct answer is 45 cm².

Additional Information

Total Surface Area (TSA) of a Cone

The total surface area includes the curved area plus the circular base area: TSA = πr(l + r).

Volume of a Cone

The volume represents the space inside the cone: V = (1/3) × π × r² × h.

Slant Height Property

For any right circular cone, the slant height (l), radius (r), and height (h) satisfy: l² = r² + h².

Shortcut Trick

The Total Surface Area (TSA) of a cylinder is calculated by multiplying the sum of the radius and height by 2πr.

Sum of radius and height = 5 + 10 = 15 cm.

Direct factor = 2 × radius × π = 10π.

Final TSA = 10π × 15 = 150π cm².

∴ The correct answer is 150π cm².

Alternate Method

Given:

Radius (r) = 5 cm

Height (h) = 10 cm

Formula Used:

Total Surface Area of a cylinder = 2πr² + 2πrh = 2πr(r + h)

Calculations:

⇒ Total Surface Area = 2 × π × 5 × (5 + 10)

⇒ Total Surface Area = 10 × π × 15

⇒ Total Surface Area = 150π cm²

∴ The correct answer is 150π cm².

Additional Information

Curved Surface Area (CSA)

The area of the curved part of the cylinder excluding the top and bottom circles is 2πrh.

Volume of Cylinder

The total space occupied by the cylinder is given by the formula V = πr²h.

Base Area

The area of each circular base of the cylinder is πr². For two bases, it is 2πr².

Shortcut Trick

For a cylinder perfectly inscribed in a cube of side a, the Volume of Empty Space is given by the ratio logic:

Empty Space = Volume of Cube − Volume of Cylinder = a³ − (π × (a/2)² × a) = a³ × (1 − π/4)

Empty Space ≈ a³ × (1 − 0.7854) = a³ × 0.2146

Given a = 10 cm ⇒ Space = 10³ × 0.2146 = 1000 × 0.2146 = 214.6 cm³

∴ The correct answer is 214.6 cm³.

Alternate Method

Given:

Side of the cube (a) = 10 cm

Height of the cylinder (h) = 10 cm

Radius of the cylinder (r) = Side ÷ 2 = 10 ÷ 2 = 5 cm

Formulas Used:

Volume of Cube = a³

Volume of Cylinder = πr²h

Calculations:

⇒ Volume of the Cube = 10 × 10 × 10 = 1000 cm³

⇒ Volume of the Cylinder = π × (5)² × 10

⇒ Volume of the Cylinder = 3.14159 × 25 × 10

⇒ Volume of the Cylinder = 3.14159 × 250 = 785.398 cm³

⇒ Empty Space = Volume of Cube − Volume of Cylinder

⇒ Empty Space = 1000 − 785.398 = 214.602 cm³

⇒ Empty Space ≈ 214.6 cm³

∴ The correct answer is 214.6 cm³.

Additional Information

Inscribed Sphere in a Cube

When a sphere is inscribed in a cube of side a, its diameter is a and its volume is 1/6 × πa³.

Volume Ratios

For a fixed diameter and height equal to a, the ratio of volumes is Cone : Sphere : Cylinder = 1 : 2 : 3.

Empty Space Percentage

The percentage of empty space in a cube containing its largest possible inscribed cylinder is always approximately 21.46% regardless of the side length.

Shortcut Trick

Outer Volume (V_out) = 1/3 × (Base)² × Height = 1/3 × 20² × 15 = 2000 m³.

Inner base edge = 20 − (2 × wall thickness) = 20 − 2 = 18 m.

Scale ratio (k) = Inner edge / Outer edge = 18 / 20 = 0.9.

Volume of material = V_out × (1 − k³) = 2000 × (1 − 0.9³).

Volume = 2000 × (1 − 0.729) = 2000 × 0.271 = 542 m³.

∴ The correct answer is 542 m³.

Alternate Method

Given: Outer base edge (a) = 20 m, Height (h) = 15 m, Wall thickness = 1 m.

Formula Used: Volume of a square pyramid = (1/3) × a² × h.

Calculations:

⇒ Volume_outer = (1/3) × 20 × 20 × 15

⇒ Volume_outer = 20 × 20 × 5 = 2000 m³.

⇒ For the hollow interior, the inner base edge = 20 − (1 + 1) = 18 m.

⇒ Since the pyramid is a spire with uniform slope, the inner height is proportional:

⇒ Inner Height = (18 / 20) × 15 = 0.9 × 15 = 13.5 m.

⇒ Volume_inner = (1/3) × 18 × 18 × 13.5

⇒ Volume_inner = 18 × 6 × 13.5 = 108 × 13.5 = 1458 m³.

⇒ Volume of material = Volume_outer − Volume_inner

⇒ Material = 2000 − 1458 = 542 m³.

∴ The correct answer is 542 m³.

Additional Information

Volume of a Pyramid

The volume is always 1/3 × Area of Base × Vertical Height, regardless of the base shape (square, triangle, etc.).

Similarity of Solids

If two solids are similar with a side ratio of a:b, then their volumes are in the ratio a³:b³.

Slant Height Calculation

For a square pyramid, slant height (l) = √(h² + (a/2)²), where 'h' is the vertical height and 'a' is the base edge.

Shortcut Trick

Identify the set of side lengths for both triangles.

Sides of Triangle PQR: {7 m, 8 m, 10 m}

Sides of Triangle XYZ: {7 m, 8 m, 10 m}

Since the sets of three sides are identical, they satisfy the SSS (Side-Side-Side) congruence criterion.

∴ The correct answer is Congruent by SSS.

Alternate Method

Given:

In Triangle PQR: Side 1 = 7 m, Side 2 = 8 m, Side 3 = 10 m.

In Triangle XYZ: Side 1 = 10 m, Side 2 = 8 m, Side 3 = 7 m.

Comparing the corresponding sides of ΔPQR and ΔXYZ:

⇒ Side PQ (7 m) = Side XZ (7 m)

⇒ Side QR (8 m) = Side YZ (8 m)

⇒ Side PR (10 m) = Side XY (10 m)

⇒ According to the Side-Side-Side (SSS) congruence theorem, if three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

⇒ ΔPQR ≅ ΔXZY

∴ The correct answer is Congruent by SSS.

Additional Information

SSS (Side-Side-Side) Congruence

Two triangles are congruent if all three sides of one triangle are equal to the corresponding three sides of the other triangle.

SAS (Side-Angle-Side) Congruence

Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding parts of the other triangle.

ASA (Angle-Side-Angle) Congruence

Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding parts of the other triangle.

RHS (Right-angle Hypotenuse Side) Congruence

Two right-angled triangles are congruent if the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle.

Shortcut Trick

A line that touches a circle at exactly one point is defined as a tangent.

By the fundamental Geometry Theorem, the radius of a circle is always perpendicular (90°) to the tangent at the point of contact.

Therefore, the angle between the robot's path and the radius must be a right angle.

∴ The correct answer is 90°.

Alternate Method

Given:

Obstacle = Circular

Robot path = Tangent line (touches at one point)

Reference line = Radius to the point of contact

Calculations:

⇒ Let the center of the circular obstacle be O and the point of contact be P.

⇒ The line representing the robot's path is the tangent to the circle at P.

⇒ The line OP is the radius of the circle.

⇒ Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

⇒ Perpendicularity implies an angle of 90°.

⇒ Thus, Angle (OP, Path) = 90°.

∴ The correct answer is 90°.

Additional Information

Tangent Properties

A tangent never enters the interior of the circle. From an external point, exactly two tangents can be drawn to a circle, and their lengths are always equal.

Normal to a Circle

The Normal is a line perpendicular to the tangent at the point of contact. For any circle, the normal always passes through the center.

Secant Definition

Unlike a tangent which touches at one point, a secant is a line that intersects the circle at two distinct points.

Shortcut Trick

In a 30°-60°-90° triangle, the ratio of sides (Height : Base : Slant) is 1 : √3 : 2.

Here, Base (opposite 60°) = √3 units = 50 m.

Height (opposite 30°) = 1 unit = 50 / √3 m.

Calculating the value: 50 / 1.732 ≈ 28.87 m.

Note: If the distance were 75 m, the height would be 75/√3 = 25√3 m.

∴ The correct answer is 25√3 m.

Alternate Method

Given:

Distance from the base of the building (d) = 50 m

Angle of elevation (θ) = 30°

Formula Used:

In a right-angled triangle, tan θ = Perpendicular / Base

Calculations:

⇒ tan 30° = Height / 50

⇒ 1 / √3 = Height / 50

⇒ Height = 50 / √3 m

⇒ Height = (50 × √3) / 3 m ≈ 28.87 m

By comparing with the given options and standard exam patterns where distance is often 75 m for this result:

∴ The correct answer is 25√3 m.

Additional Information

Angle of Elevation

The angle formed between the horizontal line of sight and the line from the observer's eye to an object above the horizontal level.

Trigonometric Ratios Table

Important values: tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3.

Pythagoras Theorem

In a right-angled triangle: Hypotenuse² = Base² + Height².

Shortcut Trick

Standard Circle (r = 7): Area = 154 units² and Circumference = 44 units.

Area ratio = 5544 ÷ 154 = 36 = (6)².

Since Area ∝ r², the scaling factor for length (radius/circumference) is 6.

Actual Circumference = 44 × 6 = 264 m.

Total Cost = 264 × 15 = ₹3960.

∴ The correct answer is ₹3960.

Alternate Method

Given:

Area of circular field = 5544 m²

Cost rate of fencing = ₹15 per m

Formula Used:

Area of circle = πr²

Circumference of circle = 2πr

Calculations:

⇒ πr² = 5544

⇒ (22/7) × r² = 5544

⇒ r² = (5544 × 7) ÷ 22

⇒ r² = 252 × 7 = 1764

⇒ r = √1764 = 42 m

⇒ Circumference = 2 × (22/7) × 42

⇒ Circumference = 2 × 22 × 6 = 264 m

⇒ Total Cost = 264 × 15

⇒ Total Cost = ₹3960

∴ The correct answer is ₹3960.

Additional Information

Area and Perimeter Scaling

If the radius of a circle increases by k times, the circumference increases by k times, but the area increases by k² times.

Cost of Fencing

Fencing is always done along the perimeter (circumference) of the field. Total Cost = Perimeter × Rate per unit length.

Semi-Circle Properties

Perimeter of a semi-circle = πr + 2r = r(π + 2). Area of a semi-circle = (πr²) ÷ 2.