📘 Quiz

Solution: Step 1: Let h_A be the height of tower A and h_B be the height of tower B. Step 2: Let 'x' be the horizontal distance from the observation point on the ground to the common base of the two towers. Step 3: For tower A, the angle of elevation is 30°. In the right triangle formed: tan(30°) = h_A / x. Step 4: So, h_A = x * tan(30°) = x / sqrt(3). Step 5: For tower B, the angle of elevation is 45°. In the right triangle formed: tan(45°) = h_B / x. Step 6: So, h_B = x * tan(45°) = x * 1 = x. Step 7: To find the ratio of heights h_A : h_B, divide the expression for h_A by h_B: h_A / h_B = (x / sqrt(3)) / x. Step 8: Simplify the ratio: h_A / h_B = 1 / sqrt(3). Step 9: The ratio of the heights of A and B is 1 : sqrt(3).

Solution: Step 1: Let the interior angles of ΔABC be ∠CAB = x, ∠CBA = y, and ∠ACB = z. We know x + y + z = 180°. Step 2: The bisector of the exterior angle at A meets BC produced at D. The exterior angle at A is (180° - x). The angle formed by this bisector with AC is ∠CAD = (180° - x)/2 = 90° - x/2. Step 3: The bisector of the exterior angle at B meets AC produced at E. The exterior angle at B is (180° - y). The angle formed by this bisector with BC is ∠EBC = (180° - y)/2 = 90° - y/2. Step 4: Given AB = AD. This implies ΔABD is an isosceles triangle. Since D is on BC produced, ∠ABD refers to ∠ABC (y), and thus ∠ADB = ∠ABD = y. Step 5: In ΔABD, the sum of angles is 180°: ∠BAD + ∠ABD + ∠ADB = 180°. Step 6: ∠BAD = ∠CAB + ∠CAD = x + (90° - x/2) = 90° + x/2. Step 7: Substitute into the sum of angles: (90° + x/2) + y + y = 180°. This simplifies to x/2 + 2y = 90°, or x + 4y = 180° (Equation 1). Step 8: Given AB = BE. This implies ΔABE is an isosceles triangle. Since E is on AC produced, ∠BAE refers to ∠CAB (x), and thus ∠BEA = ∠BAE = x. Step 9: In ΔABE, the sum of angles is 180°: ∠ABE + ∠BAE + ∠BEA = 180°. Step 10: ∠ABE = ∠CBA + ∠CBE = y + (90° - y/2) = 90° + y/2. Step 11: Substitute into the sum of angles: (90° + y/2) + x + x = 180°. This simplifies to y/2 + 2x = 90°, or y + 4x = 180° (Equation 2). Step 12: Solve the system of linear equations: (1) x + 4y = 180° (2) 4x + y = 180° Step 13: Multiply Equation (1) by 4: 4x + 16y = 720°. Step 14: Subtract Equation (2) from the modified Equation (1): (4x + 16y) - (4x + y) = 720° - 180°. Step 15: 15y = 540°, so y = 36°. Step 16: Substitute y = 36° into Equation (1): x + 4(36°) = 180° ⇒ x + 144° = 180° ⇒ x = 36°. Step 17: Now find ∠ACB (z): z = 180° - (x + y) = 180° - (36° + 36°) = 180° - 72° = 108°.

Solution: Step 1: Let PQ be the height of the tree (h). Step 2: Let M be Mohan's initial position and N be his final position after moving back 10 cm. Q is the base of the tree. Step 3: Let MQ be the initial distance from the tree. Then NQ = MQ + 10 cm. Step 4: The initial angle of elevation (∠PMQ) is 45°. Step 5: The final angle of elevation (∠PNQ) is 30°. Step 6: In right-angled ΔPMQ: tan(45°) = PQ/MQ ⇒ 1 = h/MQ ⇒ MQ = h. Step 7: In right-angled ΔPNQ: tan(30°) = PQ/NQ. Step 8: Substitute NQ = MQ + 10 and MQ = h: 1/√3 = h/(h + 10). Step 9: Cross-multiply: h + 10 = h√3. Step 10: Rearrange terms to solve for h: 10 = h√3 - h ⇒ 10 = h(√3 - 1). Step 11: h = 10 / (√3 - 1) cm.

Solution: Step 1: Given ΔABC is an isosceles right-angled triangle at B. This means ∠ABC = 90°, and ∠BAC = ∠BCA = 45°. Step 2: AE is perpendicular to AC (as Q is the foot of the perpendicular from D to AC), so ΔAQD is a right-angled triangle at Q. Step 3: AP is perpendicular to AB (as P is the foot of the perpendicular from D to AB), so ΔAPD is a right-angled triangle at P. Step 4: Calculate ∠DAQ: ∠DAQ = ∠BAC - ∠BAD = 45° - 15° = 30°. Step 5: In right-angled ΔAQD: - We know AQ = b and ∠DAQ = 30°. - Using cosine relation: cos(∠DAQ) = Adjacent/Hypotenuse = AQ/AD. - cos(30°) = b/AD. - √3/2 = b/AD. - Solve for AD: AD = 2b/√3. Step 6: Now consider right-angled ΔAPD: - We know AP = a. - We need to find sin 75°. Notice that ∠ADP = 180° - ∠APD - ∠PAD = 180° - 90° - 15° = 75°. - Using sine relation in ΔAPD: sin(∠ADP) = Opposite/Hypotenuse = AP/AD. - So, sin(75°) = AP/AD. Step 7: Substitute the values AP = a and AD = 2b/√3 from Step 5: - sin(75°) = a / (2b/√3). Step 8: Simplify the expression: sin(75°) = a * (√3 / 2b) = (a√3) / (2b). Step 9: Therefore, the value of sin 75° is (a√3) / (2b).

Solution: Step 1: Recall the property of similar triangles: The ratio of their areas is equal to the square of the ratio of their corresponding altitudes (heights). Step 2: Let A1 be the area of the smaller triangle and h1 be its height. A1 = 12 cm², h1 = 2.1 cm. Step 3: Let A2 be the area of the larger triangle and h2 be its height. A2 = 48 cm². Step 4: Set up the ratio according to the property: A1 / A2 = (h1 / h2)². Step 5: Substitute the given values: 12 / 48 = (2.1 / h2)². Step 6: Simplify the area ratio: 1/4 = (2.1 / h2)². Step 7: Take the square root of both sides: √(1/4) = √(2.1 / h2)². Step 8: 1/2 = 2.1 / h2. Step 9: Solve for h2: h2 = 2.1 * 2 = 4.2 cm. Step 10: The corresponding height of the bigger triangle is 4.2 cm.

Solution: Step 1: Rewrite sec(30° + α) as 1/cos(30° + α). Step 2: The given equation becomes sin α / cos(30° + α) = 1. Step 3: This implies sin α = cos(30° + α). Step 4: Recall the identity: If sin A = cos B, then A + B = 90° (for acute angles). Step 5: Apply this identity: α + (30° + α) = 90°. Step 6: Solve for α: 2α + 30° = 90° ⇒ 2α = 60° ⇒ α = 30°. Step 7: Substitute α = 30° into the expression sin α + cos 2α. Step 8: sin 30° + cos (2 × 30°) = sin 30° + cos 60°. Step 9: Recall the special angle values: sin 30° = 1/2 and cos 60° = 1/2. Step 10: The value is 1/2 + 1/2 = 1.

Solution: Step 1: Visualize the path taken. Traveling east and then north forms two perpendicular sides of a right-angled triangle. - The eastward journey (12 km) forms one leg. - The northward journey (5 km) forms the other leg. - The shortest distance from L to M is the hypotenuse of this right triangle. Step 2: Apply the Pythagorean theorem (`a^2 + b^2 = c^2`). - Let `a = 12` km and `b = 5` km. - `LM^2 = 12^2 + 5^2` - `LM^2 = 144 + 25` - `LM^2 = 169` Step 3: Calculate the square root to find the distance LM. - `LM = sqrt(169) = 13` km. Step 4: The shortest distance from L to M is 13 km.

Solution: Step 1: Let the height of the chimney be AB = h. Step 2: Let the initial position of the observer be D, and the position after walking 'x' meters be C. Step 3: So, the distance CD = x. Let the distance BC = y. Step 4: The initial distance from the chimney's base is BD = BC + CD = y + x. Step 5: The initial angle of elevation from D to A is 30°, so angle ADB = 30°. Step 6: The final angle of elevation from C to A is 60°, so angle ACB = 60°. Step 7: In right triangle ABC (right-angled at B), tan(60°) = AB/BC => √3 = h/y => y = h/√3 --- (Equation 1). Step 8: In right triangle ABD (right-angled at B), tan(30°) = AB/BD => 1/√3 = h/(y + x) => y + x = h√3 --- (Equation 2). Step 9: Substitute the value of 'y' from Equation 1 into Equation 2: (h/√3) + x = h√3. Step 10: Isolate x: x = h√3 - h/√3. Step 11: Find a common denominator for the right side: x = (h * 3 - h) / √3 = 2h / √3. Step 12: Solve for h: h = x√3 / 2. Therefore, the height of the chimney is x√3 / 2.

Solution: Step 1: Let the measures of the four angles of the quadrilateral be x, 2x, 4x, and 5x, based on the given ratio. Step 2: Recall that the sum of the interior angles of a quadrilateral is 360°. Step 3: Set up an equation using the sum of the angles: x + 2x + 4x + 5x = 360°. Step 4: Combine the terms: 12x = 360°. Step 5: Solve for x: x = 360 / 12 = 30°. Step 6: Now calculate the measure of each angle: - First angle: x = 30°. - Second angle: 2x = 2 × 30° = 60°. - Third angle: 4x = 4 × 30° = 120°. - Fourth angle: 5x = 5 × 30° = 150°. Step 7: Identify the largest angle from these values, which corresponds to 5x. Step 8: Final result: The biggest angle is 150°.

Solution: Step 1: Divide the numerator and denominator by cosθ to convert to tangent and secant terms: Numerator: (sinθ/cosθ - cosθ/cosθ + 1/cosθ) = tanθ - 1 + secθ. Denominator: (sinθ/cosθ + cosθ/cosθ - 1/cosθ) = tanθ + 1 - secθ. Step 2: Rewrite the expression as (tanθ + secθ - 1) / (tanθ - secθ + 1). Step 3: Use the identity sec²θ - tan²θ = 1. So, 1 can be replaced by (sec²θ - tan²θ) in the numerator. Step 4: Numerator becomes (tanθ + secθ) - (sec²θ - tan²θ). Step 5: Factor (sec²θ - tan²θ) as (secθ - tanθ)(secθ + tanθ). Step 6: Numerator becomes (tanθ + secθ) - (secθ - tanθ)(secθ + tanθ). Step 7: Factor out (tanθ + secθ) from the numerator: (tanθ + secθ) * [1 - (secθ - tanθ)]. Step 8: Numerator simplifies to (tanθ + secθ) * [1 - secθ + tanθ]. Step 9: The original expression is now [(tanθ + secθ)(1 + tanθ - secθ)] / (1 + tanθ - secθ). Step 10: Cancel out the common term (1 + tanθ - secθ) from the numerator and denominator (assuming it's not zero). Step 11: The simplified expression is tanθ + secθ.

Solution: Step 1: In an isosceles triangle ABC with AB = AC, the angles opposite to the equal sides are equal. So, angle C = angle B = 35°. Step 2: Calculate angle BAC of the triangle: angle BAC = 180° - (angle B + angle C) = 180° - (35° + 35°) = 180° - 70° = 110°. Step 3: In an isosceles triangle, the median drawn to the base is also the altitude to the base and the angle bisector of the vertex angle. Step 4: Since AD is the median to the base BC, it bisects angle BAC. Step 5: Therefore, angle BAD = angle BAC / 2 = 110° / 2 = 55°.

Solution: Step 1: Let the height of the tower be H = 100 m. Step 2: When the angle of elevation is 45°, let the shadow length be S1. Step 3: In the right-angled triangle, tan(45°) = H / S1. Step 4: Since tan(45°) = 1, we have 1 = 100 / S1, so S1 = 100 m. Step 5: When the angle of elevation is 30°, let the shadow length be S2. Step 6: In the right-angled triangle, tan(30°) = H / S2. Step 7: Since tan(30°) = 1/sqrt(3), we have 1/sqrt(3) = 100 / S2, so S2 = 100 * sqrt(3) m. Step 8: The decrease in shadow length, x, is S2 - S1. Step 9: x = 100 * sqrt(3) - 100 = 100 * (sqrt(3) - 1) meters.

Solution: Step 1: Let A be the observing point on the ground. Let AB be the horizontal distance from A to the point directly below the aeroplanes. Step 2: Let C be the higher aeroplane and D be the lower aeroplane, both vertically above B. Step 3: The height of the higher aeroplane, CB = 900 m. The angle of elevation to C, ∠CAB = 60°. Step 4: The angle of elevation to D, ∠DAB = 45°. Step 5: In right-angled ΔABC: tan(60°) = CB/AB ⇒ √3 = 900/AB ⇒ AB = 900/√3 = 300√3 m. Step 6: In right-angled ΔABD: tan(45°) = DB/AB ⇒ 1 = DB/AB ⇒ DB = AB. Step 7: So, the height of the lower aeroplane, DB = 300√3 m. Step 8: The difference in height between the two aeroplanes is CD = CB - DB = 900 - 300√3. Step 9: Substitute √3 ≈ 1.73: CD = 900 - 300 * 1.73 = 900 - 519 = 381 m.

Solution: Step 1: Let MQ be the standing part of the tree (height from the ground to the break point). Step 2: Let MN be the broken part of the tree that touches the ground at N. Step 3: Q is the foot of the tree, so ΔMNQ is a right-angled triangle at Q. Step 4: Given NQ = 10 m (distance from the foot of the tree to where the top touches the ground). Step 5: Given ∠MNQ = 30° (angle made by the fallen part with the ground). Step 6: In ΔMNQ, use the tangent function: tan(30°) = MQ/NQ. Step 7: 1/√3 = MQ/10 ⇒ MQ = 10/√3 m (Height of the stump). Step 8: In ΔMNQ, use the cosine function: cos(30°) = NQ/MN. Step 9: √3/2 = 10/MN ⇒ MN = 20/√3 m (Length of the broken part). Step 10: The original length of the tree = MQ + MN. Step 11: Total Length = 10/√3 + 20/√3 = 30/√3 m. Step 12: Rationalize the denominator: Total Length = (30√3)/3 = 10√3 m.

Solution: Step 1: Simplify the first part of the expression: (sinA / (1-cosA) + (1-cosA) / sinA). Find a common denominator: (sin²A + (1-cosA)²) / (sinA(1-cosA)) Expand (1-cosA)²: (sin²A + 1 - 2cosA + cos²A) / (sinA(1-cosA)) Use the identity sin²A + cos²A = 1: (1 + 1 - 2cosA) / (sinA(1-cosA)) = (2 - 2cosA) / (sinA(1-cosA)) Factor out 2: 2(1 - cosA) / (sinA(1-cosA)) = 2 / sinA. Step 2: Simplify the second part of the expression: (cot²A / (1+cosecA) + 1). Rewrite cot²A as cos²A/sin²A and cosecA as 1/sinA: ( (cos²A/sin²A) / (1 + 1/sinA) ) + 1 Simplify the denominator of the fraction: (1 + 1/sinA) = (sinA + 1)/sinA The fraction becomes: (cos²A/sin²A) ÷ ((sinA + 1)/sinA) = (cos²A/sin²A) × (sinA/(sinA + 1)) = cos²A / (sinA(sinA + 1)) Now add 1: cos²A / (sinA(sinA + 1)) + 1 = (cos²A + sinA(sinA + 1)) / (sinA(sinA + 1)) = (cos²A + sin²A + sinA) / (sinA(sinA + 1)) Use the identity sin²A + cos²A = 1: (1 + sinA) / (sinA(sinA + 1)) = 1 / sinA. Step 3: Divide the simplified first part by the simplified second part. (2 / sinA) ÷ (1 / sinA) = 2.

Solution: Step 1: Let ∠CAB = x. Step 2: According to the problem statement, ∠ABC = 2∠CAB, so ∠ABC = 2x. Step 3: Recall the Exterior Angle Theorem, which states that an exterior angle of a triangle is equal to the sum of its two opposite interior angles. Step 4: In ΔABC, when side BC is extended to D, ∠ACD is an exterior angle. Its opposite interior angles are ∠CAB and ∠ABC. Step 5: Apply the Exterior Angle Theorem: ∠ACD = ∠CAB + ∠ABC. Step 6: Substitute the given and defined angle values: 126° = x + 2x. Step 7: Simplify the equation: 126° = 3x. Step 8: Solve for x: x = 126° / 3 = 42°. Step 9: Therefore, ∠CAB = 42°.

Solution: Step 1: Given that ΔDAB ∽ ΔDCA (triangle DAB is similar to triangle DCA). Step 2: When two triangles are similar, the ratio of their corresponding sides is equal. Step 3: From the similarity statement, we can write the ratios of corresponding sides: DA / DC = AB / CA = DB / DA. Step 4: Substitute the given side lengths: AB = 20 cm, CA = 15 cm. AB / CA = 20 / 15 = 4 / 3. Step 5: Now, equate the ratios: DA / DC = 4 / 3 …(Equation 1) DB / DA = 4 / 3 …(Equation 2) Step 6: From Equation 1, DA = (4/3) * DC. Step 7: From Equation 2, DA = (3/4) * DB. Step 8: Equate the two expressions for DA: (4/3) * DC = (3/4) * DB. Step 9: Rearrange to find the ratio DC / DB: DC / DB = (3/4) * (3/4) = 9 / 16. Step 10: We know that D is a point on the extension of BC, so DB = BC + DC. Step 11: Substitute DB in the ratio: DC / (BC + DC) = 9 / 16. Step 12: Cross-multiply: 16 * DC = 9 * (BC + DC). Step 13: 16 * DC = 9 * BC + 9 * DC. Step 14: Subtract 9 * DC from both sides: 7 * DC = 9 * BC. Step 15: Given BC = 7 cm. Substitute this value: 7 * DC = 9 * 7. Step 16: Solve for DC: DC = 9 cm. Step 17: Therefore, the length of DC is 9 cm.

Solution: Step 1: Given that C is the midpoint of BD, AC is a median to side BD in ΔABD. Step 2: According to Apollonius' Theorem, for a triangle with sides a, b, c and a median m to side a, the theorem states b² + c² = 2 * (m² + (a/2)²). Step 3: In ΔABD, we have AB, AD as two sides and AC as the median to side BD. Let BC = CD = x. Step 4: Applying Apollonius' Theorem: AB² + AD² = 2 * (AC² + BC²). Step 5: Substitute the given values: 10² + 12² = 2 * (9² + x²). Step 6: Calculate the squares: 100 + 144 = 2 * (81 + x²). Step 7: Simplify: 244 = 162 + 2x². Step 8: Isolate 2x²: 2x² = 244 - 162 = 82. Step 9: Solve for x²: x² = 82 / 2 = 41. Step 10: Solve for x: x = √41 cm. Step 11: Since C is the midpoint of BD, BD = 2 * BC. Step 12: BD = 2 * x = 2√41 cm. Step 13: Therefore, the length of BD is 2√41 cm.

Solution: Step 1: Given angle ABC = 75°. Step 2: Given angle ACB = π/4 radians. Step 3: Convert angle ACB from radians to degrees: (π/4) * (180°/π) = 180°/4 = 45°. Step 4: The sum of angles in a triangle is 180°. So, angle BAC = 180° - angle ABC - angle ACB. Step 5: Calculate angle BAC = 180° - 75° - 45° = 60°. Step 6: Convert angle BAC from degrees to circular measure (radians): 60° * (π/180°) = π/3 radians. Step 7: The circular measure of angle BAC is π/3 radians.