Section E
Section E consists of 3 case study-based questions of 4 marks each.
Q.36. In a class, the teacher asks every student to write an example of A.P. Two boys Aryan and Roshan writes the progression as \( -5, -2, 1, 4, \ldots \) and \( 187, 184, 181, \ldots \) respectively. Now the teacher asks his various students the following questions on progression. Help the students to find answers for the following:
i. Find the sum of the common difference of two progressions.
📥 View Answer
Solution: Step 1: Aryan’s progression: -5, -2, 1, 4, … Common difference (d₁) = -2 – (-5) = -2 + 5 = 3
Step 2: Roshan’s progression: 187, 184, 181, … Common difference (d₂) = 184 – 187 = -3
Step 3: Sum of common differences = d₁ + d₂ = 3 + (-3) = 0
ii. Find the 34th term of progression written by Roshan.
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Solution: Step 1: Roshan’s AP: 187, 184, 181, … First term (a) = 187 Common difference (d) = 184 – 187 = -3
Step 2: Formula for nth term: tₙ = a + (n-1)d For n = 34: t₃₄ = 187 + (34-1)(-3)
Step 3: Calculate: t₃₄ = 187 + 33 × (-3) = 187 – 99 = 88
iii. (A) Find the sum of first 10 terms of the progression written by Aryan.
(B) Which term of the progressions will have the same value?
📥 View Answer for (A)
Solution for (A): Step 1: Aryan’s AP: -5, -2, 1, 4, … First term (a) = -5 Common difference (d) = 3 Number of terms (n) = 10
Step 2: Sum of first n terms formula: Sₙ = \(\frac{n}{2}[2a + (n-1)d]\)
Step 3: For n = 10: S₁₀ = \(\frac{10}{2}[2(-5) + (10-1)3]\) = 5[-10 + 27] = 5 × 17 = 85
📥 View Answer for (B)
Solution for (B): Step 1: Aryan’s AP nth term: tₙ = a + (n-1)d = -5 + (n-1)3 = 3n – 8
Step 2: Roshan’s AP nth term: tₙ = 187 + (n-1)(-3) = 187 – 3n + 3 = 190 – 3n
Step 3: Equate both terms: 3n – 8 = 190 – 3n Step 4: Solve for n: 3n + 3n = 190 + 8 6n = 198 n = 33
Q.37. A group of class X students goes to picnic during winter holidays. The position of three friends Aman, Kirit and Chahat are shown by the points P, Q and R. Coordinates: P(2,5), Q(4,4), R(8,3)
(i) Find the distance between P and R.
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Solution: Step 1: Distance formula: \(d = \sqrt{(x₂-x₁)^2 + (y₂-y₁)^2}\)
Step 2: P(2,5) and R(8,3) PR = \(\sqrt{(8-2)^2 + (3-5)^2}\)
Step 3: Calculate: = \(\sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40}\) = \(\sqrt{4 × 10} = 2\sqrt{10}\) units
(ii) Is Q, the midpoint of PR? Justify by finding midpoint of PR.
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Solution: Step 1: Midpoint formula: \(M = \left(\frac{x₁+x₂}{2}, \frac{y₁+y₂}{2}\right)\)
Step 2: Midpoint of P(2,5) and R(8,3): M = \(\left(\frac{2+8}{2}, \frac{5+3}{2}\right)\) = \(\left(\frac{10}{2}, \frac{8}{2}\right) = (5, 4)\)
Step 3: Compare with Q(4,4): Midpoint of PR = (5,4) Q = (4,4) Since (5,4) ≠ (4,4), Q is not the midpoint of PR.
(iii) (A) Find the point on x-axis which is equidistant from P and Q.
(B) Let S be a point which divides the line joining PQ in ratio 2:3. Find the coordinates of S.
📥 View Answer for (A)
Solution for (A): Step 1: Let point on x-axis be (x, 0) Distance from P: \(\sqrt{(x-2)^2 + (0-5)^2} = \sqrt{(x-2)^2 + 25}\) Distance from Q: \(\sqrt{(x-4)^2 + (0-4)^2} = \sqrt{(x-4)^2 + 16}\)
Step 2: Equate distances (since equidistant): \((x-2)^2 + 25 = (x-4)^2 + 16\)
Step 3: Expand and solve: \(x^2 – 4x + 4 + 25 = x^2 – 8x + 16 + 16\) \(-4x + 29 = -8x + 32\) \(4x = 3\) \(x = \frac{3}{4}\)
📥 View Answer for (B)
Solution for (B): Step 1: Section formula: If point divides line joining (x₁,y₁) and (x₂,y₂) in ratio m:n, then: \(S = \left(\frac{mx₂ + nx₁}{m+n}, \frac{my₂ + ny₁}{m+n}\right)\)
Step 2: Given: m:n = 2:3, P(2,5), Q(4,4) S = \(\left(\frac{2×4 + 3×2}{2+3}, \frac{2×4 + 3×5}{2+3}\right)\)
Step 3: Calculate: = \(\left(\frac{8 + 6}{5}, \frac{8 + 15}{5}\right)\) = \(\left(\frac{14}{5}, \frac{23}{5}\right)\)
Question 38 – India Gate Case Study
Situation: India Gate is 42m tall. Student Shreya is 1m tall. She visits India Gate for a study tour.
i. What is the angle of elevation from Shreya’s eye to the top of India Gate, if she is standing at a distance of 41m away from the India Gate?
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Visual Representation:
Height of India Gate = 42m
Shreya’s height = 1m
Distance from India Gate = 41m
Step-by-Step Solution:
Step 1: Calculate effective height
Since Shreya’s eyes are at height of 1m, we consider height from her eye level:
Effective height = Total height – Shreya’s height
= 42m – 1m = 41m
Step 2: Identify right triangle components
In the right triangle formed:
• Opposite side (height) = 41m
• Adjacent side (distance) = 41m
• Angle of elevation = θ
Step 3: Apply trigonometric ratio
tanθ = Opposite/Adjacent = 41/41 = 1
θ = tan⁻¹(1)
Step 4: Find the angle
We know: tan45° = 1
Therefore: θ = 45°
Answer: Angle of elevation = 45°
ii. If Shreya observes the angle of elevation from her eye to the top of India Gate to be 60°, then how far is she standing from the base of the India Gate?
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Given:
• Angle of elevation = 60°
• Effective height = 41m (42m – 1m)
• Distance = ?
Step-by-Step Solution:
Step 1: Set up the trigonometric equation
tan60° = Opposite/Adjacent = Height/Distance
√3 = 41/Distance
Step 2: Solve for distance
Distance = 41/√3 meters
Step 3: Rationalize the denominator
Distance = \(\frac{41}{√3} \times \frac{√3}{√3} = \frac{41√3}{3}\) meters
Step 4: Calculate approximate value
√3 ≈ 1.732
Distance ≈ \(\frac{41 \times 1.732}{3} = \frac{71.012}{3} ≈ 23.67\) meters
Answer: Distance = \(\frac{41√3}{3}\) m ≈ 23.67 m
iii. (A) If the angle of elevation from Shreya’s eye changes from 45° to 30°, when she moves some distance back from the original position. Find the distance she moves back.
OR
📐 View Complete Solution for (A)
Understanding the problem:
Shreya moves backwards, so distance increases, angle decreases from 45° to 30°.
We need to find how much she moved back.
Step-by-Step Solution:
Step 1: Find original distance (when θ = 45°)
tan45° = 41/d₁
1 = 41/d₁
d₁ = 41 meters
Step 2: Find new distance (when θ = 30°)
tan30° = 41/d₂
1/√3 = 41/d₂
d₂ = 41√3 meters
Step 3: Calculate distance moved back
Distance moved = d₂ – d₁
= 41√3 – 41
= 41(√3 – 1) meters
Step 4: Calculate approximate value
√3 ≈ 1.732
Distance moved ≈ 41(1.732 – 1)
= 41 × 0.732 ≈ 30.01 meters
Answer: Distance moved back = 41(√3 – 1) m ≈ 30.01 m
(B) If Shreya moves to a point which is at a distance of \(\frac{41}{√3}\) from the India Gate, then find the angle of elevation made by her eye to the top of India Gate.
📐 View Complete Solution for (B)
Given:
• Distance = 41/√3 meters
• Effective height = 41 meters
• Find: Angle of elevation θ
Step-by-Step Solution:
Step 1: Set up trigonometric ratio
tanθ = Opposite/Adjacent = Height/Distance
tanθ = \(\frac{41}{41/√3}\)
Step 2: Simplify the expression
tanθ = 41 × \(\frac{√3}{41}\) = √3
Step 3: Find the angle
We know: tan60° = √3
Therefore: θ = 60°
Answer: Angle of elevation = 60°
Verification: This matches part (ii) where we found that at 60° angle, distance is 41/√3 m.
The problem is symmetric: part (ii) and part (B) are inverse problems of each other.
Question 39 – Additional Practice Question
Note: Since the original paper only had 38 questions, here’s an additional practice question in the same format.
Q.39. A lighthouse is 100m tall. A ship captain observes the angle of elevation to the top of the lighthouse to be 30°. After sailing towards the lighthouse for some time, the angle of elevation becomes 60°.
(i) Find the initial distance of the ship from the lighthouse.
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Solution:
Height of lighthouse = 100m
Initial angle of elevation = 30°
tan30° = 100/distance
1/√3 = 100/distance
Distance = 100√3 meters
Answer: Initial distance = 100√3 m ≈ 173.2 m
(ii) Find the distance travelled by the ship towards the lighthouse.
📐 View Solution
Solution:
Step 1: Initial distance (θ = 30°): d₁ = 100√3 m
Step 2: Final distance (θ = 60°): tan60° = 100/d₂
√3 = 100/d₂ ⇒ d₂ = 100/√3 m
Step 3: Distance travelled = d₁ – d₂
= 100√3 – 100/√3
= \(\frac{300 – 100}{√3} = \frac{200}{√3} = \frac{200√3}{3}\) m
Answer: Distance travelled = \(\frac{200√3}{3}\) m ≈ 115.47 m
(iii) If the ship continues sailing at the same speed and the angle of elevation increases to 45°, find the additional distance it needs to travel.
📐 View Solution
Solution:
Step 1: Distance when θ = 60°: d₂ = 100/√3 m
Step 2: Distance when θ = 45°: tan45° = 100/d₃
1 = 100/d₃ ⇒ d₃ = 100 m
Step 3: Additional distance = d₂ – d₃
= 100/√3 – 100
= 100(1/√3 – 1) = 100(1 – √3)/√3
Answer: Additional distance = 100(1/√3 – 1) m ≈ 100(0.577 – 1) = -42.3 m
Note: Negative value means the ship needs to sail 42.3 m further towards the lighthouse.
📚 Trigonometric Values Reference
Angle = 30°
sin30° = 1/2
cos30° = √3/2
tan30° = 1/√3
Angle = 45°
sin45° = 1/√2
cos45° = 1/√2
tan45° = 1
Angle = 60°
sin60° = √3/2
cos60° = 1/2
tan60° = √3
💡 Key Concept: In angle of elevation problems, always identify the right triangle, label opposite (height), adjacent (distance), and use appropriate trigonometric ratio (usually tanθ = opposite/adjacent).
📝 Exam Strategy: For trigonometry application questions: 1) Draw diagram, 2) Label all given values, 3) Identify right triangle, 4) Choose correct trigonometric ratio, 5) Solve step-by-step, 6) Include units in final answer, 7) Verify if answer is reasonable.