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Some Applications of Trigonometry
CBSE 2026 Board Exam Set 1 (Code 30/7/1)
CBSE_BOARD_PYQ_2026_30_7_1
Grade 10
Question:
[Section E]
Case Study 2: Construction Work Ladder
A painter uses a ladder $OP$ resting against a vertical wall at point $P$ making an angle of $60^\circ$ with the horizontal ground $OA$. The top of the wall $P$ is at a height of $12\sqrt{3}$ m from the ground.
1. Find the length of the ladder $OP$. (1 Mark)
2. If the ladder is shifted to a new position $OR$ making an angle of $45^\circ$ with the ground, find the distance $OC$ of its foot from the wall. (1 Mark)
3. Find the total horizontal distance $AO + OC$. (2 Marks)
OR (for part 3)
3. In position $OR$, find the height $CR$ of the top of the ladder on the wall. (2 Marks)
Case Study 2: Construction Work Ladder
A painter uses a ladder $OP$ resting against a vertical wall at point $P$ making an angle of $60^\circ$ with the horizontal ground $OA$. The top of the wall $P$ is at a height of $12\sqrt{3}$ m from the ground.
1. Find the length of the ladder $OP$. (1 Mark)
2. If the ladder is shifted to a new position $OR$ making an angle of $45^\circ$ with the ground, find the distance $OC$ of its foot from the wall. (1 Mark)
3. Find the total horizontal distance $AO + OC$. (2 Marks)
OR (for part 3)
3. In position $OR$, find the height $CR$ of the top of the ladder on the wall. (2 Marks)
Step-by-Step Solution
Key Concept: Right triangles with ladder length 24 m.
1. $\sin 60^\circ = \dfrac{12\sqrt{3}}{OP} \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{12\sqrt{3}}{OP} \Rightarrow OP = 24$ m. [1.0 Mark]
2. $OR = 24$ m. $\cos 45^\circ = \dfrac{OC}{24} \Rightarrow \dfrac{1}{\sqrt{2}} = \dfrac{OC}{24} \Rightarrow OC = 12\sqrt{2}$ m. [1.0 Mark]
[Part 3 Main] $OA = 12$ m. Total horizontal distance $= 12 + 12\sqrt{2} = 12(\sqrt{2} + 1)$ m. [2.0 Marks]
-------------------------------
[Part 3 OR Choice]
$CR = OC \tan 45^\circ = 12\sqrt{2}$ m. [2.0 Marks]
2. $OR = 24$ m. $\cos 45^\circ = \dfrac{OC}{24} \Rightarrow \dfrac{1}{\sqrt{2}} = \dfrac{OC}{24} \Rightarrow OC = 12\sqrt{2}$ m. [1.0 Mark]
[Part 3 Main] $OA = 12$ m. Total horizontal distance $= 12 + 12\sqrt{2} = 12(\sqrt{2} + 1)$ m. [2.0 Marks]
-------------------------------
[Part 3 OR Choice]
$CR = OC \tan 45^\circ = 12\sqrt{2}$ m. [2.0 Marks]
Correct Answer: 1. 24 m, 2. 12\sqrt{2} m, 3. 12(\sqrt{2} + 1) m (OR 12\sqrt{2} m)
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