Not the exact question you were looking for?
Paste your question to our Mathbee AI Mentor below to get an instant step-by-step solution.
Some Applications of Trigonometry
CBSE 2026 Board Exam Set 3 (Code 30/1/3)
CBSE_BOARD_PYQ_2026_30_1_3
Grade 10
Question:
[Section E - Case Study 3]
Two vertical poles $AP$ and $BQ$ stand on level ground on opposite sides of a point $O$. The angles of elevation of the tops of the poles $A$ and $B$ from point $O$ on the ground are $60^\circ$ and $30^\circ$ respectively. The height of pole $AP$ is $6$ m and point $O$ is at a distance of $6$ m from the foot $P$ of pole $AP$.
(i) Find the distance $OB$ from point $O$ to the top of pole $B$ if the height of pole $B$ is such that $\cos 30^\circ = \dfrac{6}{OB}$. [1 Mark]
(ii) Find the distance $OA$ from point $O$ to the top of pole $A$. [1 Mark]
(iii) (a) Find the height of pole $BQ$ and the distance between the tops $A$ and $B$ of the two poles. [2 Marks]
OR
(iii) (b) Find the area of $\Delta OPB$. [2 Marks]
Two vertical poles $AP$ and $BQ$ stand on level ground on opposite sides of a point $O$. The angles of elevation of the tops of the poles $A$ and $B$ from point $O$ on the ground are $60^\circ$ and $30^\circ$ respectively. The height of pole $AP$ is $6$ m and point $O$ is at a distance of $6$ m from the foot $P$ of pole $AP$.
(i) Find the distance $OB$ from point $O$ to the top of pole $B$ if the height of pole $B$ is such that $\cos 30^\circ = \dfrac{6}{OB}$. [1 Mark]
(ii) Find the distance $OA$ from point $O$ to the top of pole $A$. [1 Mark]
(iii) (a) Find the height of pole $BQ$ and the distance between the tops $A$ and $B$ of the two poles. [2 Marks]
OR
(iii) (b) Find the area of $\Delta OPB$. [2 Marks]
Step-by-Step Solution
Key Concept: Right triangles with angles $30^\circ$ and $60^\circ$.
(i) $\cos30^\circ = \dfrac{\sqrt3}{2} = \dfrac{6}{OB} \Rightarrow OB = \dfrac{12}{\sqrt3} = 4\sqrt3$ m. [1.0 Mark]
(ii) $\cos60^\circ = \dfrac{1}{2} = \dfrac{6}{OA} \Rightarrow OA = 12$ m. [1.0 Mark]
(iii)(a) $\tan30^\circ = \dfrac{BQ}{6} \Rightarrow BQ = 2\sqrt3$ m. $\tan60^\circ = \dfrac{AP}{6} \Rightarrow AP = 6\sqrt3$ m. $AB = AP - BQ = 6\sqrt3 - 2\sqrt3 = 4\sqrt3$ m. [2.0 Marks]
OR (iii)(b) $\text{Area}(\Delta OPB) = \dfrac{1}{2} \times BP \times OP = \dfrac{1}{2} \times 2\sqrt3 \times 6 = 6\sqrt3$ m$^2$. [2.0 Marks]
(ii) $\cos60^\circ = \dfrac{1}{2} = \dfrac{6}{OA} \Rightarrow OA = 12$ m. [1.0 Mark]
(iii)(a) $\tan30^\circ = \dfrac{BQ}{6} \Rightarrow BQ = 2\sqrt3$ m. $\tan60^\circ = \dfrac{AP}{6} \Rightarrow AP = 6\sqrt3$ m. $AB = AP - BQ = 6\sqrt3 - 2\sqrt3 = 4\sqrt3$ m. [2.0 Marks]
OR (iii)(b) $\text{Area}(\Delta OPB) = \dfrac{1}{2} \times BP \times OP = \dfrac{1}{2} \times 2\sqrt3 \times 6 = 6\sqrt3$ m$^2$. [2.0 Marks]
Correct Answer: (i) 4√3 m | (ii) 12 m | (iii)(a) BQ = 2√3 m, AB = 4√3 m OR (iii)(b) 6√3 m²
Mathbee AI Mentor (Free Demo)
Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.