Q66.A horizontal park is in the shape of a triangle OAB with AB = 16 . A vertical lamp post OP is erected at the point O such that ∠PAO = ∠PBO = 15° and ∠PCO = 45° , where C is the midpoint of AB. Then (OP)2 is equal to (1) √3 32 (√3 −1) (2) √332 (2 −√3) (3) 16 (4) 16 √3 (√3 −1) √3 (2 −√3)
What This Question Tests
This problem involves applying trigonometric ratios in multiple right-angled triangles to find unknown lengths in a 3D setup, specifically calculating the square of the height of a lamp post.
Concepts Tested
Formulas Used
tan θ = Opposite/Adjacent
Pythagorean theorem
Properties of isosceles triangle
📚 NCERT Sections This Tests
9.5 — A Small Bulb Is Placed At The Bottom Of A Tank Containing Water To A
Physics Class 12 · Chapter 9
9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
9.16 — The Pin Appears Raised By 5.0 Cm. It Can Be Seen With An Explicit Ray
Physics Class 12 · Chapter 9
9.16 The pin appears raised by 5.0 cm. It can be seen with an explicit ray diagram that the answer is independent of the location of the slab (for small angles of incidence).
9.27 — (A) M = ( Fo/Fe) = 28
Physics Class 12 · Chapter 9
9.27 (a) m = ( fO/fe) = 28 f O f O (b) m = 1 + = 33.6 f e 25 349 Reprint 2025-26 Physics 9.28 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective h h = = f O 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.29 The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 –20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. 9.30 The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. 9.31 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) n = 5.09 ´ 1014Hz v = (c/n) = 2.26 × 108 m s–1, l = (v/n) = 444 nm 10.2 (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). 10.3 (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr. Therefore, the violet component of white light travels slower than the red component. 1.2 10 – 2 0.28 10 – 3 10.4 m = 600 nm 4 14. 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 350 10.8 tan–1(1.5) ~ 56.3o Reprint 2025-26 Answers
📋 Question Details
- Chapter
- Trigonometric Functions & Equations
- Topic
- Heights and Distances, Right-angled triangles
- Year
- 2022
- Shift
- 28 Jul Shift 2
- Q Number
- Q66
- Type
- MCQ
- NCERT Ref
- Class 10 Mathematics Ch 9: Some Applications of Trigonometry
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