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NEET-UG Biology

Heights and distances — practice questions

19 questions in the bank on this idea. Below are 10 of them, exactly as they appear in a test.

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  1. Question 1 · difficulty L2 · understanding

    The angle of elevation of the top P\mathrm{P} of a tower from the feet of one person standing due South of the tower is 4545^{\circ} and from the feet of another person standing due west of the tower is 3030^{\circ}. If the height of the tower is 5 meters, then the distance (in meters) between the two persons is equal to

    • A. 10
    • B. 525\frac{5}{2} \sqrt{5}
    • C. 555 \sqrt{5}
    • D. 5
  2. Question 2 · difficulty L2 · understanding

    From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60^\circ. The pole subtends an angle 30^\circ at the top of the tower. Then the height of the tower is :

    • A. 15315\sqrt 3
    • B. 20320\sqrt 3
    • C. 20 + 10310\sqrt 3
    • D. 30
  3. Question 3 · difficulty L2 · understanding

    The angle of elevation of a jet plane from a point A on the ground is 60^\circ. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30^\circ. If the jet plane is flying at a constant height, then its height is :

    • A. 360033600\sqrt 3 m
    • B. 120031200\sqrt 3 m
    • C. 180031800\sqrt 3 m
    • D. 240032400\sqrt 3 m
  4. Question 4 · difficulty L2 · understanding

    A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30 o . After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60 o . Then the time taken (in minutes) by him, from B to reach the pillar, is :

    • A. 6
    • B. 10
    • C. 20
    • D. 5
  5. Question 5 · difficulty L3 · understanding

    From the top A\mathrm{A} of a vertical wall AB\mathrm{AB} of height 30 m30 \mathrm{~m}, the angles of depression of the top P\mathrm{P} and bottom Q\mathrm{Q} of a vertical tower PQ\mathrm{PQ} are 1515^{\circ} and 6060^{\circ} respectively, B\mathrm{B} and Q\mathrm{Q} are on the same horizontal level. If C\mathrm{C} is a point on AB\mathrm{AB} such that CB=PQ\mathrm{CB}=\mathrm{PQ}, then the area (in m2\mathrm{m}^{2} ) of the quadrilateral BCPQ\mathrm{BCPQ} is equal to :

    • A. 200(33)200(3-\sqrt{3})
    • B. 300(31)300(\sqrt{3}-1)
    • C. 300(3+1)300(\sqrt{3}+1)
    • D. 600(31)600(\sqrt{3}-1)
  6. Question 6 · difficulty L3 · understanding

    The angle of elevation of the top of a tower from a point A due north of it is α\alpha and from a point B at a distance of 9 units due west of A is cos1(313)\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right). If the distance of the point B from the tower is 15 units, then cotα\cot \alpha is equal to :

    • A. 65\frac{6}{5}
    • B. 95\frac{9}{5}
    • C. 43\frac{4}{3}
    • D. 73\frac{7}{3}
  7. Question 7 · difficulty L3 · understanding

    The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is 4545^{\circ}. Let R be a point on AQ and from a point B, vertically above R\mathrm{R}, the angle of elevation of P\mathrm{P} is 6060^{\circ}. If BAQ=30,AB=d\angle \mathrm{BAQ}=30^{\circ}, \mathrm{AB}=\mathrm{d} and the area of the trapezium PQRB\mathrm{PQRB} is α\alpha, then the ordered pair (d,α)(\mathrm{d}, \alpha) is :

    • A. (10(31),25)(10(\sqrt{3}-1), 25)
    • B. (10(31),252)\left(10(\sqrt{3}-1), \frac{25}{2}\right)
    • C. (10(3+1),25)(10(\sqrt{3}+1), 25)
    • D. (10(3+1),252)\left(10(\sqrt{3}+1), \frac{25}{2}\right)
  8. Question 8 · difficulty L3 · understanding

    Let a vertical tower ABA B of height 2h2 h stands on a horizontal ground. Let from a point PP% on the ground a man can see upto height hh of the tower with an angle of elevation 2α2 \alpha. When from PP, he moves a distance dd in the direction of AP\overrightarrow{A P}, he can see the top BB of the tower with an angle of elevation α\alpha. If d=7hd=\sqrt{7} h, then tanα\tan \alpha is equal to

    • A. 52\sqrt{5}-2
    • B. 31\sqrt{3}-1
    • C. 72 \sqrt{7}-2
    • D. 73\sqrt{7}-\sqrt{3}
  9. Question 9 · difficulty L3 · understanding

    A tower PQ stands on a horizontal ground with base QQ on the ground. The point RR divides the tower in two parts such that QR=15 mQ R=15 \mathrm{~m}. If from a point AA on the ground the angle of elevation of RR is 6060^{\circ} and the part PRP R of the tower subtends an angle of 1515^{\circ} at AA, then the height of the tower is :

    • A. 5(23+3)m5(2 \sqrt{3}+3) \,\mathrm{m}
    • B. 5(3+3)m5(\sqrt{3}+3) \,\mathrm{m}
    • C. 10(3+1)m10(\sqrt{3}+1) \,\mathrm{m}
    • D. 10(23+1)m10(2 \sqrt{3}+1) \,\mathrm{m}
  10. Question 10 · difficulty L3 · understanding

    Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let π8{\pi \over 8} and θ\theta be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then tan 2 θ\theta is equal to

    • A. 3222{{3 - 2\sqrt 2 } \over 2}
    • B. 3+22{{3 + \sqrt 2 } \over 2}
    • C. 3224{{3 - 2\sqrt 2 } \over 4}
    • D. 324{{3 - \sqrt 2 } \over 4}

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