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  1. Asked: In:

    ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC²= AB² + BC² + 2 BC . BD.

    jivan kumar

    In ΔADB, by pythagoras theorem AB² = AD² + DB² ...(1) In ΔACD, by pythagoras theorem AC² = AD² + DC² ⇒AC² = AD² + (DB + BC)² ⇒AC² = AD² + BD² + BC² + 2DB BC ⇒AC² = AB² + BC² + 2DB × BC [From the equation (1)]

    In ΔADB, by pythagoras theorem
    AB² = AD² + DB² …(1)
    In ΔACD, by pythagoras theorem
    AC² = AD² + DC²
    ⇒AC² = AD² + (DB + BC)²
    ⇒AC² = AD² + BD² + BC² + 2DB BC
    ⇒AC² = AB² + BC² + 2DB × BC [From the equation (1)]

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  2. Asked: In:

    Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

    jivan kumar

    In parallelogram ABCD, Altitudes AF and DE is drawn on DC and produced BA. In ΔDEA, by pythagoras theorem, DE² + EA² = DA² = ...(1) In ΔDEB. by pythagoras theorem, DE² + EB² = DB² ⇒ DE² + (EA + AB)² = DB² ⇒ (DE² + EA²) + AB² + 2EA × AB = DB² ⇒ DA² + AB² + 2EA × AB = DB² ...(ii) In ΔADF, by PythagoraRead more

    In parallelogram ABCD, Altitudes AF and DE is drawn on DC and produced BA.
    In ΔDEA, by pythagoras theorem, DE² + EA² = DA² = …(1)
    In ΔDEB. by pythagoras theorem, DE² + EB² = DB²
    ⇒ DE² + (EA + AB)² = DB²
    ⇒ (DE² + EA²) + AB² + 2EA × AB = DB²
    ⇒ DA² + AB² + 2EA × AB = DB² …(ii)
    In ΔADF, by Pythagoras theorem, AD² = AF² + FD²
    In ΔAFC, by pythagoras theorem
    AC² = AF² + FC² = AF² + (DC – FD)² = AF² + DC² + FD² – 2DC × FD
    = (AF² + FD²) + DC² – 2DC × FD
    ⇒ AC² = AD² + DC² – 2DC × FD …(iii)
    ABCD is a parallelogram.
    Therefore
    AB = CD …(iv)
    and, BC = AD …(v)
    In ΔDEA and ΔADF,
    ∠DEA = ∠AFD [Each 90]
    ∠EAD = ∠ADF [EA II DF]
    AD = AD [Common]
    ∴ΔEAD ≅ ΔFDA [AAS congruency rule]
    ⇒ EA = DF …(vi)
    Adding equations (ii) and (iii), we have
    DA² + AB² + 2EA AB + AD² + DC² – 2DC FD = DB² + AC²
    ⇒ DA² + AB² + AD² + DC² + 2EA × AB – 2DC × FD = DB² + AC²
    ⇒ BC² + AB² + AD² + DC² + 2EA × AB – 2AB × EA = DB² + AC² [From the equation (iv) and (vi)]
    ⇒ AB² + BC² + CD² + DA² = AC² + BD²

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  3. Asked: In:

    As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

    jivan kumar

    Let AB is 75 m high lighthouse and C and D are the two ships. In ΔABC, AB/BC = tan 45° ⇒ 75/BC = 1 ⇒ BC = 75 In ΔABD, AB/BD = tan 30° ⇒ 75/(BC + CD) = 1/√3 ⇒ 75/(75 + CD) = 1/√3 ⇒ 75 + CD = 75√3 ⇒ CD = 75(√3 - 1) Hence, the distance between the two ships is 75(√3 - 1).

    Let AB is 75 m high lighthouse and C and D are the two ships.
    In ΔABC,
    AB/BC = tan 45° ⇒ 75/BC = 1
    ⇒ BC = 75
    In ΔABD,
    AB/BD = tan 30°
    ⇒ 75/(BC + CD) = 1/√3
    ⇒ 75/(75 + CD) = 1/√3
    ⇒ 75 + CD = 75√3
    ⇒ CD = 75(√3 – 1)
    Hence, the distance between the two ships is 75(√3 – 1).

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  4. Asked: In:

    A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° Find the distance travelled by the balloon during the interval.

    jivan kumar

    Let CD is 1.2m high girl and FG is the distance travelled by balloon. In ΔACE, AE/CE = tan 60° ⇒ (AF - EF)/CE = √3 ⇒ (88.2 - 1.2)/CE = √3 ⇒ 87/CE = √3 ⇒ CE = 87/√3 = 87√3/3 = 29√3 In ΔBCG, BG/GC = tan 30° ⇒ 88.2 - 1.2/CG = 1/√3 ⇒ 87/CG = 1/√3 ⇒ CG = 87√3 Distance travelled by balloon = EG = CG - CERead more

    Let CD is 1.2m high girl and FG is the distance travelled by balloon.
    In ΔACE,
    AE/CE = tan 60°
    ⇒ (AF – EF)/CE = √3
    ⇒ (88.2 – 1.2)/CE = √3
    ⇒ 87/CE = √3 ⇒ CE = 87/√3 = 87√3/3 = 29√3
    In ΔBCG,
    BG/GC = tan 30° ⇒ 88.2 – 1.2/CG = 1/√3
    ⇒ 87/CG = 1/√3 ⇒ CG = 87√3
    Distance travelled by balloon = EG = CG – CE = 87√3 – 29√3 = 58√3
    Hence, the distance travelled by balloon is 58√3m.

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  6. Asked: In:

    A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

    jivan kumar

    Let CD is highway and AB is tower. C is the initial position of the car and D is the position after 6 seconds. In ΔABD, AB/DB = tan 60° ⇒ AB/DB = √3 ⇒ BD = AB/√3 In ΔABC, AB/BC = tan 30° ⇒ AB/(BD + DC) = 1/√3 ⇒ AB√3 = BD + DC ⇒ AB = AB/√3 + DC [putting the value of BD] ⇒ AB√3 - AB/√3 = DC ⇒ DC = (3ARead more

    Let CD is highway and AB is tower. C is the initial position of the car and D is the position after 6 seconds.
    In ΔABD,
    AB/DB = tan 60° ⇒ AB/DB = √3 ⇒ BD = AB/√3
    In ΔABC,
    AB/BC = tan 30°
    ⇒ AB/(BD + DC) = 1/√3
    ⇒ AB√3 = BD + DC
    ⇒ AB = AB/√3 + DC [putting the value of BD]
    ⇒ AB√3 – AB/√3 = DC
    ⇒ DC = (3AB – AB)/√3 = 2AB/√3
    The time taken to travelled CD (= 2AB/√3) distance = 6 second
    Therefore, the speed of the car = distance/time = (2AB/√3)/6 = AB/3√3 m/s
    The time taken to travelled BD (= AB/√3) distance = distance/speed = (AB/√3)/ (AB/3√3) = 3 second.
    Hence, car will take 3 seconds to reach the foot of the car.

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  7. Asked: In:

    The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

    jivan kumar

    Let AQ is tower and C and D are the two point 4 m and 9 m away from the foot of tower. The angle of elevations are complementary. So if one angle is θ, then the other will be 90- θ. In ΔAQR, AQ/QR = tan θ ⇒ AQ/4 = tan θ ...(1) In ΔAQS, AQ/QS = tan (90 - θ) ⇒ AQ/9 = cot θ ...(2) Multiplying the equatRead more

    Let AQ is tower and C and D are the two point 4 m and 9 m away from the foot of tower.
    The angle of elevations are complementary. So if one angle is θ, then the other will be 90- θ.
    In ΔAQR,
    AQ/QR = tan θ
    ⇒ AQ/4 = tan θ …(1)
    In ΔAQS,
    AQ/QS = tan (90 – θ)
    ⇒ AQ/9 = cot θ …(2)
    Multiplying the equations (1) and (2), we have
    (AQ/4)(AQ/9) = tan θ. cot θ
    ⇒ AQ²/36 = 1
    ⇒ AQ² = 36
    ⇒ AQ = ±6
    Height can’t be negative. Hence, the height of the tower is 6 m.

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  8. Asked: In:

    From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

    jivan kumar

    Let AB is building and CD is cable tower. In ΔABD, AB/BD = tan 45° ⇒ 7/BD = 1 ⇒ BD = 7 In ΔACE, AC = BD = 7 m CE/AE = tan 60° ⇒ CE/7 = √3 ⇒ CE = 7√3 Therefore, CD = CE + ED = 7√3 + 7 = 7(√3 + 1) Hence, the height of the cable tower is 7(√3 + 1) m.

    Let AB is building and CD is cable tower.
    In ΔABD,
    AB/BD = tan 45° ⇒ 7/BD = 1 ⇒ BD = 7
    In ΔACE, AC = BD = 7 m
    CE/AE = tan 60° ⇒ CE/7 = √3 ⇒ CE = 7√3
    Therefore, CD = CE + ED = 7√3 + 7 = 7(√3 + 1)
    Hence, the height of the cable tower is 7(√3 + 1) m.

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  9. Asked: In:

    From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

    jivan kumar

    Let BC is Building and AB is transmission tower. In ΔBCD, BC/CD = tan 45° ⇒ 20/CD = CD = 1 In ΔACD, AC/ CD = tan 60° ⇒ AB+BC/CD = √3 ⇒ (AB+20)/20 = √3 ⇒ AB + 20 = 20√3 ⇒ AB = 20(√3- 1) Hence, the height of the transmission towwer is 20(√3- 1)m.

    Let BC is Building and AB is transmission tower.
    In ΔBCD,
    BC/CD = tan 45° ⇒ 20/CD = CD = 1
    In ΔACD,
    AC/ CD = tan 60° ⇒ AB+BC/CD = √3 ⇒ (AB+20)/20 = √3
    ⇒ AB + 20 = 20√3 ⇒ AB = 20(√3- 1)
    Hence, the height of the transmission towwer is 20(√3- 1)m.

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  11. Asked: In:

    A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

    jivan kumar

    Let BC is Pedestal and AB is 1.6 m high statue. In ΔBCD, BC/CD = tan 45° ⇒ BC/CD = 1 ⇒ BC = CD In ΔACD, AB+BC/CD = tan 60° ⇒ AB+BC/BC = √3 ⇒ 1.6 + BC/BC = √3 ⇒ BC + 1.6 = BC√3 ⇒ 1.6 = BC (√3 - 1) ⇒ BC = 1.6/(√3-1) × (√3+1)/(√3+1) = 1.6(√3+1)/2 = 0.8(√3+1) Hence, the height of the pedestal is 0.8(√3+Read more

    Let BC is Pedestal and AB is 1.6 m high statue.
    In ΔBCD,
    BC/CD = tan 45° ⇒ BC/CD = 1 ⇒ BC = CD
    In ΔACD,
    AB+BC/CD = tan 60° ⇒ AB+BC/BC = √3
    ⇒ 1.6 + BC/BC = √3
    ⇒ BC + 1.6 = BC√3
    ⇒ 1.6 = BC (√3 – 1)
    ⇒ BC = 1.6/(√3-1) × (√3+1)/(√3+1) = 1.6(√3+1)/2
    = 0.8(√3+1)
    Hence, the height of the pedestal is 0.8(√3+1)m.

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  12. Asked: In:

    The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

    jivan kumar

    Let CD is 50m high tower and AB is building. In ΔCDB, CD/BD = tan 60° ⇒ 50/BD = √3 ⇒ BD = 50/√3 In ΔABD, AB/BD = tan 30° ⇒ AB/(50√3) = 1/√3 ⇒ AB = 50/√3 × 1/√3 = 50/ 3 = 16 (2/3) Hence, the height of the building is 16(2/3)m.

    Let CD is 50m high tower and AB is building.
    In ΔCDB,
    CD/BD = tan 60° ⇒ 50/BD = √3 ⇒ BD = 50/√3
    In ΔABD,
    AB/BD = tan 30° ⇒ AB/(50√3) = 1/√3 ⇒ AB = 50/√3 × 1/√3 = 50/ 3 = 16 (2/3)
    Hence, the height of the building is 16(2/3)m.

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