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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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