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    from the top of a 7m high building the angle of elevation

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    From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60^o and the angle of depression of its foot is 45^o . The height of the tower in metre is

    Click here👆to get an answer to your question ✍️ From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60^o and the angle of depression of its foot is 45^o . The height of the tower in metre is

    From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60

    Question o

    and the angle of depression of its foot is 45

    o

    . The height of the tower in metre is 

    A7(

    3 ​ −1)

    B7

    3 ​

    C7+

    3 ​

    D7(

    3 ​ +1) Medium Open in App

    Updated on : 2022-09-05

    Solution Verified by Toppr

    Correct option is D)

    In ΔABE, tan60 o = AB h ​ h=AB 3 ​ ....(i) In ΔABC, tan45 o = AB 7 ​ AB=7   ...(ii)

    By equation (i) and equation (ii)

    h=7 3 ​

    So, height of tower is =h+7=7(

    3 ​ +1)

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    स्रोत : www.toppr.com

    From the top of 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.

    From the top of 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.

    Byju's Answer From the top ... Question From the top of 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.

    Open in App Solution

    Solve for the height of the tower :

    Given, Height of building =7 m .

    2. Angle of elevation of the top of a cable tower

    =60° .

    3. Angle of depression to the foot of a cable tower

    =45° . From the figure, tan45°=ABBC ⇒ 1=ABBC ⇒AB=BC=7 m tan60°=EDAD ⇒ 3=ED7 [∵AD=BC] ⇒ ED=73 m Height of tower =ED+CD =73+7 ∵AB=CD ⇒ Height of tower =19.12 m

    Hence, height of the tower is

    19.12 m

    .

    Suggest Corrections 39

    SIMILAR QUESTIONS

    Q. 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.

    स्रोत : byjus.com

    Ex 9.1, 12

    Ex 9.1 , 12 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. Let building be AB & tower be CE Given height of building = AB = 7m From the top of building, angle

    Check sibling questions

    Ex 9.1, 12 - Chapter 9 Class 10 Some Applications of Trigonometry (Term 2)

    Last updated at March 16, 2023 by Teachoo

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    Transcript

    Ex 9.1 , 12 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. Let building be AB & tower be CE Given height of building = AB = 7m From the top of building, angle of elevation of top of tower = 60°. Hence, ∠EAD = 60° Angle of depression of the foot of the tower = 45° Hence, ∠CAD = 45° We need to find height of tower i.e. CE Since AB & CD are parallel, CD = AB = 7 m Also, AD & BC are parallel So, AD = BC Since tower & building are vertical to ground ∠ ABC = 90° & ∠ EDA = 90° Now, AD & BC are parallel, taking AC as transversal ∠ ACB = ∠ DAC ∠ ACB = 45° In right angle triangle ABC, tan C = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶) tan 45° = 𝐴𝐵/𝐵𝐶 1 = 𝐴𝐵/𝐵𝐶 1 = (" " 7)/𝐵𝐶 BC = 7m Since BC = AD So, AD = 7m Now, In a right angle triangle ADE, tan A = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴) tan 60° = 𝐸𝐷/𝐴𝐷 √3 = 𝐸𝐷/𝐴𝐷 √3 = 𝐸𝐷/7 7√3 = ED ED = 7√3 Height of tower = ED + DC = 7√3 + 7 =7(√3 + 1)m

    Next: Ex 9.1, 13 Important →

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    Davneet Singh

    Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

    स्रोत : www.teachoo.com

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