Question

please solve this question...fast​

Answer 1

Answer:

hey here is your answer

pls mark it as brainliest

Step-by-step explanation:

$here \: ∠ABC \: ie \: ABD = 90 \\ so \: here \: moreover \: \\ ∠ADB = 30 \\ thus \: then \\ tan \: ADB = AB/DB \\ tan \: 30 = h/x + 10 \\ so \: \: tan \: 30 = 1/ \sqrt{ 3 } \\ ie \: \: h = x + 10/ \sqrt{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)$

$similarly \: here \: \\ ∠ACB = 60 \\ thus \: then \\ tan \: ACB = AB/CB \\ tan \: 60 = h/x \\ so \: \: tan \: 60 = \sqrt{3} \\ \sqrt{3} = h/x \\ ie \: \: h = \sqrt{3} .x \: \: \: \: \: \: \: \: \: \: \: \: (2)$

$so \: equating \: (1) \: and \: (2) \\ we \: get \\ \sqrt{3} .x = x + 10/ \sqrt{3} \\ \sqrt{3} . \sqrt{3} .x = x + 10 \\ ie \: \: 3x = x + 10 \\ 2 x = 10 \\ ie \: x = 5 \\ substituting \: value \: of \: x \: in \: (2) \\ we \: get \: \\ h = \sqrt{3} x \\ = \sqrt{3} \times 5 \\ h = 5 \sqrt{3}$

$so \: thus \: then \\ h - x = 5 \sqrt{3} - 5 \\ = 5( \sqrt{3} - 1) \\ h - x = 5( \sqrt{3} - 1).metres \\ hence \: \: option (2) \: is \: ryt$

Answer 2

Answer:

your answer is 5(√3-1)m..

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