TRIGONOMETRY :
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2 trees, A&B are on the same side of a river. From a pointC in the river the distance of the trees A&Bis 250 m and 300 m ,respectively.if the the angleC is 45,find the between the trees?
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Answered by
1
Use the formula
AB² = AC² + BC² - 2 AC BC Cos C
= 250² + 300² - 2 250 300 1/√2 as cos 45 = 1/√2
= 50² (25 + 36 - 60 /√2) = 50² (61 - 30 √2)
AB = 50 ( 61 - 30 √2 )^1/2 meters
AB² = AC² + BC² - 2 AC BC Cos C
= 250² + 300² - 2 250 300 1/√2 as cos 45 = 1/√2
= 50² (25 + 36 - 60 /√2) = 50² (61 - 30 √2)
AB = 50 ( 61 - 30 √2 )^1/2 meters
Answered by
1
using Pythagoras theorem
AC² = AB² + BC²
AB² = AC² - BC²
= 250² - 300²
=62500 - 90000
=27500
AB = √27500
= 165.83
AC² = AB² + BC²
AB² = AC² - BC²
= 250² - 300²
=62500 - 90000
=27500
AB = √27500
= 165.83
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