A) 59.4m |
B) 61.4m |
C) 62.4m |
D) 63.4m |
D) 63.4m |
Given, BC=150 m
In \( \Large \angle ACB = 30 ^{\circ} \ and\ \angle DCB = 45 ^{\circ} \)
Then, AD = ?
In \( \Large \triangle ABC, \tan 30 ^{\circ} = \frac{AB}{BC} \)
=> \( \Large \frac{1}{\sqrt{3}} = \frac{AB}{150} \)
\( \Large \therefore\ AB= 86.6 m \)
In \( \Large \triangle DBC, \tan 45 ^{\circ} = \frac{DB}{BC} \)
=> \( \Large 1 = \frac{AD+AB}{BC} \)
=> BC = AD + 86.6
=> 150 - 86.6 = AD
Therefore, AD = 63.4 m