Q.
Two trees, A and B are on the same side of a river. From a point C in the river the distance of the trees A and B is 250m and 300m, respectively. If the angle C is 45∘. Then, the distance between the trees (use 2=1.44 ) is approximately
Let A and B be the trees and C be a point in the river. Then, CA=250m,CB=300m and ∠ACB=45∘
Let AB=xm
Applying cosine formula on △ACB, we get, cosC=2aba2+b2−c2 ⇒cos45∘=2×300×250(300)2+(250)2−x2 ⇒21=15000090000+62500−x2 ⇒150000152500−x2=21×22=22 ⇒152500−x2=75000×2=75000×1.44=108000 ⇒x2=152500−108000=44500 ⇒x=44500=210.95m ≈210m (approx.)