a bird sitting on the top of a tree, which is 80m high the angle of elevation of the - Maths - Some Applications of Trigonometry

Question

a bird sitting on the top of a tree, which is 80m high the angle
of elevation of the bird, from the point on the ground is
450 the bird flies away from the point of observation
horizantally and remains at a constant height after 2 seconds, the
angle of elevation of the bird from the point of observation
becomes 300 find the sped of the flying bird

Varun.Rawat · Accepted Answer

Let P and Q be the two positions of the bird and let A be the point of observation Let ABC be the horizontal line through A Given: The angle of elevations of the bird in two positions P and Q from point A are 45° and 30° ∴ ∠PAB = 45°, ∠QAB = 30° Also, PB = 80 meters

In ΔABP, we have, $\tan â¡ 45 ï¿½ = \frac{BP}{AB} â‡’ 1 = \frac{80}{AB} â‡’ AB=80 m$ In ΔACQ, we have, $\tan â¡ 30 ï¿½ = \frac{CQ}{AC} â‡’ \frac{1}{\sqrt{3}} = \frac{80}{AC} â‡’ AC=80 \sqrt{3} m$ ∴ PQ = BC = AC – AB = $80 \sqrt{3} âˆ’ 80 = 80 (\sqrt{3} âˆ’ 1) m$ Thus, The bird touch $80 (\sqrt{3} âˆ’ 1) m$ in 2 seconds Hence, Speed of the bird $= \frac{80 (\sqrt{3} âˆ’ 1)}{2} = 40 (\sqrt{3} âˆ’ 1) m/sec$ $= \frac{40 (\sqrt{3} âˆ’ 1) ï¿½ 60 ï¿½ 60}{1000} km/hr=144 (\sqrt{3} âˆ’ 1) km/hr$