Question

how to solve this prob. from application of trigo?

Answer 1

,this is the method for this question

Answer 2

Heya user !!

Here's the answer you are looking for

See the attached file for the figure (for the answer)

In ∆AED,

$\tan(45) = \frac{AE}{DE} \\ \\ 1 = \frac{80}{d} \\ \\ d = 80m$

Now, in ∆BCD
$\tan(30) = \frac{BC}{DC} \\ \\ \frac{1}{ \sqrt{3} } = \frac{80}{d + x} \\ \\ \frac{1}{ \sqrt{3} } = \frac{80}{80 + x} \\ \\ 80 + x = 80 \sqrt{3} \\ \\ x = 80 \sqrt{3} - 80 \\ \\ x = 80( \sqrt{3} - 1) \\ \\ x = 80(1.732 - 1) = 80(0.732) = 58.56m$

So, in 2 sec it covered 58.56m

Thus, it's speed is

$= \frac{distance \: travelled}{time \: taken} \\ \\ = \frac{58.56}{2} \\ \\ = 29.28m/s$

Therefore, the speed of the bird is 29.28m/s



★★ HOPE THAT HELPS ☺️ ★★