Question

A crow starts off in his nest near the ground. It leaves its nest, flying along the ground, due east for 60m. The crow then travels north for 80m, still flying along the ground. Then the crow flies straight up into the sky for 30 meters.

a. Draw a 3D diagram that represents the crow’s flight path. 

b. At this point, how far is the crow from his nest? 

c. What is angle of elevation that a baby crow would have to look up to see the adult crow from

the nest? 

Answer 1

$\text{From the diagram |BC| represents the height the crow travelled after flying 80m north ward } \text{|OA|=60m; |AB|=80m |BC|=30m} \text{Note: Triangle OBC is right angled } \text{In triangle OAB using pythagoras' theorem we have }\\ |OB|^2=|OA|^2+|AB|^2 \\ |OB|=\sqrt{|OA|^2+|AB|^2}=\sqrt{60^2+80^2}=\sqrt{3600+6400}=\sqrt{10000}=100\\ |OB|=100m\\ (ii)\text{ The distance of the crow from the nest}=|OC|\\ \text{In triangle OBC using pythagoras' theorem we have:}\\ |OC|^2=|OB|^2+|BC|^2\\ |OC|=\sqrt{|OB|^2+|BC|^2}=\sqrt{100^2+30^2}=\sqrt{10000+900}=\sqrt{10900}=104.40306509\\ \text{The distance of the crow from its nest is 104.40306509m}\approx104m\\ (iii)\text{The angle of elevation the baby crow will look is < BOC}\\ \text{In triangle OBC } \tan(<BOC) =\frac{|BC|}{|OB|}=\frac{30}{100}=0.3\\ <BOC=\tan^{-1}(0.3)=16.699^o\approx17^o\\ \text{Therefore the baby crow must look }17^o\text{ upward to see the adult crow}$