At the foot of a mountain the angles of elevation of its summit is 45 degree, after ascending 1 km forwards at an inclination of 30 degree, the elevation changes to 60 degree ,what is the height of the mountain?

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Answer 1 · 2018-02-23T14:18:01

The height of the mountain is #(sqrt(3)+1)/2# km.

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Let AB be the height of the mountain and above be the figure as described in the question.

In rt #DeltaEFC#,

#rarrsin30°=y/1#

#rarry=1/2 km# and

#rarrcos30°=a/1#

#rarra=sqrt(3)/2km#

In rt #Delta#ADE,

#rarrtan60°=x/z#

#rarrsqrt(3)=x/z#

#rarrz=x/sqrt(3)#

In rt#Delta#ABC,

#tan45°=(AB)/(BC)=(x+y)/(a+z)#

#rarr1=(x+y)/(a+z)#

#rarra+z=x+y#

#rarrsqrt(3)/2+x/sqrt(3)=x+1/2#

#rarrx-x/sqrt(3)=(sqrt(3)-1)/2#

#rarrx(1-1/sqrt(3))=(sqrt(3)-1)/2#

#rarrx*cancel((sqrt(3)-1))/sqrt(3)=cancel((sqrt(3)-1))/2#

#rarrx=sqrt(3)/2km#

Now, the height of the mountain#=AB=x+y=sqrt(3)/2+1/2=(sqrt(3)+1)/2# km.