trigonometry
Can someone please help me with this problem? I really can't get it right.
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From the top of the tower, the angle of depression of the base of the flagpole is 51 degrees, while from 2 floors below, the angle of
depression of the base of the flagpole is 47 degrees. If each floor is
12 m high, what is the height of the tower?
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Please. I really need help.
BF is the ground level.
Write EG in terms of angle $50^\circ$.Find $x$. Then the height will be $x\tan 46^\circ + 10$
Draw altitudes $AM$ & $BN$ from the points $A$ & $B$ respectively on the horizontal plane & let the horizontal distance of the point $A$ from the observer $O$ be $d$. Now if $h$ is the height of tower then we have
In right $\Delta AMO$ $$\tan 30^o=\frac{AM}{OM}=\frac{h}{d}$$ $$\implies d=h\cot 30^o=h\sqrt{3}$$ In right $\Delta BNO$ $$\tan 40^o=\frac{BN}{ON}=\frac{h}{d-50}$$ $$\implies d-50=h\cot 40^o \implies h\sqrt{3}-50=h\cot 40^o$$ $$\implies \color{blue}{h=\frac{50}{\sqrt{3}-\cot 40^o}\approx 92.54165784 \space ft}$$
Best Answer
As pointed out, draw a diagram.
Let the height of the second observation point (above the ground) be h m.
Then, that of the first observation point is h + 2*12 m.
Assume that the flagpole is x m away from the tower.
You now have two right angled triangles with sufficient data to let you setup 2 equations to eliminate x and get h.
Don't forget the required height is h + 2*12 m.