trigonometry
A ray of sunlight casts a shadow of a flagpole on the ground at an angle of depression of 58 degrees. If the length of the shadow is 3m shorter than the height of the flagpole, find the height of the flagpole.
I'm confused about the given problem above because I can't determine yet the length of the shadow, and also I'm stuck at the angle of depression, in which I supposed that it has to be the angle of elevation instead.
My solution:
$\sin\left(58^{\circ }\right)=\frac{x}{x-3}$
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
First, we draw a flagpole, $x$m long. Then we draw a shadow that is $x-3$m long. Then the sun ray which casts the shadow connects the TOP of the pole and TOP of the shadow. This means the angle formed by the hypotenuse (the sun ray) and the shadow is $58^o$. Now we have $\tan 58^o=\frac{x}{x-3}$, giving us $x\tan 58^o-3\tan 58^o=x$. We bring $x$ to one side, having $x\left(\tan 58^o-1\right)=3\tan 58^o$. $x=7.997...=8.00\left(3sf\right)$.