Heights and Distances Quiz

$\tan(30^\circ) = H/100 \Rightarrow 1/\sqrt{3} = H/100 \Rightarrow H = 100/\sqrt{3}$.

$\tan \theta = 6 / 2\sqrt{3} = 3 / \sqrt{3} = \sqrt{3}$. Since $\tan 60^\circ = \sqrt{3}$, $\theta = 60^\circ$.

Angle of depression = Angle of elevation = $45^\circ$. $\tan 45^\circ = 50 / X \Rightarrow 1 = 50 / X \Rightarrow X = 50$.

$\tan \theta = H / \sqrt{3}H = 1/\sqrt{3}$. Since $\tan 30^\circ = 1/\sqrt{3}$, $\theta = 30^\circ$.

$\tan30^\circ = h/50 \Rightarrow 1/\sqrt3 = h/50 \Rightarrow h = 50/\sqrt3$ m.

Angle of depression $=$ angle of elevation $=45^\circ$. $\tan45^\circ = 40/d \Rightarrow 1=40/d \Rightarrow d=40$ m.

$\tan60=h/d\Rightarrow h=d\sqrt3$. $\tan30=h/(d+20)\Rightarrow 1/\sqrt3 = d\sqrt3/(d+20)\Rightarrow d+20=3d\Rightarrow d=10$. $h=10\sqrt3$ m.

$\tan60^\circ = (20+h)/20 \Rightarrow \sqrt3 = (20+h)/20 \Rightarrow 20+h = 20\sqrt3 \Rightarrow h = 20(\sqrt3-1)$ m.

$\tan60^\circ = h/40 \Rightarrow \sqrt3 = h/40 \Rightarrow h = 40\sqrt3$ m.

$\sin60^\circ = h/10 \Rightarrow \sqrt3/2 = h/10 \Rightarrow h = 5\sqrt3$ m.