It can be observed that AC and PR are the slides for younger and elder children respectively.
In \(\triangle ABC\),
\(\frac{AB}{ AC} = sin 30°\)
\(\frac{1.5}{ AC} = \frac{1}2\)
\(AC = 3m\)
In \(\triangle PQR,\)
\(\frac{PQ}{PR} = sin60°\)
\(\frac{3}{ PR} = \frac{\sqrt 3}2\)
\(PR = \frac{6}{\sqrt3}\)
\(PR = 2\sqrt3 m\)
Therefore, the lengths of these slides are \(3 \,m\) and \(2\sqrt3 \,m\).
The shadow of a tower on level ground is $30\ \text{m}$ longer when the sun's altitude is $30^\circ$ than when it is $60^\circ$. Find the height of the tower. (Use $\sqrt{3}=1.732$.)
In the adjoining figure, \(PQ \parallel XY \parallel BC\), \(AP=2\ \text{cm}, PX=1.5\ \text{cm}, BX=4\ \text{cm}\). If \(QY=0.75\ \text{cm}\), then \(AQ+CY =\)
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD. 