Step 1: Let the height of the tower be \( h \) m and the height of the flagstaff be \( x \) m.
Total height of tower + flagstaff = \( h + x \). Distance from the tower = \( 10 \, \text{m} \).
Step 2: Use trigonometric ratios for each angle of elevation.
For the top of the tower: \[ \tan 45^\circ = \dfrac{h}{10} \Rightarrow 1 = \dfrac{h}{10} \Rightarrow h = 10 \, \text{m} \] For the top of the flagstaff: \[ \tan 60^\circ = \dfrac{h + x}{10} \Rightarrow \sqrt{3} = \dfrac{h + x}{10} \Rightarrow h + x = 10\sqrt{3} \] Step 3: Substitute the value of \( h \).
\[ 10 + x = 10\sqrt{3} \Rightarrow x = 10(\sqrt{3} - 1) \] \[ x = 10(1.732 - 1) = 7.32 \, \text{m} \] Step 4: Conclusion.
Hence, the length of the flagstaff is approximately 7.32 m.
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be:
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 figure, the angles of depression of point \( O \) as seen from points \( A \) and \( P \) are: 