The area of a right-angled triangle is 165 sq metres.

Let the base be $x \mathrm{~m}$.

Therefore, the altitude will be $(x+7) \mathrm{m}$.

Area of a triangle $=\frac{1}{2} \times$ Base $\times$ Altitude

$\therefore \frac{1}{2} \times x \times(x+7)=165$

$\Rightarrow x^{2}+7 x=330$

$\Rightarrow x^{2}+7 x-330=0$

$\Rightarrow x^{2}+(22-15) x-330=0$

$\Rightarrow x^{2}+22 x-15 x-330=0$

$\Rightarrow x(x+22)-15(x+22)=0$

$\Rightarrow(x+22)(x-15)=0$

$\Rightarrow x=-22$ or $x=15$

The value of $x$ cannot be negative.

Therefore, the base is $15 \mathrm{~m}$ and the altitude is $\{(15+7)=22 \mathrm{~m}\}$.