It tan A = cot (A + 10°), where (A + 10°) is acute then find ∠A.

Given: tanA = cot(A + 10°)

$\tan A=\cot \left(A+10^{\circ}\right)$

$\Rightarrow \cot \left(90^{\circ}-A\right)=\cot \left(A+10^{\circ}\right) \quad\left(\because \tan \theta=\cot \left(90^{\circ}-\theta\right)\right)$

$\Rightarrow 90^{\circ}-A=A+10^{\circ}$

$\Rightarrow 90^{\circ}-10^{\circ}=A+A$

$\Rightarrow 2 A=80^{\circ}$

$\Rightarrow A=\frac{80^{\circ}}{2}$

$\Rightarrow A=40^{\circ}$

Hence, $\angle A=40^{\circ}$.