Question

The measurement of h is ______.

Options

• \[\sqrt{8 \times 6} \mathrm{~cm}\]

• \[\sqrt{\frac{4}{3}} \mathrm{~cm}\]

• \[(8 - 6) \mathrm{cm}\]

• \[2 \sqrt{2} \mathrm{~cm}\]

Answer 1

The measurement of h is \[\underline{\boldsymbol{\sqrt{8 \times 6} \mathrm{~cm}}}\].

Explanation:

We know that,

\[\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\]

From figure,

In \[\triangle ABC,\]

\[\Rightarrow \tan 50^\circ = \frac{AB}{BC}\]

\[\Rightarrow \tan 50^\circ = \frac{h}{6}\]

\[\Rightarrow h = 6 \tan 50^\circ\]

\[\Rightarrow h = 6 \tan (90^\circ - 40^\circ)\]

\[\Rightarrow h = 6 \cot 40^\circ\]    ...(1)

In \[\triangle ABD,\]

\[\Rightarrow \tan 40^\circ = \frac{AB}{BD}\]

\[\Rightarrow \tan 40^\circ = \frac{h}{8}\]

\[\Rightarrow h = 8 \tan 40^\circ\]    ...(2)

Multiplying equation (1) and (2), we get:

\[\Rightarrow h \times h = 6 \cot 40^\circ \times 8 \tan 40^\circ\]

\[\Rightarrow h^2 = 8 \times 6 \times \cot 40^\circ \times \frac{1}{\cot 40^\circ}\]

\[\Rightarrow h^2 = 8 \times 6\]

\[\Rightarrow h = \sqrt{8 \times 6}.\]