Question

A man is standing on the fourth step of a staircase. Each step is 40 cm wide as shown in the figure. Find the distance between the man and the ground if the inclination of the staircase from the ground is ${50}^{\circ }[tan{50}^{\circ }=1.19]$

• A. 181.2 cm
• B. 175. 5 cm
• C. 190.6 cm
• D. 210. 6 cm

Answer 1

The correct option is C

190.6 cm

Draw $BC{\perp }^{r}AC$

We can see from the figure that AC is equal to four times the width of stairs

$AC=4\times 40=160m$

In right-angled $\Delta ABC$

$tan{50}^{\circ }=\frac{BC}{AC}1.19=\frac{BC}{160}\Rightarrow BC=190.6cm$

Hence, man standing at a height of 190.6 cm from the ground.