Sum

If the elevation of the sun changed from 30º to 60º, then find the difference between the lengths of shadows of a pole 15 m high, made at these two positions

Advertisement Remove all ads

#### Solution

When AB = 15m, θ = 30º,

`\text{then }\frac{AC}{AB}=\text{tan}30^\text{o}`

`\Rightarrow AC=\frac{15}{\sqrt{3}}m.`

When AB = 15m, θ = 60º,

`\text{then }\frac{AC}{AB}=\text{tan6}0^\text{o}`

⇒ AC = 15√3 m.

∴ Diff. in lengths of shadows

`=( 15\sqrt{3}-\frac{15}{\sqrt{3}} )`

`=\frac{30}{\sqrt{3}}=10\sqrt{3}m`

Concept: Heights and Distances

Is there an error in this question or solution?

Advertisement Remove all ads