Question

From the top of the hill , the the angles of depression of two points due east are found to be 30° and 45°. If the distance between these points is 1 km , then find the height of the hill.

Answer 1

Answer:

Let the distance between the nearer kilometre stone and the hill be 'x' km. 

So, the distance between the farther kilometre stone and the hill is '1+x' km since both are on the same side of the hill..

In triangle APB,

tan450=xh

       ⇒1=xh

       ⇒h=x

In triangle AQB,

 tan300=1+xh

 ⇒31=1+xh

⇒1+x=3h

From equation 1,

1+h=3h⇒1=3h−h

⇒h=3−11 

⇒h=1.365km

Answer 2

Answer:

[tex]Let \: the \: distance \: between \: the \: nearer \: kilometre \\ stone \: and \: the \: hill \: be \: 'x' \: km.  \\ So, \: the \: distance \: between \: the \: farther \: kilometre \\ stone \: and \: the \: hill \: is \: '1+x' \: km \: since \: both \: are \: on m\: the \\ same \: side \: of \: the \: hill. \\

In \: triangle \: APB, \\ \tan( {40}^{0} ) = \frac{h}{x} \\ ⇒1 = \frac{h}{x} \\⇒ h = x \\ In triangle AQB, \\ \tan( {30}^{0} ) = \frac{h}{1 + x} \\ ⇒\frac{1}{ \sqrt{3} } = \frac{h}{1 + x} \\ ⇒1 + x = \sqrt{3h} \\ ⇒ From \: equation \: 1, \\ 1 + h = \sqrt{3h} ⇒1 = \sqrt{3h} - h \\ ⇒h = \frac{1}{ \sqrt{3} - 1} \\ ⇒h = 1.365 \: km [\tex]

☆Hope It Is Right Answer.

☆If It's Wrong Then Report It.

Hey Friends Please Please Report It.

I can't change in Code.

Please Report This Please.