Question

Can anyone please try to solve this problem...... ​

Answer 1

QUESTION:

A tree standing on a horizontal plane is leaning towards east. At two points situated at distance a and b exactly due west on it, the angles of elevation of the top are respectively α and β. Prove that the height of the top from the ground is [ (b − a)tan α tan β ] /  tan α − tan β

GIVEN:

• A tree standing on a horizontal plane is leaning towards east.

• At two points situated at distance a and b exactly due west on it, the angles of elevation of the top are respectively α and β.

TO PROVE:

• The height of the top from the ground is [ (b − a) tan α tan β ] /  tan α − tan β .

DIAGRAM:

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(6,2)\qbezier(2,0)(2,0)(6,2)\qbezier(4,0)(4,0)(6,2)\qbezier(6,0)(6,0)(6,2)\qbezier(0,0)(0,0)(6,0)\qbezier(2.4,0.22)(3,0.2)(2.5,0)\qbezier(0.7,0)(1,0.2)(0.7,0.26)\put(6,2.1){ \bf D }\put(6,-0.3){ \bf O }\put(4,-0.3){ \bf C }\put(2,-0.3){ \bf A }\put(0,-0.3){ \bf B }\put(2.8,0.1){ \boldsymbol \alpha }\put(1,0.1){ \boldsymbol \beta }\put(6.1,1){ \bf h }\put(2,-0.5){\vector(1,0){2}}\put(3.9,-0.5){\vector(-1,0){1.9}}\put(6,-0.5){\vector(-1,0){2}}\put(4,-0.5){\vector(1,0){2}}\put(0,-0.7){\vector(1,0){4}}\put(4,-0.7){\vector(-1,0){4}}\put(5,-0.3){ \bf x }\put(3,-0.3){ \bf a }\put(2,-1){ \bf b }\end{picture}$

PROOF:

Take ΔAOD

$\boxed{\bold{\large{\gray{Tan \ \theta = \dfrac{Opposite\ side}{Adjacent\ side}}}}}$

$\sf Tan \ \alpha = \dfrac{OD}{OA}$

OD = h

OA = AC + OC

AC = a

OC = x

OA = a + x

$\sf Tan \ \alpha = \dfrac{h}{a+x}$

$\sf a+x= \dfrac{h}{Tan \ \alpha}$

$\boxed{\bold{\large{\gray{Cot \ \theta = \dfrac{1}{Tan\ \theta}}}}}$

$\sf a+x= h\ Cot \ \alpha$

$\sf x= h\ Cot \ \alpha - a$

Take ΔBOD

$\boxed{\bold{\large{\gray{Tan \ \theta = \dfrac{Opposite\ side}{Adjacent\ side}}}}}$

$\sf Tan \ \beta = \dfrac{OD}{OB}$

OD = h

OB = BC + OC

BC = b

OC = x

OB = b + x

$\sf Tan \ \beta = \dfrac{h}{b+x}$

$\sf b+x= \dfrac{h}{Tan \ \beta}$

$\boxed{\bold{\large{\gray{Cot \ \theta = \dfrac{1}{Tan\ \theta}}}}}$

$\sf b+x= h\ Cot \ \beta$

$\sf x= h\ Cot \ \beta - b$

Equate both the x values

$\sf h\ Cot \ \beta - b= h\ Cot \ \alpha - a$

$\sf h\ Cot \ \beta -h\ Cot \ \alpha=b - a$

$\sf h( Cot \ \beta -Cot \ \alpha)=b - a$

$\sf h\left( \dfrac{1}{Tan \ \beta} -\dfrac{1}{Tan \ \alpha}\right)=b - a$

$\sf h\left( \dfrac{Tan \ \alpha-Tan \ \beta}{Tan \ \alpha\ Tan \ \beta}\right)=b - a$

$\sf h( Tan \ \alpha-Tan \ \beta)=b - a(Tan \ \alpha\ Tan \ \beta)$

$\sf h=\dfrac{(b - a)Tan \ \alpha\ Tan \ \beta}{Tan \ \alpha-Tan \ \beta}$

HENCE PROVED.