Question

the angle of elevation of a hill from a point c is alpha .after walking to some point D at a distance a metres from c on a slope of inclination bita, the angle of elevation was found to be Gama, then prove that h = a sin alpha .sin (gama minus bita) /sin(gama minus alpha) ​

Answer 1

Proved below.

Step-by-step explanation:

Given:

As shown in the figure, Let BC be the hill and C be the top of hill.

∠ CAB = α

PA = a

From right angle ΔPAD,

$sin \beta = \frac{Opposite}{Hypotenuse}$

$sin \beta = \frac{PD}{PA}$

$sin \beta = \frac{b_2}{a}$

$a sin \beta = b_2$                 [1]

also from Δ CQP,

$sin \gamma = \frac{Opposite}{Hypotenuse}$

$sin \gamma = \frac{QC}{PC}$

$sin \gamma = \frac{b_1}{PC}$

$PC sin \gamma = b_1$             [2]

∠ PCA = γ − α

$\frac{a}{sin(\gamma - \alpha)} =\frac{PC}{sin(\alpha - \beta)}$

$PC = \frac{bsin (\alpha - \beta)}{sin (\gamma - \alpha)}$         [3]

Putting Eq (3) in Eq (2), we get

$\frac{bsin (\alpha - \beta)}{sin (\gamma - \alpha)} sin \gamma = b_1$    [4]

Let h be the height of hill, so

Height of hill = $\frac { a }{ \sin { \left( \gamma -a \right) } } \left[ \sin { \beta } \sin { \left( \gamma -a \right) } +\sin { \left( \alpha -\beta \right) } \sin { \left( \gamma -a \right) } \sin { \gamma } \right]$

                       = $\frac{asin \alpha}{sin (\gamma- \alpha)} [sin \gamma cos \beta - sin \beta cos \gamma]$

                    h = $\frac{a sin \alpha sin (\gamma- \beta)}{sin(\gamma - \beta)}$