Question

plz give me solutions
solve the D ONE​

Answer 1

Solution for (c) part

LHS

$\large\sf { \frac{1+ sin \theta }{ cos \theta} + \frac{ cos \theta}{ 1 + sin \theta}}$

$\large\sf { = \frac{(1+sin \theta)^{2} + cos^{2}}{cos \theta (1+ sin \theta}}$

$\large\sf { = \frac{ 1+ ( sin^{2} \theta + cos^{2} \theta) + 2 \ sin \theta}{ cos \theta (1+ sin \theta)}}$

$\large\sf { = \frac{2( 1 + sin \theta)}{ cos \theta ( 1 + sin \theta}}$

$\large\sf { = 2 sec \theta}$

$\large\rm{ \therefore }$ LHS = RHS

Solution for (d) part ( for fig. refer to Attachment)

given height of tower $\large\sf { CD = 50 m}$

let the height of building , $\large\sf { AB = h}$

in right angled $\large\sf { \triangle BDC}$ ,

$\large\sf { tan \ 60^{ \circ} = \frac{CD}{BD}}$

$\large\sf { \implies \sqrt{3} = \frac{50}{BD}}$

$\large\sf { \therefore BD = \frac{50}{ \sqrt{3}} m }$

In right angled △ABD,

$\large\sf { tan \ 30^{ \circ} = \frac{AB}{BD}}$

$\large\sf { = \frac{1}{ \sqrt{3}} = \frac{h}{ \frac{50}{\sqrt{3}}} }$

$\large\sf { \therefore h = \frac{50}{3}}$

$\large\sf{ h = 16.66 m}$