Question

from a lighthouse the angle of depression of two ships on opposite side of the lighthouse observed to be 30 degree and 45 degree. the height of the lighthouses 4 metres, the distance between the ships is:-​

Answer 1

Step-by-step explanation:

May this will help you in understanding

Answer 2

Step-by-step explanation:

Answer:

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(8,1){\line(1,0){4}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\qbezier(12,1)(11,1.4)(8,2.9)\put(7.3,2){\sf{\large{80 m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\qbezier(11,1)(10.8,1.2)(11.1,1.4)\put(9.4,1.2){\sf\large{45^{\circ} }}\put(10.5,1.2){\sf\large{30^{\circ} }}\put(11.9,.7){\sf\large D}\put(7.9,3){\sf\large A}\put(10.4,.7){\sf\large B}\put(7.9,.7){\sf\large C}\end{picture}$

• In Triangle ABC :

$:\implies\sf tan(\theta)=\dfrac{Perpendicular}{Base}\\\\\\:\implies\sf tan(45^{\circ})=\dfrac{80}{BC}\\\\\\:\implies\sf 1=\dfrac{80}{BC}\\\\\\:\implies\sf BC=\dfrac{80}{1}\\\\\\:\implies\sf BC=80\qquad -eq.(I)$

$\rule{180}{1.5}$

• In Triangle ACD :

$:\implies\sf tan(\theta)=\dfrac{Perpendicular}{Base}\\\\\\:\implies\sf tan(30^{\circ})=\dfrac{80}{CD}\\\\\\:\implies\sf \dfrac{1}{ \sqrt{3} }=\dfrac{80}{BC+BD}\\\\\\:\implies\sf BC+BD=\dfrac{80}{\frac{1}{ \sqrt{3} }}\\\\\\:\implies\sf 80+BD=80 \sqrt{3}\\\\\\:\implies\sf BD = 80 \sqrt{3} - 80\\\\\\:\implies\underline{\boxed{\sf BD = 80( \sqrt{3} - 1) \:metres}}$