Question

trigonometry

why did you report it last time i just had a doubt

plz dont report

i dont find the necessity to

Answer 1

SolutioN :

Let,

• ∠MON be θ.

Now, We have given.

• MN = 900 m.
• NO = 1200 m.
• ∠MNO = 90°

We know that,

• Sinθ / cosθ = tanθ.

$\tt \bullet \: \: \: \: \: tan \,\theta = \dfrac{Perpendicular(P)}{Base(B)}$

Here,

• Perpendicular ( P ) = 900 m.
• Base ( B ) = 1200 m.

$\tt \mapsto \: \: \: \: \: tan \,\theta = \dfrac{Perpendicular(P)}{Base(B)}$

$\tt \mapsto \: \: \: \: \: tan \,\theta = \dfrac{900}{1200}$

$\tt \mapsto \: \: \: \: \: tan \,\theta = \dfrac{9 \cancel{0} \cancel{0}}{12 \cancel{0} \cancel{0}}$

$\tt \mapsto \: \: \: \: \: tan \,\theta = \dfrac{ \cancel9}{ \cancel{12}}$

$\tt \mapsto \: \: \: \: \: tan \,\theta = \dfrac{3}{4}$

• Note :

• tan 37° = 3 / 4.

Diagram for ( 37° )

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 3}\put(2.8,.3){\large\bf 4}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf 37\degree }\end{picture}$

$\tt \mapsto \: \: \: \: \: tan \,\theta = tan \,37\degree$

$\tt \mapsto \: \: \: \: \: \theta = 37\degree$

Therefore, the value of ∠MON is 37°

Answer 2

Please see the attached image.

Thanks !