Question

a pole make an angle of evaluation with the moon find this angle if the length of the shadow of pole is twice it's height​ options 1)44° 2)26.56° 3) 55° 4) 35°

Answer 1

Answer:

Referring figure, if AB is pole and AC is length

of the shadow of pole then AB=x, AC=2x and let ∠ACB = θ

Then, tanθ =

AC

AB

⇒tanθ=

2AB

AB

=

2

1

∴ Value of θ is neither 30

o

, 60

o

nor 45

o

.

Answer 2

Given :

• Length of a pole is twice the length of the shadow of pole

To Find :

• The angle made by the pole with the moon

Figure :

$\setlength{\unitlength}{2cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.6,2){\sf{\large{a}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\put(9.4,1.2){\sf\large{ \theta }}\put(7.9,3){\sf\large X}\put(10.4,.7){\sf\large Z}\put(7.9,.7){\sf\large Y}\put(9,0.7){\sf\large 2a}\end{picture}$

Solution :

Let ,

• XY be the length of the shadow
• θ be the angle of elevation
• YZ be the length of the shadow

Let the length of the pole (XY) be ' a '

Then the length of the shadow = 2( Length of the pole)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀= 2(a) = 2a

Since the image forms right-angled triangle . Tan(θ) is given by,

$\boxed {\rm{ \tan( \theta) = \frac{opposite \: side}{adjacent \: side} }}$

Here ,

• opposite side (to θ) = XY = a
• Adjacent side (to θ) = YZ = 2a

Now ,

$:\implies\rm \tan(\theta) =\frac{XY}{YZ}\\ \\: \implies \rm \tan( \theta) = \frac{a}{2a} \\ \\ : \implies \rm \tan( \theta) = \frac{ \cancel{a}}{2 \cancel{a}} \\ \\ : \implies \rm \: \tan( \theta) = \frac{1}{2} \\ \\ : \implies \rm \: \tan( \theta) = 0.5 \\ \\ : \implies \rm \: \theta = { \tan}^{ - 1} (0.5) \\ \\ : \implies \rm \theta = 26.56 {} \: ^ { \circ}$

∴ The Angle made by the pole with moon is 26.56°

Hence , option(2) is correct