Question

. A child is watching a bird on a tree.The angle

of elevation of his/her sight is 450

. The child is

1.5 m tall.Find the height of the tree when the

distance between the foot of the tree and child

on the ground is 30m.​

Answer 1

Answer:

$\bigstar{\bold{Height\:of\:the\:tree=31.5\:m}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Height of the child = 1.5 m
• Angle of elevation  = 45°
• Distance between foot of tree and the child = 30 m

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The height of the tree

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Let AC be the height of the tree

→ Let EC be the height of the child = 1.5 m

→ Let AB be x m.

→ Hence

  Height of the tree = AB + BC

  Height of the tree = x + 1.5 m (∵BC = ED)

→ Now consider ΔAEB

  tan 45 = AB/EB

  tan 45 = x/30 (∵ EB = DC = 30 m)

  1 = x/30

  x = 30 m

→ Hence

  Height of the tree = x + 1.5m

  Height of the tree = 30 + 1.5

  Height of the tree = 31.5 m

$\boxed{\bold{Height\:of\:the\:tree=31.5\:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

sin A = opposite/hypotenuse

cos A = adjacent/hypotenuse

tan A = opposite/adjacent

$\begin{gathered}\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}\end{gathered}$

Answer 2

Answer :-

★ Height of the tree = 31.5 m

★ Concept :

The concept used here is of Heights and Distances in trigonometry using angle ratios. Here, according this , if two sides in a right triangle are taken in ratios, they will surely give some value, which corresponds to the angle.

★ Solution :

Note* , here we are using the diagram for referencing the points.

• Given,

=> Angle of elevation to the bird = 45°

=> Height of the child = 1.5 m

=> Distance between child and foot of tree = 30m

Now,

By applying the diagram points with values, we get,

• AB = EC = 30 m

• EA = CB = 1.5 m

• < DEC = 45°

• DC = x m (let)

We know that,

Height of tree = DC + CB = ( x + 1.5 ) m

Now using the triangle ∆ DEC,

• < DEC = 45°

• EC = 30 m

• DC = x m

We know that,

tan 45° = 1 .... (i)

And,

tan <E = Perpendicular / Base

= DC /EC ..(ii)

Now, using eq. (i) and eq. (ii), we get,

=> (DC / EC) = 1

=> DC = EC

By applying the value EC, we get,

=> EC = DC = 30 m

So, the value of DC = 30 m

Now,

Height of the tree = DC + CB

= (30 + 1.5)

m = 31.5 m

✏ Hence, height of the tree = 31.5 m

★ More to Know :-

=> Trigonometric Ratios :

• sin ø = Perpendicular / Hypotenuse

• cos ø = Base / Hypotenuse

• tan ø = Perpendicular / Base

• cosine ø = 1 / (sin ø)

• cosec ø = 1 / (cos ø)

• cotangent ø = 1 / (tan ø)

• Trigonometry is the branch of mathematics which deals with the ratio of angles and sides in order to calculate values of different measures.

It has been derived from latin word,

'Trigon' = Triangle

'Metron' = Measure

which collectively means, measure of triangle.