Question

bow with semicircle tunnel a person with height 152cm is standing 36cm away from one end of bow what is the breadth of the bow
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Answer 1

The width of the bow is 677.7 cm

Step-by-step explanation:

Given as :

A bow with semicircular tunnel

The height of person = h = 152 cm

The distance of person from one end of bow = 36 cm

Let The distance of person from one end of bow = x cm

Let The width of bow = w  = (x + 36) cm

According to question

From figure

Height of person = h = AO = 152 cm

The distance of person from one end of bow = BA = 36 cm

In ΔAOB

OB² = OA² + AB²

y² = (152 cm)² + (36 cm)²

Or, y² = 23104 cm² + 1296 cm²

Or, y² = 24400            .........1

∴    y = $\sqrt{24400}$

i.e  y = 156.2 cm

Again

In ΔAOC

OC² = OA² + AC²

z² = (152 cm)² + (x cm)²

Or, z² = 23104 cm² + x² cm²         ,,,,,,,,,2

And

In ΔCOB

BC² = OC² + OB²

(x + 36 cm)² = (z cm)² + (y cm)²

Or, x² + 36² + 72 x = z² + y²        ..........3

From eq 1 , 2 , 3

x² + 1296 + 72 x = 23104  + x²  + 24400.

Solving the equation

( x² - x² ) + 72 x = 47504 - 1296

Or, 0 + 72 x = 46208

∴     x = $\dfrac{46208}{72}$

i.e  x = 641.7 cm

Now,

The width of bow = CB

i.e   CB = CA + AB

Or,   w = x cm + 36 cm

Or,   w = 641.7 cm + 36 cm

∴     w = 677.7 cm

So, The width of the bow = w = 677.7 cm

Hence, The width of the bow is 677.7 cm Answer