Tag: trigonometry

Questions Related to trigonometry

The angle measuring $\displaystyle \frac{\pi ^{c}}{4}$ when expressed in centesimal system is ___ 

  1. $\displaystyle 50^{g}$

  2. $\displaystyle 60^{g}$

  3. $\displaystyle 75^{g}$

  4. $\displaystyle 100^{g}$


Correct Option: A
Explanation:

In $\text{Centesimal System}$, an angle is measured in grades, minutes and seconds.
Given angle $ = \dfrac {{\pi}^c}{4} = \dfrac {{180}^{0}}{4} = {45}^{0} $

We know that $ {1}^{0} = {(\dfrac {10}{9})}^{g} $


$ \Rightarrow {45}^{0} = {\dfrac {10}{9}} \times 45^{g}  ={50}^{g} $

$\displaystyle 30^{\circ}$ in centesimal measure is _____

  1. $\displaystyle \frac{50^{g}}{3}$

  2. $\displaystyle \frac{100^{g}}{3}$

  3. $\displaystyle \frac{160^{g}}{3}$

  4. $\displaystyle \frac{200^{g}}{3}$


Correct Option: B
Explanation:

In $\text{Centesimal System}$, an angle is measured in grades, minutes and seconds.
In centesimal system, we know that $ {1}^{0} = {(\dfrac {10}{9})}^{g} $
$ \Rightarrow  {30}^{0} =  {(\dfrac {10}{9})}^{g} \times 30 = {(\dfrac {100}{3})}^{g} $

When the sun is $30^o$ above the horizon, what is the length of the shadow cast by a building $40$ m high?

  1. $50.23$ m

  2. $70.24$ m

  3. $68.25$ m

  4. $69.28$ m


Correct Option: D
Explanation:

$\tan 30^o = \dfrac{40m}{shadow}$

Shadow $= \dfrac{40}{\tan 30^o}$

$= \dfrac{40}{\frac{1}{\sqrt{3}}}$= $\dfrac{40}{0.57735}$

The shadow of the building is $69.28$ m.

So, option D is correct.

From the tower $30$ m above the sea, the angle of depression of a boat is $68^o$. How far is the boat from the tower?

  1. $12.12$ m

  2. $11.11$ m

  3. $10.10$ m

  4. $9.99$ m


Correct Option: A
Explanation:
$\tan 68^o = \dfrac{30}{distance}$

Distance $=\dfrac{30}{\tan 68^o}$

Distance = $\dfrac{30}{2.475087}$

Distance $= 12.12$ m

So, option A is correct.

When the sun is $50^o$ above the horizon, how long is the shadow cast by a building $16$ m high?

  1. $23$ m

  2. $13.42$ m

  3. $43.42$ m

  4. $23.42$ m


Correct Option: B
Explanation:

$\tan 50^o = \dfrac{16m}{shadow}$

Shadow $= \dfrac{16}{\tan 50^o}$= $\dfrac{16}{1.191754}$= $13.42$

The shadow of the building is $13.42$ m

So, option B is correct.