Tag: trigonometric ratios of some specific angles

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Questions Related to trigonometric ratios of some specific angles

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

The distance between $A ( \cos \theta , \sin \theta )$ and $B ( - \sin \theta , \cos \theta )$ is

  1. 1
  2. $2 + 2 \sin \theta$
  3. $1 + \sin \theta$
  4. $\sqrt { 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $A(\cos \theta,\sin \theta)$ and $B(-\sin \theta,\cos \theta)$

Distance between $AB=\sqrt{(\cos \theta+\sin \theta)^2+(\sin \theta-\cos \theta)^2}=\sqrt{\cos^2 \theta+\sin ^2 \theta+2\sin \theta\cos \theta+\cos^2\theta+\sin ^2 \theta-2\sin \theta\cos \theta}=\sqrt{1+1}=\sqrt{2}$

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

$\displaystyle3\tan^2{30^\circ}+\frac{4}{3}\cos^2{30^\circ}-2\sin^2{45^\circ}-\frac{1}{3}\sin^2{60^\circ}$ is equal to__________________.

  1. $\displaystyle\frac{1}{4}$
  2. $\displaystyle\frac{3}{4}$
  3. $1$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given 


$3\tan ^230+\dfrac 43\cos ^230-2\sin ^245-\dfrac 13\sin ^260$

$=3\left(\dfrac 1{\sqrt 3}\right)^2+\dfrac 43\left(\dfrac {\sqrt 3}2 \right)^2-2\left(\dfrac 1{\sqrt 2}\right)^2-\dfrac 13\left(\dfrac {\sqrt 3}2\right)^2$

$=1+1-1-\dfrac 14$

$= \dfrac 34$

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

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. $\displaystyle 100^{g}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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} $

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

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

  1. $\displaystyle \frac{50^{g}}{3}$
  2. $\displaystyle \frac{100^{g}}{3}$
  3. $\displaystyle \frac{160^{g}}{3}$
  4. $\displaystyle \frac{200^{g}}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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} $

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.