Trigonometry Questions and Answers

We know that $sintheta = cos (90degree - theta)$

$therefore sin 2theta = cos 3theta$

$Rightarrow cos (90degree - 2theta) = cos 3theta$

$Rightarrow 2theta = 90degree - 3theta$

$Rightarrow theta = 18degree$

Given $dfrac{1}{cos 11theta} = dfrac{1}{sin 7theta} = dfrac{1}{cos (90- 7theta) }$

$Rightarrow 11theta = 90 - 7theta$

$Rightarrow theta = 5degree$

By definition, for Maxima or Minima,

$frac{d}{dtheta} (sin theta cos theta ) = 0$

$Rightarrow cos^2 theta - sin^2 theta = 0$

$Rightarrow cos theta = sin theta Rightarrow theta = 45$

$therefore text{max } (sin theta cos theta )$

$=sin 45^circ cos 45^circ = frac{1}{2}$

$sin^3 15 - cos^3 15 = (sin 15 - cos 15)(1 + sin 15 cos 15)$

$= (sin 15 - cos 15) Big(1 + frac{1}{2}sin 30Big)$

$=frac{5}{4}(sin 15 - cos 15)$

$=frac{5}{4}Big[pmsqrt{1 - sin 30}Big]$

$=pmfrac{5}{4sqrt{2}} = - frac{5}{4sqrt{2}} ; text{(Neglecting +ve sign)}$

We know that $tan 3theta =frac{3 tan theta - tan^3 theta}{1 - 3 tan^2theta}$

Here, $theta = 20degree$

$therefore frac{3 tan 20degree - tan^3 20degree }{1 - 3 tan^2 20degree}$

$=tan 3theta = tan 60degree = sqrt{3}$

The given equation can be written as

$2 - 2 sin^2 theta + 11 sin theta - 7 = 0$

$Rightarrow 2 sin^2 theta - 11 sin theta + 5 = 0$

$Rightarrow sin theta = frac{11 pm sqrt{121-40}}{4}$

$=frac{11 pm 9}{4} = 5 text{ or } frac{1}{2}$

Since $sin theta$ can not be greater than 1,

$therefore sin theta = frac{1}{2}$

The angle between two consecutive hour marks  $=frac{360}{12} = 30degree$
By question, minute hand is at 3 and hour hand slightly ahead of 2.

$therefore$ In 15 minutes hour hand moves by an angle of $frac{30}{60} times 15 = 7frac{1}{2}degree$

$therefore$ Required angle $= 30degree - 7frac{1}{2}degree =22frac{1}{2}degree$

$begin{aligned} text{Let} quad text{AD} &= h therefore quad text{BD} &= (3000 - h) text{Now, } tan 45degree &= frac{text{BD}}{text{CB}} therefore quad text{ BD} &= text{CB} = (3000 - h) text{m} end{aligned}$

Again

$quad text{tan } 60degree = frac{text{AB}}{text{BC}}$

$Rightarrow quad text{V}_3 = frac{3000}{3000 - h}$

$Rightarrow quad h =frac{3000}{1} - frac{3000}{sqrt{3}} =frac{3000sqrt{3} - 3000}{sqrt{3}}$

$= 1000 (3 - sqrt3)$

$begin{aligned} text{Let CD } &= hm text{ thus} tan 45degree &= frac{text{CD}}{text{BC}} Rightarrow 1 = frac{h}{text{BC}} Rightarrow quad text{BC } &= text{CD} = h text{Now, } tan 30degree &= frac{text{CD}}{text{AC}} end{aligned}$

$begin{aligned} Rightarrow quad frac{1}{sqrt3} &= frac{h}{h + 20} Rightarrow quad h + 20 &= sqrt3 h therefore quad sqrt3 h - h &= 20 therefore quad h(sqrt3 - 1) &= 20 therefore quad h &= frac{20 times (sqrt3 + 1)}{sqrt3 - 1 times(sqrt3 + 1)} &= frac{20(sqrt3 + 1)}{2} &= 10(sqrt3 + 1) end{aligned}$

Let $quad text{AB} = h$
then $sin 30degree = frac{text{AB}}{sqrt12}$

$Rightarrow quad frac{1}{2} = frac{h}{sqrt12}$

$therefore quad h = frac{sqrt12}{2} = frac{2sqrt3}{2} = sqrt3$