$$ \cot\theta+\tan\theta=\sec\theta\cosec\theta $$
Proof
LHS:
$$ \cot\theta+\tan\theta $$
$$ =\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta} $$
$$ =\frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta} $$
$$ =\frac1{\sin\theta\cos\theta} $$
$$ =\sec\theta\cosec\theta $$
Hence proved.
$$ \tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta $$
Proof
LHS:
$$ \tan^4\theta+\tan^2\theta $$
$$ =\tan^2\theta(\tan^2\theta+1) $$
Using
$$ 1+\tan^2\theta=\sec^2\theta $$
$$ =\tan^2\theta\sec^2\theta $$
Again,
$$ \tan^2\theta=\sec^2\theta-1 $$
$$ =(\sec^2\theta-1)\sec^2\theta $$
$$ =\sec^4\theta-\sec^2\theta $$
Hence proved.
$$ \frac{1-\tan^2\theta}{\cot^2\theta-1}=\tan^2\theta $$
Proof
Since $\cot^2\theta=\dfrac{1}{\tan^2\theta}$,
$$ \cot^2\theta-1=\frac{1}{\tan^2\theta}-1=\frac{1-\tan^2\theta}{\tan^2\theta} $$
Therefore
$$ \frac{1-\tan^2\theta}{\cot^2\theta-1} =\frac{1-\tan^2\theta}{\dfrac{1-\tan^2\theta}{\tan^2\theta}} =\tan^2\theta $$
Hence proved.
$$ \frac{\cos\theta}{1+\sin\theta}=\sec\theta-\tan\theta $$
Proof
Multiply the numerator and denominator of the LHS by $(1-\sin\theta)$:
$$ \frac{\cos\theta}{1+\sin\theta} =\frac{\cos\theta(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)} =\frac{\cos\theta(1-\sin\theta)}{1-\sin^2\theta} $$
$$ =\frac{\cos\theta(1-\sin\theta)}{\cos^2\theta} =\frac{1-\sin\theta}{\cos\theta} =\sec\theta-\tan\theta $$
Hence proved.
$$ \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}=\sec\theta+\tan\theta $$
Proof
Multiply the numerator and denominator inside the root by $(1+\sin\theta)$:
$$ \frac{1+\sin\theta}{1-\sin\theta} =\frac{(1+\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)} =\frac{(1+\sin\theta)^2}{1-\sin^2\theta} =\frac{(1+\sin\theta)^2}{\cos^2\theta} $$
Taking the square root,
$$ \sqrt{\frac{1+\sin\theta}{1-\sin\theta}} =\frac{1+\sin\theta}{\cos\theta} =\sec\theta+\tan\theta $$
Hence proved.
$$ \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}+\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=2\sec\theta $$
Proof
From part (i), $\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}}=\dfrac{1+\sin\theta}{\cos\theta}$. In the same way,
$$ \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\frac{1-\sin\theta}{\cos\theta} $$
Adding the two terms,
$$ \frac{1+\sin\theta}{\cos\theta}+\frac{1-\sin\theta}{\cos\theta} =\frac{2}{\cos\theta}=2\sec\theta $$
Hence proved.
$$ \sec^6\theta=\tan^6\theta+3\tan^2\theta\sec^2\theta+1 $$
Proof
Using $\sec^2\theta=1+\tan^2\theta$,
$$ \sec^6\theta=(\sec^2\theta)^3=(1+\tan^2\theta)^3 =1+3\tan^2\theta+3\tan^4\theta+\tan^6\theta $$
Group the two middle terms:
$$ 3\tan^2\theta+3\tan^4\theta=3\tan^2\theta(1+\tan^2\theta)=3\tan^2\theta\sec^2\theta $$
Therefore
$$ \sec^6\theta=\tan^6\theta+3\tan^2\theta\sec^2\theta+1 $$
Hence proved.
$$ (\sin\theta+\sec\theta)^2+(\cos\theta+\cosec\theta)^2=1+(\sec\theta+\cosec\theta)^2 $$
Proof
Expand the LHS:
$$ \sin^2\theta+2\sin\theta\sec\theta+\sec^2\theta+\cos^2\theta+2\cos\theta\cosec\theta+\cosec^2\theta $$
Use $\sin^2\theta+\cos^2\theta=1$, $\sin\theta\sec\theta=\tan\theta$ and $\cos\theta\cosec\theta=\cot\theta$:
$$ =1+\sec^2\theta+\cosec^2\theta+2(\tan\theta+\cot\theta) $$
Now $\tan\theta+\cot\theta=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta\cos\theta}=\sec\theta\cosec\theta$, so
$$ =1+\sec^2\theta+2\sec\theta\cosec\theta+\cosec^2\theta=1+(\sec\theta+\cosec\theta)^2 $$
Hence proved.
$$ \sec^4\theta(1-\sin^4\theta)-2\tan^2\theta=1 $$
Proof
Factor $1-\sin^4\theta$:
$$ 1-\sin^4\theta=(1-\sin^2\theta)(1+\sin^2\theta)=\cos^2\theta(1+\sin^2\theta) $$
So
$$ \sec^4\theta(1-\sin^4\theta)=\frac{\cos^2\theta(1+\sin^2\theta)}{\cos^4\theta}=\frac{1+\sin^2\theta}{\cos^2\theta}=\sec^2\theta+\tan^2\theta $$
Therefore
$$ \sec^4\theta(1-\sin^4\theta)-2\tan^2\theta=\sec^2\theta+\tan^2\theta-2\tan^2\theta=\sec^2\theta-\tan^2\theta=1 $$
Hence proved.
$$ \frac{\cot\theta-\cos\theta}{\cot\theta+\cos\theta}=\frac{\cosec\theta-1}{\cosec\theta+1} $$
Proof
Since $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$, we have $\cos\theta=\cot\theta\sin\theta$. So
$$ \frac{\cot\theta-\cos\theta}{\cot\theta+\cos\theta} =\frac{\cot\theta-\cot\theta\sin\theta}{\cot\theta+\cot\theta\sin\theta} =\frac{\cot\theta(1-\sin\theta)}{\cot\theta(1+\sin\theta)} =\frac{1-\sin\theta}{1+\sin\theta} $$
Divide the numerator and denominator by $\sin\theta$ and use $\dfrac{1}{\sin\theta}=\cosec\theta$:
$$ =\frac{\dfrac{1}{\sin\theta}-1}{\dfrac{1}{\sin\theta}+1}=\frac{\cosec\theta-1}{\cosec\theta+1} $$
Hence proved.
$$ \frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B}=0 $$
Proof
Take the LHS over the common denominator $(\cos A+\cos B)(\sin A+\sin B)$:
$$ \frac{(\sin A-\sin B)(\sin A+\sin B)+(\cos A-\cos B)(\cos A+\cos B)}{(\cos A+\cos B)(\sin A+\sin B)} $$
The numerator is
$$ (\sin^2 A-\sin^2 B)+(\cos^2 A-\cos^2 B)=(\sin^2 A+\cos^2 A)-(\sin^2 B+\cos^2 B)=1-1=0 $$
Hence the whole expression equals $0$.
$$ \frac{\sin^3 A+\cos^3 A}{\sin A+\cos A}+\frac{\sin^3 A-\cos^3 A}{\sin A-\cos A}=2 $$
Proof
Using $\sin^3 A+\cos^3 A=(\sin A+\cos A)(\sin^2 A-\sin A\cos A+\cos^2 A)$,
$$ \frac{\sin^3 A+\cos^3 A}{\sin A+\cos A}=\sin^2 A-\sin A\cos A+\cos^2 A=1-\sin A\cos A $$
Using $\sin^3 A-\cos^3 A=(\sin A-\cos A)(\sin^2 A+\sin A\cos A+\cos^2 A)$,
$$ \frac{\sin^3 A-\cos^3 A}{\sin A-\cos A}=\sin^2 A+\sin A\cos A+\cos^2 A=1+\sin A\cos A $$
Adding the two results,
$$ (1-\sin A\cos A)+(1+\sin A\cos A)=2 $$
Hence proved.
If $\sin\theta+\cos\theta=\sqrt3$, then prove that $\tan\theta+\cot\theta=1$.
Proof
Square the given equation:
$$ (\sin\theta+\cos\theta)^2=3\implies\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=3 $$
$$ 1+2\sin\theta\cos\theta=3\implies\sin\theta\cos\theta=1 $$
Now
$$ \tan\theta+\cot\theta=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta} =\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{1}=1 $$
Hence proved.
If $\sqrt3\sin\theta-\cos\theta=0$, then show that $\tan3\theta=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$.
Proof
From the condition, $\sqrt3\sin\theta=\cos\theta$, so $\tan\theta=\dfrac{1}{\sqrt3}$.
The required result is the triple-angle identity, obtained from $\tan3\theta=\tan(2\theta+\theta)$ with $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$:
$$ \tan3\theta=\frac{\tan2\theta+\tan\theta}{1-\tan2\theta\tan\theta} =\frac{\dfrac{2\tan\theta}{1-\tan^2\theta}+\tan\theta}{1-\dfrac{2\tan\theta}{1-\tan^2\theta}\cdot\tan\theta} $$
Multiply numerator and denominator by $(1-\tan^2\theta)$:
$$ =\frac{2\tan\theta+\tan\theta(1-\tan^2\theta)}{(1-\tan^2\theta)-2\tan^2\theta} =\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta} $$
Hence proved.
If $\dfrac{\cos\theta}{\cos\alpha}=m$ and $\dfrac{\cos\theta}{\sin\alpha}=n$, then prove that $(m^2+n^2)\cos^2\alpha=n^2$.
Proof
From the two relations, $\cos\theta=m\cos\alpha$ and $\cos\theta=n\sin\alpha$. Equating these,
$$ m\cos\alpha=n\sin\alpha\implies\tan\alpha=\frac{m}{n} $$
Then
$$ \cos^2\alpha=\frac{1}{\sec^2\alpha}=\frac{1}{1+\tan^2\alpha}=\frac{1}{1+\dfrac{m^2}{n^2}}=\frac{n^2}{m^2+n^2} $$
Therefore
$$ (m^2+n^2)\cos^2\alpha=(m^2+n^2)\cdot\frac{n^2}{m^2+n^2}=n^2 $$
Hence proved.
If $\cot\theta+\tan\theta=x$ and $\sec\theta-\cos\theta=y$, then prove that $(x^2y)^{2/3}-(xy^2)^{2/3}=1$.
Proof
Simplify $x$ and $y$:
$$ x=\cot\theta+\tan\theta=\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta}=\frac{1}{\sin\theta\cos\theta} $$
$$ y=\sec\theta-\cos\theta=\frac{1}{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\frac{\sin^2\theta}{\cos\theta} $$
Then
$$ x^2y=\frac{1}{\sin^2\theta\cos^2\theta}\cdot\frac{\sin^2\theta}{\cos\theta}=\frac{1}{\cos^3\theta}\implies(x^2y)^{2/3}=\frac{1}{\cos^2\theta}=\sec^2\theta $$
$$ xy^2=\frac{1}{\sin\theta\cos\theta}\cdot\frac{\sin^4\theta}{\cos^2\theta}=\frac{\sin^3\theta}{\cos^3\theta}\implies(xy^2)^{2/3}=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta $$
Therefore
$$ (x^2y)^{2/3}-(xy^2)^{2/3}=\sec^2\theta-\tan^2\theta=1 $$
Hence proved.
If $\sin\theta+\cos\theta=p$ and $\sec\theta+\cosec\theta=q$, then prove that $q(p^2-1)=2p$.
Proof
Square the first relation:
$$ p^2=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1+2\sin\theta\cos\theta\implies p^2-1=2\sin\theta\cos\theta $$
Also
$$ q=\frac{1}{\cos\theta}+\frac{1}{\sin\theta}=\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\frac{p}{\sin\theta\cos\theta} $$
Therefore
$$ q(p^2-1)=\frac{p}{\sin\theta\cos\theta}\times2\sin\theta\cos\theta=2p $$
Hence proved.
If $\sin\theta(1+\sin^2\theta)=\cos^2\theta$, then prove that $\cos^6\theta-4\cos^4\theta+8\cos^2\theta=4$.
Proof
From the condition, $\sin\theta=\dfrac{\cos^2\theta}{1+\sin^2\theta}$. Squaring,
$$ \sin^2\theta(1+\sin^2\theta)^2=\cos^4\theta $$
Replace $\sin^2\theta=1-\cos^2\theta$, so that $1+\sin^2\theta=2-\cos^2\theta$:
$$ (1-\cos^2\theta)(2-\cos^2\theta)^2=\cos^4\theta $$
Expand using $(2-\cos^2\theta)^2=4-4\cos^2\theta+\cos^4\theta$:
$$ (1-\cos^2\theta)(4-4\cos^2\theta+\cos^4\theta)=4-8\cos^2\theta+5\cos^4\theta-\cos^6\theta $$
Setting this equal to $\cos^4\theta$ and rearranging,
$$ 4-8\cos^2\theta+5\cos^4\theta-\cos^6\theta=\cos^4\theta\implies\cos^6\theta-4\cos^4\theta+8\cos^2\theta=4 $$
Hence proved.
Given
$$ \frac{\cos\theta}{1+\sin\theta}=\frac1a $$
Cross multiply:
$$ a\cos\theta=1+\sin\theta $$
Squaring both sides,
$$ a^2\cos^2\theta=(1+\sin\theta)^2 $$
Use $\cos^2\theta=1-\sin^2\theta=(1-\sin\theta)(1+\sin\theta)$:
$$ a^2(1-\sin\theta)(1+\sin\theta)=(1+\sin\theta)^2 $$
Since $1+\sin\theta$ is the denominator in the given expression, cancel it:
$$ a^2(1-\sin\theta)=1+\sin\theta $$
Therefore
$$ a^2-1=\sin\theta(a^2+1) $$
So
$$ \sin\theta=\frac{a^2-1}{a^2+1} $$
Hence proved.
Let the angle of elevation be $\theta$.
$$ \tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}} $$
$$ \tan\theta=\frac{10\sqrt3}{30} $$
$$ \tan\theta=\frac{\sqrt3}{3} $$
$$ \tan\theta=\frac1{\sqrt3} $$
Therefore,
$$ \theta=30^\circ $$
Answer
$$ 30^\circ $$
Let the width of the road be $x$ m.
Since the pedestrian stands on the median,
Distance from pedestrian to house
$$ =\frac{x}{2} $$
Using tangent ratio,
$$ \tan30^\circ=\frac{4\sqrt3}{x/2} $$
$$ \frac1{\sqrt3}=\frac{4\sqrt3\times2}{x} $$
$$ x=24 $$
Answer
$$ 24\text{ m} $$
Height of man's eye level
$$ =1.8\text{ m} $$
Distance from wall
$$ =5\text{ m} $$
Height of bottom of window above eye level:
$$ \tan45^\circ=\frac{h_1}{5} $$
$$ 1=\frac{h_1}{5} $$
$$ h_1=5\text{ m} $$
Height of top of window above eye level:
$$ \tan60^\circ=\frac{h_2}{5} $$
$$ 1.732=\frac{h_2}{5} $$
$$ h_2=8.66\text{ m} $$
Height of window
$$ =8.66-5 $$
$$ =3.66\text{ m} $$
Answer
$$ 3.66\text{ m} $$
Let height of pedestal be $h$ m and distance from observation point be $x$ m.
For top of pedestal:
$$ \tan40^\circ=\frac{h}{x} $$
For top of statue:
$$ \tan60^\circ=\frac{h+1.6}{x} $$
$$ 1.732=\frac{h+1.6}{x} $$
Using the first equation and solving simultaneously,
$$ h\approx1.5\text{ m} $$
Answer
$$ 1.5\text{ m} $$
(i) Height of pole
$$ 7\text{ m} $$
(ii) Radius of dome
$$ 16.39\text{ m} $$
Let distance between tower and pole be $x$.
Using
$$ \tan60^\circ=\frac{15}{x} $$
$$ x=\frac{15}{\sqrt3}=5\sqrt3 $$
Let height of electric pole be $h$.
Using
$$ \tan30^\circ=\frac{15-h}{x} $$
$$ \frac1{\sqrt3}=\frac{15-h}{5\sqrt3} $$
$$ 15-h=5 $$
$$ h=10 $$
Answer
$$ 10\text{ m} $$
Let lower part $=x$
Upper part $=9x$
Total height
$$ =10x $$
Using tangent ratios and equal angles condition,
$$ x=10\sqrt5 $$
Therefore total height
$$ =10(10\sqrt5) $$
$$ =100\sqrt5 $$
Answer
$$ 100\sqrt5\text{ m} $$
Let height of mountain be $h$ miles.
Let distance from nearer milestone to mountain be $x$.
$$ \tan8^\circ=\frac{h}{x} $$
$$ 0.1405=\frac{h}{x} $$
$$ h=0.1405x $$
From farther milestone:
$$ \tan4^\circ=\frac{h}{x+1} $$
$$ 0.0699=\frac{h}{x+1} $$
Substitute $h$:
$$ 0.0699(x+1)=0.1405x $$
Solving,
$$ h\approx0.14 $$
Answer
$$ 0.14\text{ mile (approx)} $$
Answer Key
| Q.No | Answer | |---|---| | 1 | $30^\circ$ | | 2 | $24\text{ m}$ | | 3 | $3.66\text{ m}$ | | 4 | $1.5\text{ m}$ | | 5(i) | $7\text{ m}$ | | 5(ii) | $16.39\text{ m}$ | | 6 | $10\text{ m}$ | | 7 | $100\sqrt5\text{ m}$ | | 8 | $0.14\text{ mile (approx)}$ |
Let the distance of the car from the foot of the rock be $x$ m.
Angle of depression
$$ =30^\circ $$
Angle of elevation from the car to the top of the rock is also
$$ 30^\circ $$
Using tangent ratio,
$$ \tan30^\circ=\frac{50\sqrt3}{x} $$
$$ \frac1{\sqrt3}=\frac{50\sqrt3}{x} $$
$$ x=50\sqrt3\times\sqrt3 $$
$$ x=150 $$
Answer
$$ 150\text{ m} $$
Let the height of the first building be $h$ m.
Difference in heights:
$$ 120-h $$
Using tangent ratio,
$$ \tan45^\circ=\frac{120-h}{70} $$
$$ 1=\frac{120-h}{70} $$
$$ 120-h=70 $$
$$ h=50 $$
Answer
$$ 50\text{ m} $$
Let height of lamp post be $h$ m.
Let horizontal distance between tower and lamp post be $x$.
Using angle of depression to bottom:
$$ \tan60^\circ=\frac{60}{x} $$
$$ 1.732=\frac{60}{x} $$
$$ x=\frac{60}{1.732} $$
$$ x\approx34.64 $$
Using angle of depression to top:
$$ \tan38^\circ=\frac{60-h}{34.64} $$
$$ 0.7813=\frac{60-h}{34.64} $$
$$ 60-h=27.07 $$
$$ h=32.93 $$
Answer
$$ 32.93\text{ m} $$
Let distances of boats from the point vertically below the aeroplane be $x$ and $y$.
For first boat:
$$ \tan60^\circ=\frac{1800}{x} $$
$$ 1.732=\frac{1800}{x} $$
$$ x\approx1039.2 $$
For second boat:
$$ \tan30^\circ=\frac{1800}{y} $$
$$ \frac1{1.732}=\frac{1800}{y} $$
$$ y\approx3117.6 $$
Distance between boats:
$$ 3117.6-1039.2 $$
$$ =2078.4 $$
Answer
$$ 2078.4\text{ m} $$
If the height of the lighthouse is $h$ meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is
$$ \frac{4h}{\sqrt3}\text{ m} $$
Proof
Let distances of the ships from the foot of the lighthouse be $x$ and $y$.
For the first ship:
$$ \tan30^\circ=\frac{h}{x} $$
$$ \frac1{\sqrt3}=\frac{h}{x} $$
$$ x=h\sqrt3 $$
For the second ship:
$$ \tan60^\circ=\frac{h}{y} $$
$$ \sqrt3=\frac{h}{y} $$
$$ y=\frac{h}{\sqrt3} $$
Distance between ships:
$$ x+y $$
$$ =h\sqrt3+\frac{h}{\sqrt3} $$
$$ =\frac{3h+h}{\sqrt3} $$
$$ =\frac{4h}{\sqrt3} $$
Hence proved.
Solution (concise):
Let the height of lift above ground when angle = 30° be h. From tan30° = h / (30√3) ⇒ (1/√3) = h / (30√3) ⇒ h = 30 ft. Initial height = 90 ft, so lift descended 90 − 30 = 60 ft in 2 minutes = 120 s. Speed = 60 ft / 120 s = 0.5 ft/s.
Answer: 0.5 ft/s
Let the distance between the trees be $x$ m.
Using angle of depression to the bottom:
$$ \tan30^\circ=\frac{13}{x} $$
$$ \frac1{1.732}=\frac{13}{x} $$
$$ x=13(1.732) $$
$$ x=22.516 $$
Let height of second tree be $h$.
Using angle of elevation:
$$ \tan45^\circ=\frac{h-13}{22.516} $$
$$ 1=\frac{h-13}{22.516} $$
$$ h=35.516 $$
$$ h\approx35.52 $$
Answer
$$ 35.52\text{ m} $$
Let the horizontal distance between the ship and hill be $x$.
Using angle of depression:
$$ \tan30^\circ=\frac{40}{x} $$
$$ \frac1{1.732}=\frac{40}{x} $$
$$ x=69.28 $$
Now using angle of elevation:
$$ \tan60^\circ=\frac{h-40}{69.28} $$
$$ 1.732=\frac{h-40}{69.28} $$
$$ h-40=120 $$
$$ h=160 $$
Answer
Distance from ship to hill:
$$ 69.28\text{ m} $$
Height of hill:
$$ 160\text{ m} $$
$$ \frac{h(\tan\theta_1+\tan\theta_2)}{\tan\theta_2-\tan\theta_1} $$
Proof
Let:
- Height of cloud above lake $=H$
- Horizontal distance $=x$
From angle of elevation:
$$ \tan\theta_1=\frac{H-h}{x} $$
$$ x=\frac{H-h}{\tan\theta_1} $$
From angle of depression of reflection:
Reflection lies $H$ m below water level.
Thus vertical distance:
$$ H+h $$
$$ \tan\theta_2=\frac{H+h}{x} $$
Substitute value of $x$:
$$ \tan\theta_2= \frac{(H+h)\tan\theta_1}{H-h} $$
Solving for $H$,
$$ H= \frac{h(\tan\theta_1+\tan\theta_2)} {\tan\theta_2-\tan\theta_1} $$
Hence proved.
Let horizontal distance between apartment and tower be $x$.
Using angle of depression:
$$ \tan30^\circ=\frac{50}{x} $$
$$ \frac1{\sqrt3}=\frac{50}{x} $$
$$ x=50\sqrt3 $$
Let height of tower be $h$.
Using angle of elevation:
$$ \tan60^\circ=\frac{h}{50\sqrt3} $$
$$ \sqrt3=\frac{h}{50\sqrt3} $$
$$ h=150 $$
Since:
$$ 150>120 $$
the tower satisfies radiation norms.
Answer
Height of tower:
$$ 150\text{ m} $$
Radiation norms:
$$ \text{Yes} $$
Let horizontal distance be $x$.
Using angle of depression:
$$ \tan30^\circ=\frac{66}{x} $$
$$ \frac1{1.732}=\frac{66}{x} $$
$$ x=114.31 $$
Let height of lamp post be $h$.
Using angle of elevation:
$$ \tan60^\circ=\frac{h-66}{114.31} $$
$$ 1.732=\frac{h-66}{114.31} $$
$$ h-66=198 $$
$$ h=264 $$
Answer
(i) Height of lamp post
$$ 264\text{ m} $$
(ii) Difference in heights
$$ 198\text{ m} $$
(iii) Distance between lamp post and apartment
$$ 114.31\text{ m} $$
Solution:
(i) Let vertical height between A and B be h1. tan20° = h1 / 8 ⇒ h1 = 8·tan20° = 8·0.3640 = 2.912 ≈ 2.91 km.
(ii) Let vertical height between B and C be h2. tan30° = h2 / 12 ⇒ h2 = 12·tan30° = 12·(1/√3) ≈ 12·0.57735 = 6.928 ≈ 6.93 km.
Answers: (i) 2.91 km, (ii) 6.93 km
1/(1 + tan²θ) = 1/sec²θ = cos²θ, so the value = sin²θ + cos²θ = 1. (B) 1.
tan θ(cosec²θ − 1) = tan θ·cot²θ = (sin θ/cos θ)(cos²θ/sin²θ) = cos θ/sin θ = cot θ. (D) cot θ.
Expanding: (sin²α + cos²α) + 4 + (cosec²α + sec²α) = 1 + 4 + (2 + cot²α + tan²α) = 7 + tan²α + cot²α. So k = 7. (B) 7.
a² − 1 = 2 sin θ cos θ, and b = (sin θ + cos θ)/(sin θ cos θ) = a/(sin θ cos θ). So b(a² − 1) = 2a. (A) 2a.
25x² − 25/y² = (5x)² − (5/y)² = sec²θ − tan²θ = 1. (B) 1.
sin θ = cos θ ⇒ θ = 45°. 2 tan²45° + sin²45° − 1 = 2(1) + 1/2 − 1 = 3/2. (B) 3/2.
tan²θ = x²/a² and sec²θ = y²/b². Since sec²θ − tan²θ = 1, we get y²/b² − x²/a² = 1. (A) y²/b² − x²/a² = 1.
Writing in terms of sin/cos and simplifying gives 2. (C) 2.
p² − q² = (cot²θ − cosec²θ)(a² − b²) = (−1)(a² − b²) = b² − a². (B) b² − a².
tan(elevation) = height/shadow = √3, so the angle = 60°. (D) 60°.
If d is the horizontal distance, height = d·tan30° = d/√3, and tan60° = b/d ⇒ d = b/√3. So height = (b/√3)(1/√3) = b/3. (B) b/3.
x = 60(cot30° − cot45°) = 60(√3 − 1) ≈ 60(0.732) = 43.92 m. (B) 43.92 m.
Let H be the height and d the distance. tan60° = H/d and tan30° = (H − 20)/d give H = 30 m and d = 10√3 m. (D) 30, 10√3.
Let the shorter height be a (taller = 2a). tan α = a/(x/2) and cot α = 2a/(x/2); multiplying gives x/(2a) = 4a/x ⇒ x² = 8a² ⇒ a = x/(2√2). (B) x/(2√2).
Taking the cloud height H above the lake with horizontal distance d: tan β = (H − h)/d and tan45° = (H + h)/d. Eliminating d gives H = h(1 + tan β)/(1 − tan β). (A) h(1 + tan β)/(1 − tan β).
We prove
$$ \cot^2 A\cdot\frac{\sec A-1}{1+\sin A}+\sec^2 A\cdot\frac{\sin A-1}{1+\sec A}=0 $$
Let the two terms be $T_1$ and $T_2$.
$$ T_1=\frac{\cos^2 A}{\sin^2 A}\cdot\frac{\dfrac{1-\cos A}{\cos A}}{1+\sin A} =\frac{\cos A(1-\cos A)}{\sin^2 A(1+\sin A)} $$
Since $\sin^2 A=(1-\cos A)(1+\cos A)$,
$$ T_1=\frac{\cos A}{(1+\cos A)(1+\sin A)} $$
Also
$$ T_2=\frac1{\cos^2 A}\cdot\frac{\sin A-1}{1+\dfrac1{\cos A}} =\frac{\sin A-1}{\cos A(1+\cos A)} $$
Adding,
$$ T_1+T_2=\frac1{1+\cos A}\left(\frac{\cos A}{1+\sin A}+\frac{\sin A-1}{\cos A}\right) $$
$$ =\frac1{1+\cos A}\cdot \frac{\cos^2 A+(\sin A-1)(1+\sin A)}{\cos A(1+\sin A)} $$
Now $(\sin A-1)(1+\sin A)=\sin^2 A-1=-\cos^2 A$, so the numerator is $0$.
Hence the given expression is $0$.
We prove
$$ \frac{\tan^2\theta-1}{\tan^2\theta+1}=1-2\cos^2\theta $$
Using $\tan^2\theta=\dfrac{\sin^2\theta}{\cos^2\theta}$,
$$ \frac{\tan^2\theta-1}{\tan^2\theta+1} =\frac{\dfrac{\sin^2\theta}{\cos^2\theta}-1}{\dfrac{\sin^2\theta}{\cos^2\theta}+1} =\frac{\sin^2\theta-\cos^2\theta}{\sin^2\theta+\cos^2\theta} $$
Since $\sin^2\theta+\cos^2\theta=1$,
$$ =\sin^2\theta-\cos^2\theta =(1-\cos^2\theta)-\cos^2\theta =1-2\cos^2\theta $$
Hence proved.
Let
$$ R=\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta} $$
We first show that
$$ R=\frac{1-\cos\theta}{\sin\theta} $$
Cross multiplication gives
$$ \sin\theta(1+\sin\theta-\cos\theta) =(1-\cos\theta)(1+\sin\theta+\cos\theta) $$
The right hand side is
$$ 1+\sin\theta-\sin\theta\cos\theta-\cos^2\theta =\sin^2\theta+\sin\theta-\sin\theta\cos\theta $$
$$ =\sin\theta(1+\sin\theta-\cos\theta) $$
So
$$ R=\frac{1-\cos\theta}{\sin\theta} $$
Now square both sides:
$$ R^2=\frac{(1-\cos\theta)^2}{\sin^2\theta} $$
Using $\sin^2\theta=1-\cos^2\theta=(1-\cos\theta)(1+\cos\theta)$,
$$ R^2=\frac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)} =\frac{1-\cos\theta}{1+\cos\theta} $$
Hence proved.
$$ x\sin^3\theta+y\cos^3\theta=\sin\theta\cos\theta $$
and
$$ x\sin\theta=y\cos\theta $$
then prove that
$$ x^2+y^2=1 $$
Proof
Given:
$$ x\sin\theta=y\cos\theta $$
$$ x=y\cot\theta $$
Substitute in
$$ x\sin^3\theta+y\cos^3\theta=\sin\theta\cos\theta $$
$$ y\cot\theta\sin^3\theta+y\cos^3\theta = \sin\theta\cos\theta $$
$$ y\sin^2\theta\cos\theta+y\cos^3\theta = \sin\theta\cos\theta $$
$$ y\cos\theta(\sin^2\theta+\cos^2\theta) = \sin\theta\cos\theta $$
$$ y\cos\theta=\sin\theta\cos\theta $$
$$ y=\sin\theta $$
and
$$ x=y\cot\theta $$
$$ x=\sin\theta\cdot\frac{\cos\theta}{\sin\theta} $$
$$ x=\cos\theta $$
Therefore,
$$ x^2+y^2 = \cos^2\theta+\sin^2\theta = 1 $$
Hence proved.
$$ a\cos\theta-b\sin\theta=c $$
then prove that
$$ a\sin\theta+b\cos\theta = \pm\sqrt{a^2+b^2-c^2} $$
Proof
Square the given equation:
$$ (a\cos\theta-b\sin\theta)^2=c^2 $$
$$ a^2\cos^2\theta+b^2\sin^2\theta -2ab\sin\theta\cos\theta = c^2 $$
Now consider
$$ (a\sin\theta+b\cos\theta)^2 $$
$$ = a^2\sin^2\theta+b^2\cos^2\theta +2ab\sin\theta\cos\theta $$
Adding the two expressions:
$$ (a\cos\theta-b\sin\theta)^2 + (a\sin\theta+b\cos\theta)^2 = a^2+b^2 $$
$$ c^2+(a\sin\theta+b\cos\theta)^2 = a^2+b^2 $$
$$ (a\sin\theta+b\cos\theta)^2 = a^2+b^2-c^2 $$
$$ a\sin\theta+b\cos\theta = \pm\sqrt{a^2+b^2-c^2} $$
Hence proved.
Initially,
$$ \tan45^\circ=\frac{80}{x} $$
$$ x=80 $$
After flying,
$$ \tan30^\circ=\frac{80}{y} $$
$$ \frac1{1.732}=\frac{80}{y} $$
$$ y=138.56 $$
Distance travelled:
$$ 138.56-80=58.56 $$
Time:
$$ 2\text{ s} $$
Speed:
$$ \frac{58.56}{2} = 29.28 $$
Answer
$$ 29.28\text{ m/s} $$
Let initial horizontal distance be $x$.
$$ \tan37^\circ=\frac{600}{x} $$
$$ 0.7536=\frac{600}{x} $$
$$ x=796.18 $$
Let final horizontal distance be $y$.
$$ \tan53^\circ=\frac{600}{y} $$
$$ 1.327=\frac{600}{y} $$
$$ y=452.15 $$
Distance travelled:
$$ 796.18-452.15 = 344.03 $$
Speed:
$$ 175\text{ m/s} $$
Time:
$$ \frac{344.03}{175} = 1.97 $$
Answer
$$ 1.97\text{ seconds (approx)} $$
(i)
North component:
$$ 30\sin55^\circ $$
$$ =30(0.8192) $$
$$ =24.58 $$
(ii)
West component:
$$ 30\cos55^\circ $$
$$ =30(0.5736) $$
$$ =17.21 $$
(iii)
North component:
$$ 32\cos48^\circ $$
$$ =32(0.6691) $$
$$ =21.41 $$
(iv)
East component:
$$ 32\sin48^\circ $$
$$ =32(0.7431) $$
$$ =23.78 $$
Answer
(i)
$$ 24.58\text{ km (approx)} $$
(ii)
$$ 17.21\text{ km (approx)} $$
(iii)
$$ 21.41\text{ km (approx)} $$
(iv)
$$ 23.78\text{ km (approx)} $$
Let height of lighthouse be $h$.
Distances from lighthouse:
$$ x=\frac{h}{\tan60^\circ} = \frac{h}{\sqrt3} $$
$$ y=\frac{h}{\tan45^\circ} = h $$
Total distance:
$$ \frac{h}{\sqrt3}+h = 200 $$
$$ h\left(\frac{1+\sqrt3}{\sqrt3}\right) = 200 $$
$$ h=\frac{200\sqrt3}{1+\sqrt3}\approx126.79 $$
Answer
$$ 126.79\text{ m} $$
Let height of building be $h$.
Using angle of depression:
$$ \tan34^\circ=\frac{h}{35} $$
$$ 0.6745=\frac{h}{35} $$
$$ h=23.61 $$
Let height of statue be $H$.
Using angle of elevation:
$$ \tan24^\circ=\frac{H-h}{35} $$
$$ 0.4452=\frac{H-23.61}{35} $$
$$ H-23.61=15.58 $$
$$ H=39.19 $$
Answer
$$ 39.19\text{ m} $$
Answer Key
| Q.No | Answer | |---|---| | 5 | $29.28\text{ m/s}$ | | 6 | $1.97\text{ seconds (approx)}$ | | 7(i) | $24.58\text{ km (approx)}$ | | 7(ii) | $17.21\text{ km (approx)}$ | | 7(iii) | $21.41\text{ km (approx)}$ | | 7(iv) | $23.78\text{ km (approx)}$ | | 8 | $200\text{ m}$ | | 9 | $39.19\text{ m}$ |
Ace Grade 10 Maths.
Revise this Samacheer Class 10 Maths topic, then continue with the Revision Challenge.