tan 20° + tan 40°
Question: $\tan 20^{\circ}+\tan 40^{\circ}+\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}$ is equal to (a) $\frac{\sqrt{3}}{4}$ (b) $\frac{\sqrt{3}}{2}$ (c) $\sqrt{3}$ (d) 1 Solution: (c) $\sqrt{3}$ $\tan 20^{\circ}+\tan 40^{\circ}+\sqrt{3} \tan 20^{\circ} \tan 40^{\circ}$ $=\tan 60^{\circ}\left(1-\tan 20^{\circ} \tan 40^{\circ}\right)+\tan 60^{\circ} \tan 20^{\circ} \tan 40^{\circ}$ $\left[\right.$ Using $\tan 60^{\circ}=\frac{\tan 20+\tan 40}{1-\tan 20 \tan 40}$ and $\left.\tan 60^{\circ}=\sqrt{3}\r...
Read More →Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.
Question: Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights. Solution: Given: Two isosceles triangles have equal vertical angles and their areas are in the ratio of 36:25. To find: Ratio of their corresponding heights. Suppose $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$ are two isosceles triangles with $\angle \mathrm{A}=\angle \mathrm{P}$. Now, AB = AC and PQ = PR $\therefore \mathrm{ABAC}=\mathrm...
Read More →Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.
Question: Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights. Solution: Given: Two isosceles triangles have equal vertical angles and their areas are in the ratio of 36:25. To find: Ratio of their corresponding heights. Suppose $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$ are two isosceles triangles with $\angle \mathrm{A}=\angle \mathrm{P}$. Now, AB = AC and PQ = PR $\therefore \mathrm{ABAC}=\mathrm...
Read More →If A + B + C = π, then sec A (cos B cos C − sin B sin C) is equal to
Question: IfA+B+C= , then secA(cosBcosC sinBsinC) is equal to (a) 0 (b) 1 (c) 1 (d) None of these Solution: (b) 1 $\pi=180^{\circ}$ $\sec A(\cos B \cos C-\sin B \sin C)=\frac{\cos B \cos (\pi-(A+B))-\sin B \sin (\pi-(A+B))}{\cos A}$ We know that, $\cos (\pi-\theta)=-\cos \theta$ and $\sin (\pi-\theta)=\sin \theta$, $\therefore \sec A(\cos B \cos C-\sin B \sin C)=\frac{\cos B \cos (A+B)-\sin B \sin (A+B)}{\cos A}$ Now, using the identities $\cos (A+B)=\cos A \cos B-\sin A \sin B$ and $\sin (A+B)=...
Read More →If A + B + C = π, then sec A (cos B cos C − sin B sin C) is equal to
Question: IfA+B+C= , then secA(cosBcosC sinBsinC) is equal to (a) 0 (b) 1 (c) 1 (d) None of these Solution: (b) 1 $\pi=180^{\circ}$ $\sec A(\cos B \cos C-\sin B \sin C)=\frac{\cos B \cos (\pi-(A+B))-\sin B \sin (\pi-(A+B))}{\cos A}$ We know that, $\cos (\pi-\theta)=-\cos \theta$ and $\sin (\pi-\theta)=\sin \theta$, $\therefore \sec A(\cos B \cos C-\sin B \sin C)=\frac{\cos B \cos (A+B)-\sin B \sin (A+B)}{\cos A}$ Now, using the identities $\cos (A+B)=\cos A \cos B-\sin A \sin B$ and $\sin (A+B)=...
Read More →For each of the differential equations given below, indicate its order and degree (if defined).
Question: For each of the differential equations given below, indicate its order and degree (if defined). (i) $\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x$ (ii) $\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x$ (iii) $\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0$ Solution: (i)The differential equation is given as: $\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x$ $\Rightarrow \frac{d^{2} y}{d x...
Read More →The areas of two similar triangles are 169 cm2 and 121 cm2 respectively.
Question: The areas of two similar triangles are $169 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$ respectively. If the longest side of the larger triangle is $26 \mathrm{~cm}$, find the longest side of the smaller triangle. Solution: Given: The area of two similar triangles is $169 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$ respectively. The longest side of the larger triangle is $26 \mathrm{~cm}$. To find: Longest side of the smaller triangle We know that the ratio of areas of two similar tr...
Read More →The value of sin
Question: The value of $\sin ^{2} \frac{5 \pi}{12}-\sin ^{2} \frac{\pi}{12}$ is (a) $\frac{1}{2}$ (b) $\frac{\sqrt{3}}{2}$ (c) 1 (d) 0 Solution: (b) $\frac{\sqrt{3}}{2}$ $\frac{5 \pi}{12}=75^{\circ}, \frac{\pi}{12}=15^{\circ}$ $\sin ^{2} 75^{\circ}-\sin ^{2} 15^{\circ}$ $=\sin ^{2} 75^{\circ}-\cos ^{2} 75^{\circ} \quad\left[\sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$ Now, $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)$ $=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin ...
Read More →The integrating factor of the differential equation.
Question: The integrating factor of the differential equation. $\left(1-y^{2}\right) \frac{d x}{d y}+y x=a y(-1y1)$ is A. $\frac{1}{y^{2}-1}$ B. $\frac{1}{\sqrt{y^{2}-1}}$ C. $\frac{1}{1-y^{2}}$ D. $\frac{1}{\sqrt{1-y^{2}}}$ Solution: The given differential equation is: $\left(1-y^{2}\right) \frac{d x}{d y}+y x=a y$ $\Rightarrow \frac{d y}{d x}+\frac{y x}{1-y^{2}}=\frac{a y}{1-y^{2}}$ This is a linear differential equation of the form: $\frac{d x}{d y}+p y=Q$ (where $p=\frac{y}{1-y^{2}}$ and $Q=...
Read More →The areas of two similar triangles are 81 cm2 and 49 cm2 respectively.
Question: The areas of two similar triangles are $81 \mathrm{~cm}^{2}$ and $49 \mathrm{~cm}^{2}$ respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians? Solution: Given: The area of two similar triangles is 81cm2and 49cm2respectively. To find: (1) Ratio of their corresponding heights. (2) Ratio of their corresponding medians. (1) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding ...
Read More →Prove that
Question: Prove that $(2 \sqrt{3}+3) \sin x+2 \sqrt{3} \cos x$ lies between $-(2 \sqrt{3}+\sqrt{15})$ and $(2 \sqrt{3}+\sqrt{15})$. Solution: Let $f(x)=(2 \sqrt{3}+3) \sin x+2 \sqrt{3} \cos x$ We know that, $-\sqrt{(2 \sqrt{3}+3)^{2}+(2 \sqrt{3})^{2}} \leq f(x) \leq \sqrt{(2 \sqrt{3}+3)^{2}+(2 \sqrt{3})^{2}}$ $\Rightarrow-\sqrt{12+9+12 \sqrt{3}+12} \leq f(x) \leq \sqrt{12+9+12 \sqrt{3}+12}$ $\Rightarrow-\sqrt{33+12 \sqrt{3}} \leq f(x) \leq \sqrt{33+12 \sqrt{3}}$ Disclaimer : Instead of $-(2 \sqr...
Read More →Show that sin
Question: Show that sin 100 sin 10 is positive. Solution: Let $f(\theta)=\sin 100^{\circ}-\sin 10^{\circ}$ Multiplying and dividing by $\sqrt{1^{2}+1^{2}}$, i.e. by $\sqrt{2}$, we get: $\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin 100^{\circ}-\frac{1}{\sqrt{2}} \sin 10^{\circ}\right)$ $=\sqrt{2}\left(\cos 45^{\circ} \sin \left(90^{\circ}+10^{\circ}\right)-\sin 45^{\circ} \sin 10^{\circ}\right)$ $=\sqrt{2}\left(\cos 45^{\circ} \cos 10^{\circ}-\sin 45^{\circ} \sin 10^{\circ}\right)$ $=\sqrt{2} \cos \left...
Read More →Reduce each of the following expressions to the sine and cosine of a single expression:
Question: Reduce each of the following expressions to the sine and cosine of a single expression: (i) $\sqrt{3} \sin x-\cos x$ (ii) cosx sinx (iii) 24 cosx+ 7 sinx Solution: (i) Let $f(x)=\sqrt{3} \sin x-\cos x$ Dividing and multiplying by $\sqrt{3+1}$, i. e. by 2 , we get : $f(x)=2\left(\frac{\sqrt{3}}{2} \sin x-\frac{1}{2} \cos x\right)$ $\Rightarrow f(x)=2\left(\cos \frac{\pi}{6} \sin x-\sin \frac{\pi}{6} \cos x\right)$ $\Rightarrow f(x)=2 \sin \left(x-\frac{\pi}{6}\right)$ Again, $f(x)=2\lef...
Read More →In the given figure, $ riangle mathrm{ACB} sim riangle mathrm{APQ}$.
Question: In the given figure, $\triangle \mathrm{ACB} \sim \triangle \mathrm{APQ}$. If $\mathrm{BC}=10 \mathrm{~cm}, \mathrm{PQ}=5 \mathrm{~cm}, \mathrm{BA}=6.5 \mathrm{~cm}$ and $\mathrm{AP}=2.8 \mathrm{~cm}$, find $\mathrm{CA}$ and $\mathrm{AQ}$. Also, find the area $(\triangle \mathrm{ACB}):$ area $(\triangle \mathrm{APQ})$. Solution: Given: ΔACB is similar to ΔAPQ. BC = 10 cm, PQ = 5cm, BA = 6.5cm and AP = 2.8 cm TO FIND: (1) CA and AQ (2) Area of ΔACB : Area of ΔAPQ (1) It is given that ΔA...
Read More →The front compound wall of a house is decorated by wooden spheres of diameter 21 cm,
Question: The front compound wall of a house is decorated by wooden spheres of diameter $21 \mathrm{~cm}$, placed on small supports as shown in the figure. Eight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius $1.5 \mathrm{~cm}$ and height $7 \mathrm{~cm}$ and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per $\mathrm{cm}^{2}$. Solution: Wooden sphere radius = 21/2 = 10.5 cm Surface area of a woode...
Read More →The integrating factor of the differential equation
Question: The integrating factor of the differential equation $x \frac{d y}{d x}-y=2 x^{2}$ is A. $e^{-x}$ B. $e^{-y}$ C. $\frac{1}{x}$ D. $x$ Solution: The given differential equation is: $x \frac{d y}{d x}-y=2 x^{2}$ $\Rightarrow \frac{d y}{d x}-\frac{y}{x}=2 x$ This is a linear differential equation of the form: $\frac{d y}{d x}+p y=Q\left(\right.$ where $p=-\frac{1}{x}$ and $\left.Q=2 x\right)$ $e^{\int p d x}$ $\therefore$ I.F $=e^{\int-\frac{1}{x} d x}=e^{-\log x}=e^{\log \left(x^{-1}\righ...
Read More →Triangles ABC and DEF are similar.
Question: Triangles ABC and DEF are similar. (i) If area $(\triangle \mathrm{ABC})=16 \mathrm{~cm}^{2}$, area $(\Delta \mathrm{DEF})=25 \mathrm{~cm}^{2}$ and $\mathrm{BC}=2.3 \mathrm{~cm}$, find $\mathrm{EF}$. (ii) If area $(\Delta \mathrm{ABC})=9 \mathrm{~cm}^{2}$, area $(\Delta \mathrm{DEF})=64 \mathrm{~cm}^{2}$ and $\mathrm{DE}=5.1 \mathrm{~cm}$, find $\mathrm{AB}$. (iii) If $A C=19 \mathrm{~cm}$ and $D F=8 \mathrm{~cm}$, find the ratio of the area of two triangles. (iv) If area $(\triangle \...
Read More →Find the maximum and minimum values of each of the following trigonometrical expressions:
Question: Find the maximum and minimum values of each of the following trigonometrical expressions: (i) 12 sinx 5 cosx (ii) 12 cosx+ 5 sinx+ 4 (iii) $5 \cos x+3 \sin \left(\frac{\pi}{6}-x\right)+4$ (iv) sinx cosx+ 1 Solution: (i) $\operatorname{Le} t f(x)=12 \sin x-5 \cos x$ We know that $-\sqrt{12^{2}+(-5)^{2}} \leq 12 \sin x-5 \cos x \leq \sqrt{12^{2}+(-5)^{2}}$ $-\sqrt{144+25} \leq 12 \sin x-5 \cos x \leq \sqrt{144+25}$ $-13 \leq 12 \sin x-5 \cos x \leq 13$ Hence the maximum and minumun value...
Read More →Find the equation of a curve passing through the point (0, 2)
Question: Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5. Solution: Let $F(x, y)$ be the curve and let $(x, y)$ be a point on the curve. The slope of the tangent to the curve at $(x, y)$ is $\frac{d y}{d x}$. According to the given information: $\frac{d y}{d x}+5=x+y$ $\Rightarrow \frac{d y}{d x}-y=x-5$ This is a linear differential equa...
Read More →A hemispherical dome of a building needs to be painted.
Question: A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is $17.6 \mathrm{~cm}$, find the cost of painting it, given the cost of painting is Rs.5 per $100 \mathrm{~cm}^{2}$ Solution: Given that only the rounded surface of the dome to be painted, we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done. Now, circumference of the dome = 17.6 cm Therefore2r = 17.6 2 22/7 r = 17.6 m So, ...
Read More →In the given figure, ∆ABC is right angled at C and DE ⊥ AB.
Question: In the given figure, $\triangle \mathrm{ABC}$ is right angled at $\mathrm{C}$ and $\mathrm{DE} \perp \mathrm{AB}$. Prove that $\triangle \mathrm{ABC} \sim \triangle \mathrm{ADE}$ and hence find the length of $\mathrm{AE}$ and $\mathrm{DE}$. Solution: It is given that $A C B$ is right angle triangle and $\angle C=90^{\circ}$ We have to prove that $\triangle A B C \sim \triangle A D E$ and find the lengths of $A E$ and $D E$. In $\triangle A B C \sim \triangle A D E$ $\angle \mathrm{A}=\...
Read More →If α and β are two solutions of the equation a tan x + b sec x = c,
Question: If and are two solutions of the equationatanx + bsecx=c, then find the values of sin ( + ) and cos ( + ). Solution: $a \tan x+b \sec x=c$ $\Rightarrow(c-a \tan x)=b \sec x$ $\Rightarrow(c-a \tan x)^{2}=(b \sec x)^{2}$ $\Rightarrow c^{2}+a^{2} \tan ^{2} x-2 a c \tan x=b^{2} \sec ^{2} x$ $\Rightarrow c^{2}+a^{2} \tan ^{2} x-2 a c \tan x=b^{2}\left(1+\tan ^{2} x\right)$ $\Rightarrow\left(a^{2}-b^{2}\right) \tan ^{2} x-2 a c \tan x+\left(c^{2}-b^{2}\right)=0$ This is a quadratic in $\tan x...
Read More →The diameter of the moon is approximately one-fourth of the diameter of the earth.
Question: The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas. Solution: Let the diameter of the earth be d Then, Diameter of moon will bed/4 Radius of earth = d/2 Radius of moon $=\frac{\frac{\mathrm{d}}{2}}{4}=\frac{\mathrm{d}}{8}$ Surface area of moon $=4 \pi(\mathrm{d} / 8)^{2}$ Surface area of earth $=4 \pi(d / 2)^{2}$ Required Ratio $=\frac{4 \pi\left(\frac{\mathrm{d}}{8}\right)^{2}}{4 \pi\left(\frac{\mathrm{d}}{2}\right)...
Read More →If tan
Question: If $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$, then show that $\sin \alpha+\cos \alpha=\sqrt{2} \cos \theta$.[NCERT EXEMPLER] Solution: $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$ Dividing numerator and denominator on the RHS by $\cos \alpha$, we get $\tan \theta=\frac{\frac{\sin \alpha}{\cos \alpha}-1}{\frac{\sin \alpha}{\cos \alpha}+1}$ $\Rightarrow \tan \theta=\frac{\tan \alpha-\tan \frac{\pi}{4}}{1+\tan \alpha \tan \frac{\pi}{4}}...
Read More →Find the equation of a curve passing through the origin given that the slope
Question: Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x,y) is equal to the sum of the coordinates of the point. Solution: LetF(x,y) be the curve passing through the origin. At point $(x, y)$, the slope of the curve will be $\frac{d y}{d x}$. According to the given information: $\frac{d y}{d x}=x+y$ $\Rightarrow \frac{d y}{d x}-y=x$ $\frac{d y}{d x}=x+y$ $\Rightarrow \frac{d y}{d x}-y=x$ This is a linear differential equat...
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