If sin
Question: If sin 1+ sin 2+ sin 3= 3, then write the value of cos 1+ cos 2+ cos 3. Solution: Sine function can take the maximum value of 1. If, $\sin \theta_{1}+\sin \theta_{2}+\sin \theta_{3}=3$, then we have: $\sin \theta_{1}=1$ $\Rightarrow \theta_{1}=\frac{\pi}{2}$ Similarly, $\theta_{2}=\theta_{3}=\frac{\pi}{2}$ $\Rightarrow \cos \theta_{1}=\cos \theta_{2}=\cos \theta_{3}=0$ $\Rightarrow \cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}=0$...
Read More →If sin
Question: If sin 1+ sin 2+ sin 3= 3, then write the value of cos 1+ cos 2+ cos 3. Solution: Sine function can take the maximum value of 1. If, $\sin \theta_{1}+\sin \theta_{2}+\sin \theta_{3}=3$, then we have: $\sin \theta_{1}=1$ $\Rightarrow \theta_{1}=\frac{\pi}{2}$ Similarly, $\theta_{2}=\theta_{3}=\frac{\pi}{2}$ $\Rightarrow \cos \theta_{1}=\cos \theta_{2}=\cos \theta_{3}=0$ $\Rightarrow \cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}=0$...
Read More →∆ABC is an isosceles triangle in which ∠C = 90. If AC = 6 cm, then AB =
Question: $\triangle \mathrm{ABC}$ is an isosceles triangle in which $\angle \mathrm{C}=90$. If $\mathrm{AC}=6 \mathrm{~cm}$, then $\mathrm{AB}=$ (a)62cm(b) 6 cm(c)26cm(d)42cm Solution: Given: In an isosceles $\triangle \mathrm{ABC}, \angle C=90^{\circ}, \mathrm{AC}=6 \mathrm{~cm}$. To find: AB In an isosceles $\triangle \mathrm{ABC}, \angle C=90^{\circ}$. Therefore, BC = AC = 6 cm Applying Pythagoras theorem in ΔABC, we get $\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}$ $\mathrm{AB}^{2}=6^{2...
Read More →If sin x + sin
Question: If sinx+ sin2x= 1, then write the value of cos8x+ 2 cos6x+ cos4x. Solution: We have: $\sin x+\sin ^{2} x=1$ $\Rightarrow \sin x=1-\sin ^{2} x$ $\Rightarrow \sin x=\cos ^{2} x$ ...(2) Now, taking square of $(1):$ $\Rightarrow\left(\sin x+\sin ^{2} x\right)^{2}=(1)^{2}$ $\Rightarrow(\sin x)^{2}+\left(\sin ^{2} x\right)^{2}+2(\sin x)\left(\sin ^{2} x\right)=1$ $\Rightarrow(\sin x)^{2}+(\sin x)^{4}+2(\sin x)^{3}=1$ $\Rightarrow(\sin x)^{2}+2(\sin x)^{3}+(\sin x)^{4}=1$ $\Rightarrow\left(\c...
Read More →Find the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm.
Question: Find the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm. Solution: It is given that: Slant height of cone (l) = 60 cm Radius of the base of the cone(r) = 21 cm Curved Surface Area (C. S. A) = ? C.S.A $=\pi r l=22 / 7 * 21 * 60=3960 \mathrm{~cm}^{2}$ Therefore the curved surface area of the right circular cone is $3960 \mathrm{~cm}^{2}$...
Read More →In an isosceles triangle ABC if AC = BC and AB2 = 2AC2, then ∠C =
Question: In an isosceles triangle $\mathrm{ABC}$ if $\mathrm{AC}=\mathrm{BC}$ and $\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$, then $\angle \mathrm{C}=$ (a) $30^{\circ}$ (b) $45^{\circ}$ (C) $90^{\circ}$ (d) $60^{\circ}$ Solution: Given: In Isosceles $\triangle \mathrm{ABC}, \mathrm{AC}=\mathrm{BC}$ and $\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$. To find: Measure of angle C In Isosceles ΔABC,AC = BCB=A (Equal sides have equal angles opposite to them) $\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$ $\mathrm{AB}^{2}=\mathr...
Read More →In a bank, principal increases continuously at the rate of
Question: In a bank, principal increases continuously at the rate of $r \%$ per year. Find the value of $r$ if Rs 100 doubles itself in 10 years (log $_{e} 2=0.6931$ ). Solution: Letp,t,andrrepresent the principal, time, and rate of interest respectively. It is given that the principal increases continuously at the rate ofr% per year. $\Rightarrow \frac{d p}{d t}=\left(\frac{r}{100}\right) p$ $\Rightarrow \frac{d p}{p}=\left(\frac{r}{100}\right) d t$ Integrating both sides, we get: $\int \frac{d...
Read More →If sin x + sin
Question: If sinx+ sin2x= 1, then write the value of cos12x+ 3 cos10x+ 3 cos8x+ cos6x. Solution: We have: $\sin x+\sin ^{2} x=1$ .....(1) $\Rightarrow \sin x=1-\sin ^{2} x$ $\Rightarrow \sin x=\cos ^{2} x$ ..(2) Now, taking cube of (1): $\sin x+\sin ^{2} x=1$ $\Rightarrow\left(\sin x+\sin ^{2} x\right)^{3}=(1)^{3}$ $\Rightarrow(\sin x)^{3}+\left(\sin ^{2} x\right)^{3}+3(\sin x)^{2}\left(\sin ^{2} x\right)+3(\sin x)\left(\sin ^{2} x\right)^{2}=1$ $\Rightarrow(\sin x)^{3}+(\sin x)^{6}+3(\sin x)^{4...
Read More →If ∆ABC ∼ ∆DEF such that AB = 9.1 cm and DE = 6.5 cm.
Question: If ∆ABC ∆DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of ∆DEF is 25 cm, then the perimeter of ∆ABC is (a) 36 cm(b) 30 cm(c) 34 cm(d) 35 cm Solution: Given: ΔABC is similar to ΔDEF such that AB= 9.1cm, DE = 6.5cm. Perimeter of ΔDEF is 25cm. To find: Perimeter of ΔABC. We know that the ratio of corresponding sides of similar triangles is equal to the ratio of their perimeters. Hence, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathr...
Read More →A well with 10 m inside diameter is dug 8.4 m deep.
Question: A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment. Solution: Given data is as follows: Inner diameter of the well = 10 m Height = 8.4 m Width of embankment = 7.5 m We have to find the height of the embankment. Given is the diameter of the well which is 10 m. Therefore, r = 5 m The outer radius of the embankment, R = Inner radius of the well + width of the embankm...
Read More →A well with 10 m inside diameter is dug 8.4 m deep.
Question: A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment. Solution: Given data is as follows: Inner diameter of the well = 10 m Height = 8.4 m Width of embankment = 7.5 m We have to find the height of the embankment. Given is the diameter of the well which is 10 m. Therefore, r = 5 m The outer radius of the embankment, R = Inner radius of the well + width of the embankm...
Read More →If sin x + cosec x = 2,
Question: If sinx+ cosecx= 2, then write the value of sinnx+ cosecnx. Solution: We have: $\sin x+\operatorname{cosec} x=2$ $\Rightarrow \sin x+\frac{1}{\sin x}=2$ $\Rightarrow \frac{\sin ^{2} x+1}{\sin x}=2$ $\Rightarrow \sin ^{2} x+1=2 \sin x$ $\Rightarrow \sin ^{2} x+1-2 \sin x=0$ $\Rightarrow(\sin x-1)^{2}=0$ $\Rightarrow \sin x-1=0$ $\Rightarrow \sin x=1$ And, $\operatorname{cosec} x=\frac{1}{\sin x}=1$ $\therefore \sin ^{n} x+\operatorname{cosec}^{n} x=1^{n}+1^{n}$ $=1+1=2$...
Read More →In an equilateral triangle ABC if AD ⊥ BC, then
Question: In an equilateral triangle ABC if AD BC, then (a) $5 \mathrm{AB}^{2}=4 \mathrm{AD}^{2}$ (b) $3 \mathrm{AB}^{2}=4 \mathrm{AD}^{2}$ (c) $4 \mathrm{AB}^{2}=3 \mathrm{AD}^{2}$ (d) $2 \mathrm{AB}^{2}=3 \mathrm{AD}^{2}$ Solution: $\triangle \mathrm{ABC}$ is an equilateral triangle and $\mathrm{AD} \perp \mathrm{BC}$. In ∆ABD,applying Pythagoras theorem, we get $\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$ $\mathrm{AB}^{2}=\mathrm{AD}^{2}+\left(\frac{1}{2} \mathrm{BC}\right)^{2}\left(\bec...
Read More →The volume of spherical balloon being inflated changes at a constant rate.
Question: The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon aftertseconds. Solution: Let the rate of change of the volume of the balloon bek(wherekis a constant). $\Rightarrow \frac{d v}{d t}=k$ $\Rightarrow \frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=k$ $\left[\right.$ Volume of sphere $\left.=\frac{4}{3} \pi r^{3}\right]$ $\Rightarrow \frac{4}{3} \pi \cdot 3 r^{2} \cdot...
Read More →Find the length of 13.2 kg of copper wire of diameter 4 mm,
Question: Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm. Solution: Given data is as follows: Weight of copperwire = 13.2 kg Diameter = 4 mm Density $=8.4 \mathrm{gm} / \mathrm{cm}^{3}$ We have to find the length of the copper wire. Given is the diameter of the wire which is 4 mm. Therefore, r = 2 mm Let us convert $\mathrm{r}$ from millimeter to centimeter, since density is in terms of $\mathrm{gm} / \mathrm{cm}^{3}$. Therefore, r = 210 cm Al...
Read More →If sin x = cos
Question: If sinx= cos2x, then write the value of cos2x(1 + cos2x). Solution: We have: $\sin x=\cos ^{2} x$ ...(1) $\therefore \cos ^{2} x\left(1+\cos ^{2} x\right)$ $=\sin x(1+\sin x) \quad[\operatorname{Using}(1)]$ $=\sin x+\sin ^{2} x$ $=\sin x+1-\cos ^{2} x$ $=\sin x+1-\sin x$ [Using (1)] = 1...
Read More →In an equilateral triangle $A B C$ if $A D perp B C$, then $A D^{2}=$
Question: In an equilateral triangle $A B C$ if $A D \perp B C$, then $A D^{2}=$ (a) $\mathrm{CD}^{2}$ (b) $2 \mathrm{CD}^{2}$ (c) $3 \mathrm{CD}^{2}$ (d) $4 \mathrm{CD}^{2}$ Solution: In an equilateral $\triangle \mathrm{ABC}, \mathrm{AD} \perp \mathrm{BC}$. In ΔADC, applying Pythagoras theorem, we get $\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}$ $\mathrm{BC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}(\because \mathrm{AC}=\mathrm{BC})$ $(2 \mathrm{DC})^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}(\because...
Read More →Write the maximum value of sin (cos x).
Question: Write the maximum value of sin (cosx). Solution: We have: $-1 \leq \cos x \leq 1$ Also, $\sin (-\theta)=-\sin \theta$ When the angle increases from 0 to $\frac{\pi}{2}$, the value of $\sin \theta$ also increases. $\therefore$ Maximum value of $\sin [\cos x]=\sin (1)$...
Read More →Write the maximum value of sin (cos x).
Question: Write the maximum value of sin (cosx). Solution: We have: $-1 \leq \cos x \leq 1$ Also, $\sin (-\theta)=-\sin \theta$ $W$ hen the angle increases from 0 to $\frac{\pi}{2}$, the value of $\sin \theta$ also increases. $\therefore$ Maximum value of $\sin [\cos x]=\sin (1)$...
Read More →If ∆ABC ∼ ∆DEF such that DE = 3 cm, EF = 2 cm,
Question: If ∆ABC ∆DEF such that DE = 3 cm, EF = 2 cm, DF = 2.5 cm, BC = 4 cm, then perimeter of ∆ABC is (a) 18 cm(b) 20 cm(c) 12 cm(d) 15 cm Solution: Given: ΔABC and ΔDEF are similar triangles such that DE = 3cm, EF = 2cm, DF = 2.5cm and BC = 4cm. To find: Perimeter of ΔABC. We know that if two triangles are similar then their corresponding sides are proportional. Hence, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}$ Substituting the values we...
Read More →Write the maximum and minimum values of sin (sin x).
Question: Write the maximum and minimum values of sin (sinx). Solution: We know : $-1 \leq \sin x \leq 1$ Also, $\sin (-\theta)=-\sin \theta$ When the angle increases from 0 to $\frac{\pi}{2}$, the value of $\sin \theta$ also increases. $\therefore$ Maximum value of $\sin [\sin (x)]=\sin (1)$ And, minimum value of $\sin [\sin (x)]=\sin (-1)=-\sin 1$...
Read More →Find the cost of sinking a tube well 280 m deep, having diameter 3 m at the rate of Rs 3.60 per cubic metre.
Question: Find the cost of sinking a tube well 280 m deep, having diameter 3 m at the rate of Rs 3.60 per cubic metre. Find also the cost of cementing its inner curved surface at Rs 2.50 per square metre. Solution: Given data is as follows: Height of the tube well = 280 m Diameter = 3 m Rate of sinking of the tube well $=$ Rs. $3.60 / \mathrm{m}^{3}$ Rate of cementing $=$ Rs. $2.50 / \mathrm{m}^{2}$ Given is the diameter of the tub well which is 3 metres. Therefore, r = 3/2 m Volume of the tube ...
Read More →At any point $(x, y)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point
Question: At any point $(x, y)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point $(-4,-3)$. Find the equation of the curve given that it passes through $(-2,1)$. Solution: It is given that (x,y) is the point of contact of the curve and its tangent. The slope $\left(m_{1}\right)$ of the line segment joining $(x, y)$ and $(-4,-3)$ is $\frac{y+3}{x+4}$. We know that the slope of the tangent to the curve is given by the relation, $...
Read More →Write the maximum and minimum values of cos (cos x).
Question: Write the maximum and minimum values of cos (cosx). Solution: We have; $-1 \leq \cos x \leq 1$ Also, $\cos (-\theta)=\cos \theta$ When the angle increases from 0 to $\frac{\pi}{2}$, the value of $\cos \theta$ decreases. $\therefore$ Maximum value of $\cos [\cos (x)]=\cos (0)=1^{\prime}$ And, minimum value of $\cos [\cos (x)]=c \cos (1)$...
Read More →If
Question: Ii $\frac{3 \pi}{4}x\pi$, then $\sqrt{\operatorname{cosec}^{2} x+2 \cot x}$ is equal to ____________ . Solution: Given $\frac{3 \pi}{4}x\pi$ $\sqrt{\operatorname{cosec}^{2} x+2 \cot x}$ $=\sqrt{2 \cot x+\cot ^{2} x+1}$ $=\sqrt{(\cot x+1)^{2}}$ $=|\cot x+1|$ Since $\frac{3 \pi}{4}x\pi$ i. e $\sqrt{\operatorname{cosec}^{2} x+2 \cot x}=-\cot x-1$...
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