The angle of elevation of a cloud from a point h metre above a lake is θ.
Question: The angle of elevation of a cloud from a point h metre above a lake is . The angle of depression of its reflection in the lake is 45. The height of the cloud is(a)htan (45 + )(b)hcot (45 )(c)htan (45 )(d)hcot (45 + ) Solution: Letbe the surface of the lake andbe the point of observation. So. The given situation can be represented as, Here,is the position of the cloud andis the reflection in the lake. Then. Letbe the perpendicular fromon. Thenand. Let,, thenand Here, we have to find the...
Read More →If the sums of n terms of two arithmetic progressions are in the ratio 2n + 5 :
Question: If the sums ofnterms of two arithmetic progressions are in the ratio 2n+ 5 : 3n+ 4, then write the ratio of theirmth terms. Solution: Given: $\frac{S_{n}}{S_{n}{ }^{1}}=\frac{2 n+5}{3 n+4}$ $\Rightarrow \frac{\frac{n}{2}\{2 a+(n-1) d\}}{\frac{n}{2}\left\{2 b+(n-1) d^{1}\right\}}=\frac{2 n+5}{3 n+4}$ $\Rightarrow \frac{2 a+(n-1) d}{2 b+(n-1) d^{1}}=\frac{2 n+5}{3 n+4}$ Ratio of their $\mathrm{m}$ terms $=\frac{a_{m}}{b_{m}}$ To find the ratio of the mthterms, replacenby 2m-1 in equation...
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Question: If $A=\left[\begin{array}{rr}2 -2 \\ 4 2 \\ -5 1\end{array}\right], B=\left[\begin{array}{rr}8 0 \\ 4 -2 \\ 3 6\end{array}\right]$, find matrix $X$ such that $2 A+3 X=5 B$. Solution: Given : $2 A+3 X=5 B$ $\Rightarrow 2\left[\begin{array}{cc}2 -2 \\ 4 2 \\ -5 1\end{array}\right]+3 X=5\left[\begin{array}{cc}8 0 \\ 4 -2 \\ 3 6\end{array}\right]$ $\Rightarrow\left[\begin{array}{cc}4 -4 \\ 8 4 \\ -10 2\end{array}\right]+3 X=\left[\begin{array}{cc}40 0 \\ 20 -10 \\ 15 30\end{array}\right]$ ...
Read More →P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of || gm ABCD.
Question: $P, Q, R, S$ are respectively the midpoints of the sides $A B, B C, C D$ and $D A$ of $\|$ gm $A B C D$. Show that $P Q R S$ is a parallelogram and also show that $\operatorname{ar}(\| \operatorname{gm} P Q R S)=\frac{1}{2} \times \operatorname{ar}(\| \operatorname{gm} A B C D)$. Solution: Given:ABCDis a parallelogram andP, Q, R and Sare the midpoints of sidesAB, BC, CD and DA, respectively. To prove: ar(parallelogram $P Q R S)=\frac{1}{2} \times$ ar (parallelogram $A B C D$ ) Proof: I...
Read More →If log 2, log
Question: If log 2, log (2x 1) and log (2x+ 3) are in A.P., write the value ofx. Solution: The numbers $\log 2, \log \left(2^{x}-1\right)$ and $\log \left(2^{x}+3\right)$ are in A.P. $\therefore \log \left(2^{x}-1\right)-\log 2=\log \left(2^{x}+3\right)-\log \left(2^{x}-1\right)$ $\Rightarrow \log \left(\frac{2^{x}-1}{2}\right)=\log \left(\frac{2^{x}+3}{2^{x}-1}\right)$ $\Rightarrow \frac{2^{x}-1}{2}=\frac{2^{x}+3}{2^{x}-1}$ $\Rightarrow\left(2^{x}-1\right)^{2}=2\left(2^{x}+3\right)$ $\Rightarro...
Read More →Two persons are a metres apart and the height of one is double that of the other.
Question: Two persons are a metres apart and the height of one is double that of the other. If from the middle point of the line joining their feet, an observer finds the angular elevation of their tops to be complementary, then the height of the shorter post is (a) $\frac{a}{4}$ (b) $\frac{a}{\sqrt{2}}$ (c) $a \sqrt{2}$ (d) $\frac{a}{2 \sqrt{2}}$ Solution: LetABandCDbe the two persons such thatABCD. Then, letAB=hso thatCD= 2h Now, the given information can be represented as, Here,Eis the midpoi...
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Question: If $A=\left[\begin{array}{ll}9 1 \\ 7 8\end{array}\right], B=\left[\begin{array}{cc}1 5 \\ 7 12\end{array}\right]$, find matrix $C$ such that $5 A+3 B+2 C$ is a null matrix. Solution: Given : $5 A+3 B+2 C=\left[\begin{array}{ll}0 0 \\ 0 0\end{array}\right]$ $\Rightarrow 5\left[\begin{array}{ll}9 1 \\ 7 8\end{array}\right]+3\left[\begin{array}{cc}1 5 \\ 7 12\end{array}\right]+2 C=\left[\begin{array}{ll}0 0 \\ 0 0\end{array}\right]$ $\Rightarrow\left[\begin{array}{cc}45 5 \\ 35 40\end{ar...
Read More →Find matrix
Question: Find matrix $A$, if $\left[\begin{array}{rrr}1 2 -1 \\ 0 4 9\end{array}\right]+A=\left[\begin{array}{rrr}9 -1 4 \\ -2 1 3\end{array}\right]$ Solution: Here, $A=\left[\begin{array}{ccc}9 -1 4 \\ -2 1 3\end{array}\right]-\left[\begin{array}{ccc}1 2 -1 \\ 0 4 9\end{array}\right]$ $\Rightarrow A=\left[\begin{array}{ccc}9-1 -1-2 4+1 \\ -2-0 1-4 3-9\end{array}\right]$ $\Rightarrow A=\left[\begin{array}{ccc}8 -3 5 \\ -2 -3 -6\end{array}\right]$...
Read More →If the sum of n terms of an AP is
Question: If the sum ofnterms of an AP is 2n2+ 3n, then write itsnth term. Solution: Given: $S_{n}=2 n^{2}+3 n$ $\Rightarrow S_{1}=2(1)^{2}+3(1)$ = 5 $S_{2}=2(2)^{2}+3(2)$ = 14 $\therefore a_{1}+a_{2}=14$ $\Rightarrow 5+a_{2}=14$ $\Rightarrow a_{2}=9$ Common difference, $d=a_{2}-a_{1}$ = 9-5 = 4 nth term $=a+(n-1) d$ $=5+(n-1) 4$ $=4 n+1$...
Read More →The height of a tower is 100 m. When the angle of elevation of the sun
Question: The height of a tower is 100 m. When the angle of elevation of the sun changes from 30 to 45, the shadow of the tower becomes x metres less. The value of x is (a) $100 \mathrm{~m}$ (b) $100 \sqrt{3} m$ (c) $100(\sqrt{3}-1) m$ (d) $\frac{100}{3} m$ Solution: The given situation can be represented as, Here, AB is the tower of heightmeters. When angle of elevation of sun changes from $\angle D=30^{\circ}$ to $\angle C=45^{\circ}, C D=x$. We assumed that $B C=y$ Here we have to find the va...
Read More →In a parallelogram ABCD, any point E is taken on the side BC.
Question: In a parallelogramABCD, any pointEis taken on the sideBC.AEandDCwhen produced meet at a pointM. Prove thatar(∆ADM) = ar(ABMC) Solution: Join BM and AC. $\operatorname{ar}(\Delta \mathrm{ADC})=\frac{1}{2} b h=\frac{1}{2} \times \mathrm{DC} \times h$ $\operatorname{ar}(\Delta \mathrm{ABM})=\frac{1}{2} \times \mathrm{AB} \times h$ AB = DC (Since ABCD is a parallelogram)So,ar(∆ADC) =ar(∆ABM)⇒ar(∆ADC) +ar(∆AMC) =ar(∆ABM) +ar(∆AMC)⇒ar(∆ADM) = ar(ABMC)Hence Proved...
Read More →Write the common difference of an A.P. the sum of whose first
Question: Write the common difference of an A.P. the sum of whose first $n$ terms is $\frac{p}{2} n^{2}+Q n$. Solution: Sum of the first $n$ terms of an A.P. $=\frac{p}{2} n^{2}+Q n$ Sum of one term of an A.P. $=S_{1}$ $\Rightarrow \frac{p}{2}(1)^{2}+Q(1)$ $\Rightarrow \frac{p}{2}+Q$ Sum of two terms of an A.P. $=S_{2}$ $\Rightarrow \frac{p}{2}(2)^{2}+Q(2)$ $\Rightarrow 2 p+2 Q$ Now, we have: $a_{1}+a_{2}=S_{2}$ $\Rightarrow \frac{p}{2}+Q+a_{2}=2 p+2 Q$ $\Rightarrow a_{2}=Q+\frac{3}{2} p$ Common...
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Question: If $X-Y=\left[\begin{array}{lll}1 1 1 \\ 1 1 0 \\ 1 0 0\end{array}\right]$ and $X+Y=\left[\begin{array}{rrr}3 5 1 \\ -1 1 4 \\ 11 8 0\end{array}\right]$, find $X$ and $Y$ Solution: Here, $X-Y+X+Y=\left[\begin{array}{lll}1 1 1 \\ 1 1 0 \\ 1 0 0\end{array}\right]+\left[\begin{array}{ccc}3 5 1 \\ -1 1 4 \\ 11 8 0\end{array}\right]$ $\Rightarrow 2 X=\left[\begin{array}{lll}1+3 1+5 1+1 \\ 1-1 1+1 0+4 \\ 1+11 0+8 0+0\end{array}\right]$ $\Rightarrow 2 X=\left[\begin{array}{lll}4 6 2 \\ 0 2 4 ...
Read More →ABCD is a parallelogram in which BC is produced to P such that CP = BC, as shown in the adjoining figure.
Question: ABCDis a parallelogram in whichBCis produced toPsuch thatCP=BC, as shown in the adjoining figure.APintersectsCDatM. If ar(DMB) = 7 cm2, find the area of parallelogramABCD. Solution: In $\triangle \mathrm{MDA}$ and $\triangle \mathrm{MCP}$, $\angle \mathrm{DMA}=\angle \mathrm{CMP}$ (Vertically opposite angles) $\angle \mathrm{MDA}=\angle \mathrm{MCP} \quad$ (Alternate interior angles) AD = CP (Since AD = CB and CB = CP) So, $\triangle \mathrm{MDA} \cong \triangle \mathrm{MCP} \quad$ (AS...
Read More →If the angle of elevation of a cloud from a point 200 m above a lake is 30°
Question: If the angle of elevation of a cloud from a point 200 m above a lake is 30 and the angle of depression of its reflection in the lake is 60, then the height of the cloud above the lake is(a) 200 m(b) 500 m(c) 30 m(d) 400 m Solution: Letbe the surface of the lake andbe the point of observation. Som. The given situation can be represented as, Here,is the position of the cloud andis the reflection in the lake. Then. Let $P M$ be the perpendicular from $P$ on $C B$. Then $\angle C P M=30^{\...
Read More →Write the common difference of an A.P.
Question: Write the common difference of an A.P. whosenth term isxn+y. Solution: We have: $a_{n}=x n+y$ $\therefore a_{1}=x+y$ $a_{2}=2 x+y$ Common difference of an A.P., $d=a_{2}-a_{1}$ $\Rightarrow(2 x+y)-(x+y)$ $\Rightarrow x$...
Read More →The angles of depression of two ships from the top of a light house are 45°
Question: The angles of depression of two ships from the top of a light house are 45 and 30 towards east. If the ships are 100 m apart. the height of the light house is (a) $\frac{50}{\sqrt{3+1}} m$ (b) $\frac{50}{\sqrt{3-1}} m$ (c) $50(\sqrt{3}-1) m$ (d) $50(\sqrt{3}+1) m$ Solution: Let=be the light house. The given situation can be represented as, It is clear thatand Again, letandm is given. Here, we have to find the height of light house. So we use trigonometric ratios. In a triangle, $\Right...
Read More →Find matrices
Question: Find matrices $X$ and $Y$, if $2 X-Y=\left[\begin{array}{rrr}6 -6 0 \\ -4 2 1\end{array}\right]$ and $X+2 Y=\left[\begin{array}{rrr}3 2 5 \\ -2 1 -7\end{array}\right]$ Solution: Given : $(2 X-Y)=\left[\begin{array}{ccc}6 -6 0 \\ -4 2 1\end{array}\right]$ ...(1) $(X+2 Y)=\left[\begin{array}{ccc}3 2 5 \\ -2 1 -7\end{array}\right]$ ...(2) Multiplying eq. (1) by eq. (2), we get $2(2 X-Y)=2\left[\begin{array}{ccc}6 -6 0 \\ -4 2 1\end{array}\right]$ $\Rightarrow 4 X-2 Y=\left[\begin{array}{c...
Read More →The sum of 4 arithmetic means between 3 and 23
Question: The sum of 4 arithmetic means between 3 and 23 is ___________. Solution: Leta= 3 b= 23 To find sum of 4 arithmetic term .............. heren= 4 Sum of 4 arithmetic terms is means between 3 and 23 $4\left(\frac{a+b}{2}\right)$ $=4\left(\frac{3+23}{2}\right)$ $=2(26)$ i.e sum of 4 arithmetic means between 3 and 23 = 52....
Read More →If the sum of n arithmetic means between 9 and 51 is 270,
Question: If the sum ofnarithmetic means between 9 and 51 is 270, then the value ofnis ___________. Solution: Leta= 9 b= 51 Given sum ofnarithmetic means between 9 and 51 is 270. LetA1,A2, ______Anbe arithmetic means betweenaandb. a,A1,A2, _____An,bare in A.P Here total number of terms isn+ 2 first term isalast term isb. $\therefore$ Sum of first $n+2$ terms $=\frac{(n+2)}{2}\{a+b\}$ $A_{1}+A_{2}+\ldots \ldots \ldots+A_{n}=\frac{n(a+b)}{2}$ i. e $270=n\left(\frac{9+51}{2}\right)$ i. e $270=n\lef...
Read More →In the adjoining figure, ∆ABC and ∆DBC are on the same base BC with A and D on opposite sides of BC such that ar(∆ABC) = ar(∆DBC).
Question: In the adjoining figure,∆ABCand ∆DBCare on the same baseBCwithAandDon opposite sides ofBCsuch that ar(∆ABC) = ar(∆DBC). Show thatBCbisectsAD. Solution: Given:∆ABCand ∆DBCare on the same baseBC.ar(∆ABC) = ar(∆DBC)To prove:BC bisects ADConstruction:Draw AL BC and DM BC.Proof:Since ∆ABCand∆DBCare on the same base BC and they have equal areas, their altitudes must be equal.i.e.,AL=DMLetADandBCintersect atO.Now, in∆ALOand∆DMO,we have:AL=DMALO=DMO= 90oAOL=DOM (Vertically opposite angles)i...
Read More →From the top of a cliff 25 m high the angle of elevation of a tower
Question: From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is(a) 25 m(b) 50 m(c) 75 m(d) 100 m Solution: Given that: height of cliff ism and angle of elevation of the tower is equal to angle of depression of foot of the tower that is. Now, the given situation can be represented as, Here,Dis the top of cliff andBEis the tower. Let $C E=h, A B=x$. Then, $A B=D C=x$ Here, we have to ...
Read More →Find
Question: Find $X$ if $Y=\left[\begin{array}{ll}3 2 \\ 1 4\end{array}\right]$ and $2 X+Y=\left[\begin{array}{rr}1 0 \\ -3 2\end{array}\right]$ Solution: Given : $2 X+Y=\left[\begin{array}{cc}1 0 \\ -3 2\end{array}\right]$ $\Rightarrow 2 X+\left[\begin{array}{ll}3 2 \\ 1 4\end{array}\right]=\left[\begin{array}{cc}1 0 \\ -3 2\end{array}\right]$ $\Rightarrow 2 X=\left[\begin{array}{cc}1 0 \\ -3 2\end{array}\right]-\left[\begin{array}{ll}3 2 \\ 1 4\end{array}\right]$ $\Rightarrow 2 X=\left[\begin{ar...
Read More →In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P.
Question: In the adjoining figure,ABCDis a quadrilateral. A line throughD, parallel toAC, meetsBCproduced inP. Prove that ar(∆ABP) = ar(quad.ABCD). Solution: We have:ar(quad.ABCD) =ar(∆ACD) +ar(∆ABC)ar(∆ABP) =ar(∆ACP) +ar(∆ABC)∆ACDand ∆ACPare on the same base and between the same parallelsACandDP. ar(∆ACD) =ar(∆ACP)By adding ar(∆ABC) on both sides, we get:ar(∆ACD)+ar(∆ABC) =ar(∆ACP) +ar(∆ABC) ⇒ ar (quad.ABCD) = ar(∆ABP)Hence, proved....
Read More →The sum of n arithmetic means between a and b
Question: The sum ofnarithmetic means betweenaandbis ___________. Solution: Letaandbbe two given numbers LetA1,A2, ______Anbe arithmetic means betweenaandb. a,A1,A2, _____An,bare in A.P Here total number of terms isn+ 2 first term isalast term isb. $\therefore$ Sum of first $n+2$ terms $=\frac{(n+2)}{2}\{a+b\}$ i. e $a+A_{1}+A_{2}+\ldots \ldots \ldots+A_{n}+b=\left(\frac{n+2}{2}\right)(a+b)$ i. e $A_{1}+A_{2}+\ldots \ldots+A_{n}=\frac{(n+2)(a+b)}{2}-a-b$ $=\frac{(n+2)(a+b)}{2}-(a+b)$ $=(a+b)\lef...
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