Assertion: If three angles of a quadrilateral are 130°, 70° and 60°, then the fourth angles is 100°.

Question: Assertion:If three angles of a quadrilateral are 130, 70 and 60, then the fourth angles is 100.Reason:The sum of all the angle of a quadrilateral is 360.(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.(c) Assertion is true and Reason is false.(d) Assertion is false and Reason is true. Solution: (a)Both Assertion and Reason are true and Reason is a correct ...

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Evaluate:

Question: Evaluate: $\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)$ Solution: $\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)=\cos \left\{\sin ^{-1}\left(\frac{3}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}+\frac{5}{13} \sqrt{1-\left(\frac{3}{5}\right)^{2}}\right)\right\}$ $\left[\because \sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right)\right]$ $=\cos \left\{\sin ^{-1}\left(\frac{3}{5} \times \frac{12}{13}+\frac{5}{13} \times \...

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An observer, 1.5 m tall, is 28.5 m away from a a tower 30 m high

Question: An observer, 1.5 m tall, is 28.5 m away from a a tower 30 m high. Determine the angle elevation of the top of the tower from his eye. Solution: Let $B E$ be the observer of $1.5 \mathrm{~m}$ tall. And $A D$ be the tower of height 30 . Here we have to find angle of elevation of the top of tower. Let $\angle A B C=\theta$ The corresponding figure is as follows $\ln \triangle A B C$, $\Rightarrow \quad \tan \theta=\frac{A C}{B C}$ $\Rightarrow \quad \tan \theta=\frac{28.5}{28.5}$ $\Righta...

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Solve the following

Question: If $S_{n}$ denotes the sum of first $n$ terms of an A.P. $a_{n}$ such that $\frac{S_{m}}{S_{n}}=\frac{m^{2}}{n^{2}}$, then $\frac{a_{m}}{a_{n}}=$ (a) $\frac{2 m+1}{2 n+1}$ (b) $\frac{2 m-1}{2 n-1}$ (c) $\frac{m-1}{n-1}$ (d) $\frac{m+1}{n+1}$ Solution: (d) $\frac{m+1}{n+1}$ $\frac{S_{m}}{S_{n}}=\frac{m^{2}}{n^{2}}$ $\Rightarrow \frac{\frac{m}{2}\{2 a+(m-1) d\}}{\frac{n}{2}\{2 a+(n-1) d\}}=\frac{m^{2}}{n^{2}}$ $\Rightarrow \frac{\{2 a+(m-1) d\}}{\{2 a+(n-1) d\}}=\frac{m}{n}$ $\Rightarrow...

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Is quad. ABCD a parallelogram?

Question: Is quad.ABCDa parallelogram?I. Its opposite sides are equal.II. Its opposite angles are equal.(a) if the question can be answered by one of the given statements alone and not by the other;(b) if the question can be answered by either statement alone;(c) if the question can be answered by both the statements together but not by any one of the two;(d) if the question cannot be answered by using both the statements together. Solution: Weknow that a quadrilateral is a parallelogram when ei...

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Is || gm ABCD a square?

Question: Is || gmABCDa square?I. Diagonals of || gmABCDare equal.II. Diagonals of || gmABCDintersect at right angles.(a) if the question can be answered by one of the given statements alone and not by the other;(b) if the question can be answered by either statement alone;(c) if the question can be answered by both the statements together but not by any one of the two;(d) if the question cannot be answered by using both the statements together. Solution: When the diagonals of a parallelogram ar...

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Sum the following series:

Question: Sum the following series: $\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{2}{9}+\tan ^{-1} \frac{4}{33}+\ldots+\tan ^{-1} \frac{2^{n-1}}{1+2^{2 n-1}}$ Solution: $\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{2}{9}+\tan ^{-1} \frac{4}{33}+\ldots+\tan ^{-1} \frac{2^{n-1}}{1+2^{2 n-1}}$ $\Rightarrow \tan ^{-1}\left(\frac{2-1}{1+2 \times 1}\right)+\tan ^{-1}\left(\frac{4-2}{1+4 \times 2}\right)+\tan ^{-1}\left(\frac{8-4}{1+8 \times 4}\right)+\ldots+\tan ^{-1}\left(\frac{2^{n}-2^{n-1}}{1+2^{n} \cdot 2^{...

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Is quadrilateral ABCD a rhombus?

Question: Is quadrilateralABCDa rhombus?I. Quad.ABCDis a || gm.II. DiagonalsACandBDare perpendicular to each other.(a) if the question can be answered by one of the given statements alone and not by the other;(b) if the question can be answered by either statement alone;(c) if the question can be answered by both the statements together but not by any one of the two;(d) if the question cannot be answered by using both the statements together. Solution: Clearly, I alone is not sufficient to answe...

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A tower subtends an angle α at a point A in the plane of its base

Question: A tower subtends an angle at a point A in the plane of its base and the angles of depression of the foot of the tower at a point b metres just above A is . Prove that the height of the tower is b tan cot . Solution: Letbe the height of tower. The towerCDsubtends an angleat a point. And the angle of depression of foot of tower at a point b meter just aboveis. Letand,. Here we have to prove height of tower is We have the corresponding figure as follows So we use trigonometric ratios. In,...

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In n A.M.'s are introduced between 3 and 17 such that the ratio of the last mean to the first mean is 3 : 1,

Question: InnA.M.'s are introduced between 3 and 17 such that the ratio of the last mean to the first mean is 3 : 1, then the value ofnis (a) 6 (b) 8 (c) 4 (d) none of these. Solution: (a) 6 Let $A_{1}, A_{2}, A_{3}, A_{4} \ldots A_{n}$ be the $n$ arithmetic means between 3 and 17 . Let $d$ be the common difference of the A.P. 3, $A_{1}, A_{2}, A_{3}, A_{4}, \ldots A_{n}$ and 17 . Then, we have: $d=\frac{17-3}{n+1}=\frac{14}{n+1}$ Now, $A_{1}=3+d=3+\frac{14}{n+1}=\frac{3 n+17}{n+1}$ And, $A_{n}=...

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Is quadrilateral ABCD a || gm?

Question: Is quadrilateralABCDa || gm?I. DiagonalsACandBDbisect each other.II. DiagonalsACandBDare equal.(a) if the question can be answered by one of the given statements alone and not by the other;(b) if the question can be answered by either statement alone;(c) if the question can be answered by both the statements together but not by any one of the two;(d) if the question cannot be answered by using both the statements together. Solution: We know that if the diagonals of a ​quadrilateral bis...

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A ladder rests against a wall at an angle α to the horizontal.

Question: A ladder rests against a wall at an angle to the horizontal. Its foot is pulled away from the wall through a distance a, so that it slides a distance b down the wall making an angle with the horizontal. Show that $\frac{a}{b}=\frac{\cos \alpha-\cos \beta}{\sin \beta-\sin \alpha}$ Solution: Letbe the ladder such that its topis on the walland bottomis on the ground. The ladder is pulled away from the wall through a distance a, so that its topslides and takes position. So $\angle O P Q=\a...

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In the adjoining figure, BDEF and AFDE are parallelograms.

Question: In the adjoining figure,BDEFandAFDEare parallelograms. IsAF=FB? Why or why not Solution: Given: ParallelogramsBDEFandAFDE.Since,BF=DE (Opposite sides of parallelogramBDEF) ...(i)And,AF=DE (Opposite sides of parallelogramAFDE) ...(ii)From (i) and (ii), we getAF = FB...

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In a quadrilateral PQRS, the diagonals PR and QS bisect each other.

Question: In a quadrilateralPQRS, the diagonalsPRandQSbisect each other. If Q= 56, determine R. Solution: Since, the diagonalsPRandQSof quadrilateralPQRSbisect each other. (Given)So,PQRSis a parallelogram. Now, $\angle Q+\angle R=180^{\circ}$ (Adjacent angles are supplementary.) $\Rightarrow 56^{\circ}+\angle R=180^{\circ}$ $\Rightarrow \angle R=180^{\circ}-56^{\circ}$ $\Rightarrow \angle R=124^{\circ}$...

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Solve the following equations for x:

Question: Solve the following equations forx: (i) $\tan ^{-1} 2 x+\tan ^{-1} 3 x=n \pi+\frac{3 \pi}{4}$ (ii) $\tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1} \frac{8}{31}$ (iii) $\tan ^{-1}(x-1)+\tan ^{-1} x \tan ^{-1}(x+1)=\tan ^{-1} 3 x$ (iv) $\tan ^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2} \tan ^{-1} x=0$, where $x0$ (v) $\cot ^{-1} x-\cot ^{-1}(x+2)=\frac{\pi}{12}, x0$ (vi) $\tan ^{-1}(x+2)+\tan ^{-1}(x-2)=\tan ^{-1}\left(\frac{8}{79}\right), x0$ (vii) $\tan ^{-1} \frac{x}{2}+\tan ^{-1} \frac{...

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If D and E are respectively the midpoints of the sides AB and BC of ∆ABC in which AB = 7.2 cm, BC = 9.8 cm and AC = 3.6

Question: IfDandEare respectively the midpoints of the sidesABandBCof ∆ABCin whichAB= 7.2 cm,BC= 9.8 cm andAC= 3.6 cm then determine the length ofDE. Solution: InΔ∆ABC,Since,DandEarerespectivelythemid-pointsofsidesABandBC. (Given) So, $D E=\frac{1}{2} A C$ (Uing mid-point theorem) ButAC= 3.6 cm (Given) $D E=\frac{1}{2}(3.6)$ or,DE=1.8cmHence, the length ofDEis 1.8 cm....

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PQ is a post of given height a, and AB is a tower at some distance.

Question: PQ is a post of given height a, and AB is a tower at some distance. If and are the angles of elevation of B, the top of the tower, at P and Q respectively. Find the height of the tower and its distance from the post. Solution: Let ABbe the tower of heightHand PQis a given post of heighta,andare angles of elevation of top of towerfromPandQ. Let PA =x.and BC =h. The corresponding figure is as follows In ∆QCB, $\Rightarrow \quad \tan \beta=\frac{h}{x}$ $\Rightarrow \quad x=\frac{h}{\tan \...

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What special name can be given to a quadrilateral whose all angles are equal?

Question: What special name can be given to a quadrilateral whose all angles are equal? Solution: Weknowthat,sumofallanglesinaquadrilateralis360.Let each angle of the quadrilateral bex.x+x+x+x=360⇒4x=360⇒x=90⇒ Allanglesofthe quadrilateralare90.Hence, given quadrilateral is a rectangle....

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Can we form a quadrilateral whose angles are 70°,

Question: Can we form a quadrilateral whose angles are 70, 115, 60 and 120? Give reasons for your answer. Solution: Since, the sum of all angles (i.e. 70 + 115 + 60 + 120 = 365).So, we cannot form a quadrilateral with these angles....

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All the angles of a quadrilateral can be obtuse. Is this statement true?

Question: All the angles of a quadrilateral can be obtuse. Is this statement true? Give reasons for your answer. Solution: No, the statement is false because if all angles are greater than 90, then the sum of four obtuse angles will be greater than 360....

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From an aeroplane vertically above a straight horizontal road,

Question: From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive mile stones on opposite sides of the aeroplane are observed to be and . Show that the height in miles of aeroplane above the road is given by $\frac{\tan \alpha \tan \beta}{\tan \alpha+\tan \beta}$ Solution: Let $h$ be the height of aero plane $P$ above the road. And $A$ and $B$ be the two consecutive milestone, then $A B=1$ mile. We have $\angle P A Q=\alpha$ and $\angle P B Q=\b...

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Sum of all two digit numbers which when divided by 4 yield unity as remainder is

Question: Sum of all two digit numbers which when divided by 4 yield unity as remainder is (a) 1200 (b) 1210 (c) 1250 (d) none of these. Solution: (b) 1210 The given series is 13, 17, 21....97. $a_{1}=13, a_{2}=17, a_{n}=97$ $d=a_{2}-a_{1}=7-3=4$ $a_{n}=97$ $\Rightarrow a+(n-1) d=97$ $\Rightarrow 13+(n-1) 4=97$ $\Rightarrow n=22$ Sum of the above series: $S_{22}=\frac{22}{2}\{2 \times 13+(22-1) 4\}$ $=11\{26+84\}$ $=1210$...

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All the angles of a quadrilateral can be right angles. Is this statement true?

Question: All the angles of a quadrilateral can be right angles. Is this statement true? Give reasons for your answer. Solution: Yes, the statement is true because all the angles of a quadrilateral such as rectangle and square are right angles....

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Sum of all two digit numbers which when divided by 4 yield unity as remainder is

Question: Sum of all two digit numbers which when divided by 4 yield unity as remainder is (a) 1200 (b) 1210 (c) 1250 (d) none of these. Solution: (b) 1210 The given series is 13, 17, 21....97. $a_{1}=13, a_{2}=17, a_{n}=97$ $d=a_{2}-a_{1}=7-3=4$ $a_{n}=97$ $\Rightarrow a+(n-1) d=97$ $\Rightarrow 13+(n-1) 4=97$ $\Rightarrow n=22$ Sum of the above series: $S_{22}=\frac{22}{2}\{2 \times 13+(22-1) 4\}$ $=11\{26+84\}$ $=1210$...

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All the angles of a quadrilateral can be acute. Is this statement true?

Question: All the angles of a quadrilateral can be acute. Is this statement true? Give reasons for your answer. Solution: No, the statement is false becauseif all the four angles of a quadrilateral are less than 90,then the sum of all four angles will be less than 360...

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