RS Aggarwal Class 7 Solutions Lines And Angles

Q1: Find the complement of each of the following angles:

(i) \( 35^{circ}\)

(ii) \( 47^{circ}\)

(iii) \( 60^{circ}\)

(iv) \( 73^{circ}\)

Sol:

(i) The given angle measures \(35^{circ}\).

Let the measure of its complement be \( x^{circ} \).

x + \(35^{circ}\) = \(90^{circ}\)

or, x = \( 90^{circ} – 35^{circ} = 55^{circ} \)

Hence, the complement of the given angle will be \( 55^{circ} \)

(ii) The given angle measures \( 47^{circ}\).

Let the measure of its complement be \( x^{circ} \).

x + \(47^{circ}\) = \(90^{circ}\)

or, x = \( 90^{circ} – 47^{circ} = 43^{circ} \)

Hence, the complement of the given angle will be \( 43^{circ} \)

(iii) The given angle measures \( 47^{circ}\).

Let the measure of its complement be \( x^{circ} \).

x + \( 60^{circ}\) = \(90^{circ}\)

or, x = \( 90^{circ} – 60^{circ} = 30^{circ} \)

Hence, the complement of the given angle will be \( 30^{circ} \)

(iv) The given angle measures \( 73^{circ}\).

Let the measure of its complement be \( x^{circ} \).

x + \( 73^{circ}\) = \(90^{circ}\)

or, x = \( 90^{circ} – 73^{circ} = 17^{circ} \)

Hence, the complement of the given angle will be \( 17^{circ} \)

Q2: Find the supplement of each of the following angles:

(i) \( 80 ^{circ}\)

(ii) \( 54 ^{circ}\)

(iii) \( 105 ^{circ}\)

(iv) \( 123 ^{circ}\)

Sol:

(i) The given angle measures \( 80 ^{circ}\).

Let the measure of its complement be \( x^{circ} \).

x + \( 80^{circ}\) = \(180^{circ}\)

or, x = \( 180^{circ} – 80^{circ} = 100^{circ} \)

Hence, the supplement of the given angle will be \( 100^{circ} \)

(ii) The given angle measures \( 54 ^{circ}\).

Let the measure of its complement be \( x^{circ} \).

x + \( 54^{circ}\) = \(180^{circ}\)

or, x = \( 180^{circ} – 54^{circ} = 126^{circ} \)

Hence, the supplement of the given angle will be \( 126^{circ} \)

(iii) The given angle measures \( 105 ^{circ}\).

Let the measure of its complement be \( x^{circ} \).

x + \( 105^{circ}\) = \(180^{circ}\)

or, x = \( 180^{circ} – 105^{circ} = 75^{circ} \)

Hence, the supplement of the given angle will be \( 75^{circ} \)

(iv) The given angle measures \( 123 ^{circ}\).

Let the measure of its complement be \( x^{circ} \).

x + \( 123^{circ}\) = \(180^{circ}\)

or, x = \( 180^{circ} – 123^{circ} = 57^{circ} \)

Hence, the supplement of the given angle will be \( 57^{circ} \)

Q3: Among two supplementary angles, the measure of the larger angle is \( 36 ^{circ}\) more than the measure of the smaller. Find their measures.

Sol:

Let the two supplementary angles be \( x^{circ} \). and \( (180 – x)^{circ} \).

Since it is given that the measure of the larger angle is \( 36 ^{circ} \) more than the smaller angle, let the larger angle be \( x^{circ} \).

Therefore, \((180 – x)^{circ} + 36^{circ} = x^{circ}\)

or 216 = 2x

Or x = 108

Thus larger angle = \(108^{circ}\)

Smaller angle = \( (108 – 36)^{circ}\)

= \( 72^{circ}\)

Q4:Find the angle which is equal to its supplement.

Sol:

Let the measure of the required angle be x.

Since it is its own supplement.

x + x = \(180^{circ}\)

or 2x = \(180^{circ}\)

or x = \( 90^{circ}\)

Therefore the required angle is \( 90^{circ}\).

Q5: Can two angles be supplementary if both of them are:

(i) acute?                          (ii) Obtuse?                          (iii) right?

Sol:

(i) No. If both the angles are acute i.e. less than \( 90^{circ}\), they cannot be supplement as their supplement will always be less than \(180 ^{circ}\).

(ii)  No. If both the angles are obtuse i.e. more than \( 90^{circ}\), they cannot be supplement as their sum will always be more than \(180 ^{circ}\).

(iii) Yes. If both the angles are right i.e. they both measure \( 90^{circ}\), then they form a supplementary pair.

\( 90^{circ} + 90^{circ} = 180^{circ}\)

Q6: In the given figure, AOB is a straight line and the ray OC stands on it. If \(angle AOC = 64^{circ}\) and \(angle BOC = x^{circ}\), find the value of x.

https://lh4.googleusercontent.com/3iYLqXrCQCwb5K9DNZPkN0F15ze7n4BNcfJK8ZOWItW3fLQJPSZ59nxS16HsvyAyrM6Ny0s6D4es16fnTmrFP4Iq__6Tkng-rMPHfIzwpcD6Oj4iE5mar9ETuM-gLYVQ10-1GnL5

Sol:

By linear pair property,

\(angle AOC + angle COB = 180^{circ}\)

\(64^{circ} + angle COB = 180^{circ}\)

\(angle COB = 180^{circ} – 64^{circ} = 116^{circ}\)

Therefore x = \( 116^{circ} \)

https://lh3.googleusercontent.com/ip5jPqMiQ5U7_jzSXA5Oxfx5JYxJ1CtincGOBl-2WAdEC5PVDt8s2qa8aErPFPMOTR_qHJ-p57UXF2Vw9RS9_CUtge6rTQI0YzhtUNolpQgRnu6XyrsnEChSKPAN_tmoIHFxCBUP

Q7: In the given figure, AOB is a straight line and the ray OC stands on it. If \(angle AOC = (2 x – 10)^{circ}\) and \(angle BOC = (3 x + 20)^{circ}\), find the value of x. Also, find \(angle AOC ;; and ;; angle BOC\).

https://lh5.googleusercontent.com/n-WjrsP8qArqyCNNfWDrnR41JP4U0aSOndN65WsyX_ssG2lLDwyoIxjufFBRzjiPnKDC4P6TgKVyZlc2yAmtFBbbBIUvronBPiy_PECkWHAd3H7Q1SA1YQcf7sdfeHVnzantQT66

Sol:

By linear property:

\(angle AOC + angle BOC = 180^{circ}\)

\((2 x – 10)^{circ} + (3x + 20) ^{circ}  = 180^{circ}\)            (Given)

Or 5x + 10 = 180

Or x = \( 34^{circ}\)

Therefore \( angle AOC = (2 x – 10) ^{circ} = (2 times 34 – 10) ^{circ} = 58^{circ} \)

\( angle BOC = (3 x + 20) ^{circ} = (3 times 34 + 20) ^{circ} = 122^{circ} \)

https://lh5.googleusercontent.com/274hC8McEQr_K1hqc1UH3FBchPpO2YkahZ9CjfKeMTOvK2hwlQeafaZ09LAiicXeHe5QDZceUb-aaaHArb1uVDvoI_rMLcn1rehOBpGdx5iNy-A53GQ8M-lREQ8GAw3jCN3ID_AK

Q8: In the given figure, AOB is a straight line and the ray OC and OD stand on it. If \(angle AOC = 65^{circ}\) , \(angle BOC = 70^{circ}\) and \(angle COD = x^{circ}\), find the value of x.

https://lh4.googleusercontent.com/WDHk_l9VAk7coWNIjxk1QmFN7aeSAIJXW3OI7XJ2HKlo8_h5t_DsN-Zyy9sRe-YDUansdGA0ig3s0K27FM3qaaOATEperuO7Bey4e-biXr4p4Oj5LmPwxfKp8TrMcys2ZJUXeTEu

Sol:

Since AOB is a straight line, we have:

\(angle AOC + angle BOD + angle COD = 180^{circ}\)

or, \(65^{circ} + 70 ^{circ} + x ^{circ} = 180^{circ}\)         (Given)

Or, \(135 ^{circ} + x ^{circ} = 180^{circ}\)

Or, \( x ^{circ} = 45^{circ}\)

Thus, the value of x is \( 45^{circ} \)

https://lh4.googleusercontent.com/ac5dY9mQMTFXZcb9t_-WZJHHDhwNTpz-j_bPJSW1Igtx6NK79121qbNF5Tkc8D4Dj24SPYwP3V4IYaMYTocZ2VVVcr7Z_QvblulfDfPl6wmoXxdZBKCftbN_NN-cqDQ77XJ1HlAy

Q9: In the given figure, two straight lines AB and CD intersect at a point O.

If \(angle AOC = 42^{circ}\), find the measure of each of the angles:

(i)  \(angle AOD \)

(ii) \(angle BOD \)

(iii) \(angle COB \)

Sol:

https://lh4.googleusercontent.com/3oklkxHAshsD0D3CBuz94pByuNCxzS9uXTLn1hnvdnsWfCjcfpULZiPp1ClmvoEMEVGBDU2EeUp_bd6kEJHBsyAZjACZRe9AMwLvBiUCyEMC70A_hc0-ZAx38S0qXQppwMtc23Tl

Sol:

AB and CD intersect at O and CD is a straight line.

(i) \(angle COA + angle AOD = 180^{circ}\)       (Linear pair)

\(42^{circ} + angle AOD = 180^{circ}\)

\( angle AOD = 138^{circ}\)

(ii) \(angle COA ;; and ;;  angle BOD \) are vertically opposite angles.

Therefore, \( angle COA = angle BOD = 42^{circ}\)    ( from (i) )

(iii) \(angle COB ;; and ;;  angle AOD \) are vertically opposite angles.

Therefore, \(angle COB = angle AOD = 138^{circ}\)    ( from (i) )

Q10: In the given figure, two straight lines PQ and RS intersect at O. If \( angle POS = 114^{circ}\), find the measure of each of the angles:

(i)  \(angle POR \)

(ii) \(angle ROQ \)

(iii) \(angle QOS \)

                  https://lh4.googleusercontent.com/KnlIi6iPU26UtdE4U8udiC3DAF219stW0E0jRONYUr6k5MiTVFpTUbYbMJl_wR0OMuSu_mcc109sN4jbFUVClNw4FcKfvUe3KPAPnt4uVD1oBNY1ldOTxkQXIgS_X7ldsJ28eSjB

Sol:

(i) \( angle POS + angle PQR = 180^{circ}\)      (Linear pair)

Or \(114^{circ} + angle PQR = 180^{circ}\)

Or \(angle PQR = 180^{circ} – 114^{circ} = 66^{circ}\)

(ii) Since \( angle POS ;; and ;; angle QOR \) are vertically opposite angles, they are equal.

Therefore, \(angle QOR = 114^{circ}\)

(iii) Since \( angle POR ;; and ;; angle QOS \) are vertically opposite angles, they are equal.

Therefore, \(angle QOS = 66^{circ}\)

https://lh5.googleusercontent.com/obbAAvrZgiB0orVNsdOLelme_hRvERRUJcJR0EJonVcbz4PVMJqug92lskbcIGsxE5ecPH8g7hitN_bdslRSu5FlobWQUGS88TvExczLF9QW7bf4MT6R3wVp5FWFR7fK8CLxQ8Qx

Q11: In the given figure, ray OA, OB, OC and OD are such that \(angle AOB = 56^{circ}\) , \(angle BOC = 100^{circ}\), \(angle COD = x^{circ}\) and \(angle DOA = 74^{circ}\), find the value of x.

https://lh3.googleusercontent.com/a1yIFeZKRKW9c0yzYMvx2FHTnjWSZyKAUwc1h79OYZN6E6Y9_dFot-y9-8VrGcgH481CQkTKwNxdwSu7A-J-NO4xnva6FDTAGpz61QIrX6O7Rqj2oZVXuXwF8XBxjTIigYnu4BSF

Sol:

https://lh6.googleusercontent.com/8FFWZ7uTrCvti5YIY3Vk7W1uibqIue1G3mmTpSCklhG99uYzT4IbDABRWpPKmMgTDdzSv_tXdc0egCBo3tqxRMuR_4f8uwYzUa9Ly_Cg6Xrx8XgXXWUTwHoICbIDfYCoJ2Fp3yBQ

Sum of all the angles around a point is \(360^{circ}\)

Therefore, \(angle AOB + angle BOC + angle COD + angle DOA = 360^{circ}\)

or, \(56^{circ} + 100^{circ} + x^{circ} + 74^{circ} = 360^{circ}\)      (Given)

or, \(230^{circ} + x^{circ} = 360^{circ}\)

or, \(x^{circ} = 360^{circ} – 230^{circ}\)

or, x = \(130^{circ}\)


Practise This Question

How many natural numbers are there between 310 to 710?