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Unit 9

Some Applications of Trigonometry

Exercise 9.1 

Q1 :

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30 °.l

Answer :

It can be observed from the figure that AB is the pole. In ΔABC,
KxYm0j T0WQPNauSdVOrzIWIzWTEm96aaVx hqjQf8uoNLSPl 7IszoX6u5Kq6lh 7TcTP2jaaFv4AN1OAPl Q iz2xNy5 UldqtfLl59wPp3IVNf u5jID 8rgy 6S7XqdBN8o

Therefore, the height of the pole is 10 m.

Q2 :

A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle 30 ° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer :

qqCMb6qi5XMAY1JZVsGeTiGjGa7jKIZN74oQ 1opXd k8ts1q2vw Bf5ndX94a2LVmqvTHmyb sLSnUsGfRbHcUIRDcoFVfxKsJuTbVDDVxRmFP4ghOrKOyiELlcczH7x72i f8

Let AC was the original tree. Due to storm, it was broken into two parts. The broken part   is making 30° with the ground.

In ,ltg3lr NRROjGf73NfS9TiEY7ViEVxsEV7I8kMyZH Ay5CGklVo6s89GHa3iMlplBxODprba3RPhWQE83CHEMlM1GyipDRQsJ qvzwFMy4kEj x 99Kq9 aNOwUEfU2IM djbM

Height of tree =  HjXjUXyyo6QPsbZWir2PNKxjgdzf4B XRUEv4oalooiu2MZ3fGeqk6qNjb65mwh6D vb3U 7yX o Ij aWdo9NJbJn F Zd 6ZSNGg95HOEQMOE3Q5qeK9yXCU3bt6WpkyB4Lbs+ BC

u Z3Srhwgz9U1LLMO8MLbbryeBWSGblm7xqV0j4Vr4fnsbZAXOFpDWR07lBVpYP1ZVVRwklbfXwe1tfsnAaIAfKar

Hence, the height of the tree is LDKJ09xvYLV JpHHTvmE9TEELlilhpjhGo29MrPba840oZcaXfcEUpTUWZgRtiGDOwUzQ8Zj7c.

Q3 :

A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30 ° to the ground, where as for the elder children she wants to have a steep side at a height of 3 m, and inclined at an angle of 60 ° to the ground. What should be the length of the slide in each case?

Answer :

It can be observed that AC and PR are the slides for younger and elder children respectively.

In ΔABC,

UyvE815RUSkpCzhVS6XtOInX1bchXiXCfc6qtZjyuVSTr4V BPvQjhKBfYx e4VUjsngnbzVb9t57bf70XB2OzoyMGeXUHzagGFNGTslSX jIPD05bOjpxp
fA8k0NldVM XN3e7wTHpmLizUvFNLo0tr1ibmTtkZiAmXP4VWybZGlr0LaLivzbE4cWsqZnnBCQfk

In ΔPQR,rDGa88zTCoEyJTtReyIp MGMpRj6WPdl1TSw5J5EL s4Sza z

Therefore, the lengths of these slides are 3 m and jrauokydrIMCq8CUm1pp7jRXCUY2VaPHCC4uQQy0 8CiNhyz9ffhMkxyFHALwu9j3Uf4q0 5ycneWYmlMmJxmz OX7 mST8tIrFkPn8clW7eDSS56Xg65TivLk m3fVGcRJClSU.

Q4 :

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.

Answer :

Let AB be the tower and the angle of elevation from point C (on ground) is 30°.

In ΔABC,kSTtr0k6 BhVVZPSTbcMuo7 Y8OpIOqzwSNYSHAWn BbfnIF0ICBk0SCRJIGAc8m6w2tHNEIEHG82vWNXlIzh8bAqWFUuhJMXpffhiGISKGWPDsH5fuVoaH2RaHG8aNHds1PeE4

Therefore, the height of the tower is LrSCtL7t18GHie WBNpU6 q4mxF iSNl6Lgiv7OWyrirmbI523Ew0CHIT FbGmv66dbTeGLrCTHyuuA9UeOSXKDQdNmY7zz552PAyCTUfr3Bm5EBeI7bDYiPwNdF9gvtUp qOBg.

*Q5 :

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer :VTkjWeW382zZ37WZvXe8hrm rsdzfVgb09raG3NWP6KK9MVrMJJJtGahAFxI9ss7tBrfmV7 AwMXhEF UEKMmNRPhbLExGIhoa307oAWjl8 OoKjgGTAGY 7qvlnMfbPm 9q Y

Let K be the kite and the string is tied to point P on the ground. In ΔKLP,

au28fLoCPZDn7dW6v93YjqS0EWWKfbyHFMecdXdk3AROnBj zBXH nUuRBwJoTbvEYCTiuDGkc7a4fqS1CD1Af9YO9iC5ZeaKHmzHCE3 Qr1chpsjw7bSwjjCTwOGsqJbmWhpeY

Hence, the length of the string is    .d3x1c4aJnPvbH47uhk7Qr8 LndUXYzgVv

Q6 :

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer :mYSeqMeT2uw8tYQESW4Vvo ELgJ0GSvSnkBYr S0EuMb6r6SoZRRvLcQlmB0UYhv2QBuykqKz66mu8D0rjAUh4kqKBglvOlIvwIoAAk6Kno4rDz7w7IgrUsSksYnJ q5fbIfGMw

Let the boy was standing at point S initially. He walked towards the building and reached at point T. It can be observed that

PR = PQ – RQ

= (30 – 1.5) m = 28.5 m = ycKSZJsPzC9QyoTIpQ71JYC NlNgFaBh spFJ5ImSSpwwBFhjc3XrrlFum5roI5Ku hrLllTCcNbnMVJvD3QC9Qf7783ha4CRcuPC48Sx0Jk h8DNsPhF CeEc8Yc 5vdIbe2bk In ΔPAR,

In ΔPRB,
1OTmWoTAYC A0Vh8qNLc89ChBpFzVU6Dv1 dpIe t1jGjavqHVDMgZv3lIBluiEe02juzGY9E56VNvZAofm2oMjgXY0SjVlRqmO6kTiVyRFtzWRZTsx4b0C N660VmDQKnbySc

kFVRhCnMgnZXVS4C9JSWO7xMubK5YhmKQp6 m19AcHsj8JDcLSlJ gimTFpeN

ST = ABGIeaEh42l0PqFHdNAqnoGBDrgjmtwMGthFDd UZoBoQCuE97gSzz30kPijjxU8Ojm g8g4AgHmt2MgqIl0yE6sF9Z2 etEX9Jk47ZjLyx3Q4MMREuNJsjRSU4e4XIP2vBoWz OU

Hence, he walked  6bXr2Fer3ySmHMXPG3r5SUNxF2zYxxYRnuPQXG3JxeKaGGXoskxQeQc6HO3aU9jhJTRgbJqmGQl YS6PuaXmXS58lGJAQQD56MQhSetimeMqJ1p4cOssaUopYz qwNog3bWPa Qtowards the building.

*Q7 :

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer :iAbGfvexgeMcf5lDAabWdOyv0lHGfW4JTxe HaVN Vz32YI06iKWr4QZFuDNxPib3UIdyaMW2PvcsLOF D3MAa3hv18v0FG hD8pEqellLDN1M4

Let BC be the building, AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured.

In ΔBCD,

In ΔACD,

tylnmmxW4z0gi oZmd otKnZWKd86u9lovpCmri8pzS4GP1MosaKKzxmoNktxW7l 2pW gDh7oEO63e1D6pekI Hk1r1SYcWKs1 4on

7

Therefore, the height of the transmission tower is    m.

Q8 :

A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45 °. Find the height of the pedestal.

Answer :
AfnLoGVDEmmI80VgRw V4ewynRcfIpjkuLkP1Ye8BGtJUbYg4e6GbNaGePNgicCp8Fxzf9mk5wTsqxoSFMcHn7EV9KbfG42734s6ksa eMjCQluUBwLCBGgqTm031roGI4cN5jA

Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.

In ΔBCD,TqOve6UvsazVBujSEScXg tXnxdNkRe5e7ko4TvyvxxVk0Ekm66AHg

In ΔACD,

Wbyhef974CWoSC5mLwIGA i5grrQkCrraAV

c6UuNr37aq9 w4bd9DHjp KPk avm 7vmT9BzyKgJHtrE54REqWobSprOeZDMnu W248Zgv9hyHEBVHtQS8KcaAsa0Cfhktc4l7Q NQWL23ikkAHAKEZhb2NDonKoaZa3eZRAFA

Therefore, the height of the pedestal is 0.8    m.

*Q9 :

The angle of elevation of the top of a building from the foot of the tower is

30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer :
EJgARK2fa2ZUVuGARHhx8 EXmFiVd5fQ0jwE3oxA2TVnXk1jlTdgZHztThi0g7TmCPc4teIo2chvLY62QbWvR7GAupW p8804w7JDzX5EMleGurntkzq vcYD3rfHfo0t4af8Wg

Let AB be the building and CD be the tower.

In ΔCDB,

In ΔABD,6kYiu2eqnteDhgN910NV uy OLlvasfs0s7He01hMhfZhAtGXSP8 UThzyQvh84sDNykIZEHZnOPFZhFLtgvJDEgEicjgZzGaOLqC0 YE1qcSg6usSBlJAwMx2dGXJMb I1yTb8

Therefore, the height of the building is V1 8lw8EIFAUUn5Z6fTlH4V3ovD6P FOEglV.

Q10 :

Two poles of equal heights are standing opposite each other an either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30º, respectively. Find the height of poles and the distance of the point from the poles.

Answer :

Let AB and CD be the poles and O is the point from where the elevation angles are measured. In ΔABO,kxS239u6QDnYVztX089EZptO2LXIrruLQ4Not1t2vw4cG1cIBE6UarKRWVvV7LtDd Q36LxHrXGVsOMFXamvS5jmKQK6H0ODxSUnb0 a1V0EKUoVKGGqv dQaaCz0stnolH3zs8

0NX IB7hBI8H w12zCYcuntfP7LG vBYFJdbbaH5o7j

In ΔCDO,

NlyF kQMWzJt0 HAj6PHTjynZ6O4SjgfY1lkHxhcwT8wiqwupHAKxufkaeryX N uVNUu0 cau80u4LuDSVoqOysrCDLtySpt1UOFo7IJ79jgyq21b8Qb2Snk 4kk MSDgSjSw

Since the poles are of equal heights, CD = AB
dSGLIyL8d9JamyKlUfMPGHOfCVqUfgHR4Uj 8 xqJm18GMQ afZ8Pf27PHFPrzlKhSwTaa

DO = BD – BO = (80 – 20) m = 60 m

Therefore, the height of poles is mspNqMi9V4bUbgIfusN6ptxPg9hi3tshMLLB5fIbdhcfKAi3GjjaFHlcand the point is 20 m and 60 m far from these poles.

Q11 :

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

BCXNcvAMLeH8BWZ2aOv28hLWH02Y0qBb0RlA9cTsocMM0LsvrVaDqpVQZMMmMEQFdqTVUlvIouZ94b8 lloSD ZlSFU9i84ao41cZSjagqwrVym0NWc398uO5 zev9EnsQWC6ys

Answer :

In ΔABC,lGeojRAz5qNO39mm5mR72UvkyythVtsGiCwJhPlk9rc0Kwb3J bFJM1FZiLeAPDw3Mt

In ΔABD,

9QStXyfhGSZodfwvY915ZHY6kxS7Kz6BuylIO9sxn92mkYVRSHrFpRfiNioxjKzl 545WuHQDnC 9nfFNGVDQui47wb9Dpsi7MGnJGP69ajhh tUsM1d3n UMgIY5DplsJt9nlw

Therefore, the height of the tower is m and the width of the canal is 10 m.

*Q12 :

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer :

Let AB be a building and CD be a cable tower. In ΔABD,

In ΔACE,PhhSGdtukGr0m8kJxPtUXCFddtM1aCU37NZDJ4y5POj4JtQtgMqpOa5RFbrjTVK 3ZuBfRUnX0KzYyL9dWhC8W aBEzO7gFSQ9lNY5EqnBR7Q1tbrstKCHYJps0uNRNGItsnc2A

AE = BD = 7 mnWNk4V7H4IKNnF9UDiqmL2noRSI0jW1b6yaK80bte5j8wnzqd2nD6uraxULksJ12i Hz6aCtiuVVUkSC6sVjOSkPhH6xjD9l4MY9yWpOFTeI9

Therefore, the height of the cable tower is

Q13 :

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer :

Let AB be the lighthouse and the two ships be at point C and D respectively. In ΔABC,ETVNB1vrAzb0LtgWBb3glpRMX9SeNNyOU31xH0J4Jd3bMSYvkWdDTOzR8QgljCqda1Sv3lDA15or Vgssll lZU

In ΔABD,meXY1RzA1YuQNPLIuDU4BZqRGjI8PTsHY6wOajg78zW TFWpajwkNtggQWQER6xwLAVyb3plaP5epiKbpkntn0KP86ZuPRnlM5Xcb0QX iQetN0sbiMilhc AygiQ13qrgAwpn8oICNfskD xa9Hn3ubTt9S4FnfkA57VTMv6cQwPqCgPG84UGqbEO3vXANS0 6HNwL8HimeJHaGOgmEpr36KH08pWnz1r rlxfGkJdr6CJe878ns A6 FbC815vd0HZTd0vW1uwm8

Therefore, the distance between the two ships is

Q14 :

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 
VHrF2Fn3eTLSMlJSrtTumbhFtscDADEOlsIAbKXXgcobVp3 nkgzLIGhKo2NluWmEgz2 9UqfEUYq7N X7JOCgPomv9VQkTlB2QIzVO3FSKFpfaM h2ft2CRHOFrtAM67WkpfFU

88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at

any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the 

distance travelled by the balloon during the interval.

Answer :LmI FlABy8dhtsyw9LgQVqWWc11lg xn gajYZ0UdGczfMiEfjcll2VUcZiIAZ7fBQHk8EWPAumXs4YjJZzjgYj71MRikkGGEJBZujgfWbn0

Let the initial position A of balloon change to B after some time and CD be the girl. In ΔACE,SyLXjKYpAghFbYz5Vx2hQrcPD Pg89kWKr4P7l9cHOC9 ZzABvDVUmyT tkC2i52Sr9FuFnGTjyuGxQ9vyz7ftxaqP0f5YRBE6lVvVYV2LDJ1 NMJrUwZeT5mft97zQkrn9z2I

In ΔBCG,
Po4GgsADMXUrSlcZ atIgxpk4G2Vag1bWBczsLI56yohOuLXuwCmXCmmkwtcsaZi

Distance travelled by balloon = EG = CG – CEUbQq6NrGGObGMS m tTcLOy xNfTXIRUG01n J1aRLdqldzE JXUc7phDE

Q15 :

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer :Nph6QI9YQawZ3paNtvj62nFIKzfAF2oDWxsSVVkDPqkLn7TB5hgVLsMh4TS5hWLkegNrA5l1rzMH5LdXNhIu1XgrmbWWMvXjBrpjwrH4V k0ICiYrNUaoRI9OfopedTHl1othBQ

Let AB be the tower.

Initial position of the car is C, which changes to D after six seconds. In ΔADB,

In ΔABC,sxIXZ0o5ZNAbX6 BCbs5YXyo5SSY AstKBJCwnmf8x9b32qmYJmbZxOizfUwydxd9WCA384ec1xSsGj5Lh0kQeEqitIoX2G RNCiD7UNkurs 19jJ63r30P6e5yCpuEZtNSjq2I

r6icXcNj5P9eTMRPU6FwewSIza1iPQAJWUzU34faHQTM8F6QjLljfhElyCSXhFllOx cA2 PiKDCB16K2DP2gjOnkP cNQ KoLW4OnIAq4p4nCa1IiZEPDfq Dymz9vYiMooY

Time taken by the car to travel distance DC= 6 secondsiO7aDB0qrGwzgHm4SZM7Ff2kqdy64EbLlenJrQwkIH27rmUcCVAamjBRx4g9s e2Y 0oivpJKS3pGq6NuLhG FiAT4jGkqlALK ysp0g0uv35IIsCHgoF5HEnhjWFzECz58ZIOQ

Time taken by the car to travel distance DBhTAMOB5t95p5ggOWTg60u8jwCDOpVT3sVeuPcMwby9 QWZp 0plOnjvkRVuqVUcK0ncvHhCq3S 2A8LBFzOUGXcSPL2TG15dARgvriiL0j4XVvHya359gsR5l5HoouoUkUF KeM

Q16 :

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m. from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer :4 rC3hgZysne3TGmRHo9DgDXpzY6qbI5SUaAYt6F9pXZKvXO2coUAlL9I04n vV1TX

Let AQ be the tower and R, S are the points 4m, 9m away from the base of the tower respectively. The angles are complementary. Therefore, if one angle is θ, the other will be 90 – θ.

In ΔAQR,9Zz6nKlYpv5XoleEL7qqL7hNanGJ6dwKWptfj2kIe6ajl4ifiGPpVa8JDAKDNkx3QWmwMdaon6afP0 UdsW1QBhcpxiMWXUstwG5oTZAdZDvXa1zyoxcykHc1HHo SPnhufFoqs
i 77mjHqk4xsYYIjGHYPBhNaWqgiMnM7p6EcvIgze397OssAjatM3ckwv1KDQpllDK8qnm3V 4

In ΔAQS,

On multiplying equations (i) and (ii), we obtain

BEuu7qpXniUQcHDS0bttnbp3DUqG95sdOgmxEmeOJ A5BLo7ptYQ435YcE LvAxSrqIabnmAp1JmfKTNxsnuEKIgUI1mkRShBA9BXCnC8nQhCozXBDsooc DfXqYQPnLZVevlZw

However, height cannot be negative. Therefore, the height of the tower is 6 m.

NCERT Solutions for Class 10 Maths Chapter 9 – Some Applications of Trigonometry

For Class 10 CBSE Maths paper, out of the 80 marks, 12 marks are assigned from the chapter “ Trigonometry”. You can expect 1 question from this chapter. The paper consists of 4 parts and each carries different marks. The questions have been assigned with 1 mark, two marks, 3 marks and 4 marks. The main topics covered in this chapter include:
9.1 Introduction
In the previous chapter, you have studied about trigonometric ratios. In this chapter, you will be studying about some ways in which trigonometry is used in the life around you. Trigonometry is one of the most ancient subjects studied by scholars all over the world. As we have studied in Chapter 8, trigonometry was invented because its need arose in astronomy. In this chapter, we will see how trigonometry is used for finding the heights and distances of various objects, without actually measuring them.
9.2 Heights and Distances
The topic discusses the line of sight, angle of deviation, angle of elevation and angle of depression. All the processes are explained by solving some problems. The numerical problems are solved with the help of trigonometric ratios.
9.3 Summary
The summary describes all the points you have studied in the chapter. It will help you to understand the important concepts that need to be focused upon further from the chapter.
List of Exercises in class 10 Maths Chapter 9 :
Exercise 9.1 Solutions – 16 Questions ( 16 long answers)

In this chapter, “ Some applications of trigonometry”, class 10 students will get to know how trigonometry is helpful in finding the height and distance of different objects without measuring. In earlier days, astronomers have used trigonometry for calculating the distance from the earth to the planets and stars. Trigonometry is mostly used in navigation and geography to locate the position in relation to the latitude and longitude.

Students will learn the applications of trigonometry with real-life examples in a better way. With the help of geometrical figures, the important terms and problems are explained and the summary is given at the end of the chapter. In NCERT solutions for class 10 maths, you are provided with step by step procedure solutions of all the questions.

Key Features of NCERT Solutions for Class 10 Maths Chapter 9 – Some Applications of Trigonometry

  • The solutions are designed by subject experts.
  • Access for chapter-wise questions and answers.
  • Valuable guidance for the students to prepare for the board exams.
  • The solutions are explained in the detailed procedure.
  • Easy access to all the exercise solutions.

Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 9

Where to download NCERT Solutions for Class 10 Maths Chapter 9?

NCERT Solutions for Class 10 Maths Chapter 9 can be downloaded at Swastik Classes website. It can be avail in free PDF. To download go to Swastik Classes NCERT Solutions website/select the Class 10/opt the subject as Maths/click on to the desired chapter that is Chapter 9 Some Applications of Trigonometry.

What are the applications of learning NCERT Solutions for Class 10 Maths Chapter 9?

Applications of learning NCERT Solutions for Class 10 Maths Chapter 9 are finding the height and distance of different objects without measuring. By learning these concepts students will be able to answer all the questions based on trigonometry as well as it may help in writing class tests and board exams.

Is NCERT Solutions for Class 10 Maths Chapter 9 enough for board exams?

Yes, you have to learn all the questions thoroughly and practice them for more time. Once you are done with solving all the questions then you can refer to other reference books and questions provided NCERT Exemplar textbooks.


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NCERT Solutions Class 10 Maths Chapters

  • Chapter 1 Real Numbers
  • Chapter 2 Polynomials
  • Chapter 3 Pair Of Linear Equations In Two Variables
  • Chapter 4 Quadratic Equations
  • Chapter 5 Arithmetic Progressions
  • Chapter 6 Triangles
  • Chapter 7 Coordinate Geometry
  • Chapter 8 Introduction To Trigonometry
  • Chapter 9 Some Applications Of Trigonometry
  • Chapter 10 Circles
  • Chapter 11 Constructions
  • Chapter 12 Areas Related To Circles
  • Chapter 13 Surface Areas And Volumes
  • Chapter 14 Statistics
  • Chapter 15 Probability

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