Vijayeemantra

Introduction to Trigonometry: CBSE Class 10 Mathematics practice questions

Practise 50 questions on Introduction to Trigonometry, free, with an explanation for each answer.

Start practising free

Sample questions

A student looks at the top of a tower. We can imagine a right triangle between the student, the foot of the tower and its top. The branch of mathematics that lets us find the height without measuring it is called trigonometry. The word comes from the Greek words tri, gon and metron. What do they mean?

A student, a tower and the line of sight forming a right triangle
  1. triangle, angle, distance
  2. three, angles, height
  3. three, sides, measure
  4. right, side, measure
Show the answer

C. three, sides, measure

Tri = three, gon = sides and metron = measure, so trigonometry literally means "measuring three-sided figures". It studies the relationships between the sides and angles of a triangle, which is exactly what lets us find the height of the tower from the distance and the angle of sight.

In △ABC\triangle ABC, right-angled at B, AB=9AB = 9 cm, BC=40BC = 40 cm and AC=41AC = 41 cm. Find sin⁡C\sin C.

Tall thin right triangle ABC with AB = 9, BC = 40, AC = 41
  1. 4041\dfrac{40}{41}
  2. 940\dfrac{9}{40}
  3. 419\dfrac{41}{9}
  4. 941\dfrac{9}{41}
Show the answer

D. 941\dfrac{9}{41}

The side opposite ∠C\angle C is AB=9AB = 9, so sin⁡C=ABAC=941\sin C = \dfrac{AB}{AC} = \dfrac{9}{41}. The trap is to reuse the answer for ∠A\angle A: 4041\dfrac{40}{41} is sin⁡A\sin A (which is also cos⁡C\cos C). 940\dfrac{9}{40} is tan⁡C\tan C and 419\dfrac{41}{9} is cosec⁡C\operatorname{cosec} C.

In △OPQ\triangle OPQ, right-angled at PP, OP=11OP = 11 cm and OQ−PQ=1OQ - PQ = 1 cm. Find cos⁡Q\cos Q.

Right triangle OPQ, right-angled at P, with OP = 11 cm and OQ one centimetre longer than PQ
  1. 1161\dfrac{11}{61}
  2. 56\dfrac{5}{6}
  3. 6162\dfrac{61}{62}
  4. 6061\dfrac{60}{61}
Show the answer

D. 6061\dfrac{60}{61}

(1+PQ)2=112+PQ2(1+PQ)^2 = 11^2 + PQ^2 gives 1+2PQ=1211 + 2PQ = 121, so PQ=60PQ = 60 cm and OQ=61OQ = 61 cm. Then cos⁡Q=PQOQ=6061\cos Q = \dfrac{PQ}{OQ} = \dfrac{60}{61}. 1161\dfrac{11}{61} is sin⁡Q\sin Q. Using OPOP instead of OP2OP^2 gives PQ=5PQ=5 and the false 56\dfrac56, and 6162\dfrac{61}{62} comes from taking the hypotenuse 6161 as PQPQ and then adding 11 again for OQOQ.

Topics covered

Start practising free

More CBSE Class 10 Mathematics chapters