Introduction to Trigonometry
3 Marks Questions
1. Given calculate all other trigonometric ratios.
Ans. Consider a triangle ABC in which A = and B =
Let AB = and BC =
Then, using Pythagoras theorem,
BC =
=
=
= =
2. If evaluate:
(i)
(ii)
Ans. Consider a triangle ABC in which A = and B =
Let AB = and BC =
Then, using Pythagoras theorem,
AC = =
= = =
(i)
= =
(ii)
= =
3. If check whether or not.
Ans. Consider a triangle ABC in which B = .
And
Let AB = and BC =
Then, using Pythagoras theorem,
AC =
=
= = =
And
Now, L.H.S.
= =
R.H.S. =
= =
L.H.S. = R.H.S.
=
4. In ABC right angles at B, if find value of:
(i)
(ii)
Ans. Consider a triangle ABC in which B = .
Let BC = and AB =
Then, using Pythagoras theorem,
AC =
=
= = =
For C, Base = BC, Perpendicular = AB and Hypotenuse = AC
(i)
= = 1
(ii)
=
5. In PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine
the values of and
Ans. In PQR, right angled at Q.
PR + QR = 25 cm and PQ = 5 cm
Let QR = cm and PR = cm
Using Pythagoras theorem,
RP2 = RQ2 + QP2
RQ = 12 cm and RP = 25 – 12 = 13 cm
And
6. If and find A and B.
Ans. (i) False, because =1
And =
(ii) True, because
0 1
It is clear, the value of increases as increases.
(iii) False, because
1 0
It is clear, the value of decreases as increases
(iv) False as it is only true for
True, because and = i.e. undefined
7. Choose the correct option. Justify your choice:
(i) =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) =
(A) 0 (B) 1 (C) 2 (D) none of these
(iii) =
(A) (B) (C) (D)
(iv) =
(A) (B) (C) (D) none of these
Ans. (i) (B) = = 9 x 1 = 9
(ii) (C)
=
=
= = 2
(iii) (D)
= = =
(iv) (D) =
= =
= =
4 Marks Questions
1. Express the trigonometric ratios and in terms of
Ans. For
By using identity
For
By using identity
For
2. Write the other trigonometric ratios of A in terms of
Ans. For
By using identity,
For
For
By using identity
For
For
3. Evaluate:
(i)
(ii)
Ans. (i)
(ii)
= =1
= =1
4. Choose the correct option. Justify your choice:
(i) =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) =
(A) 0 (B) 1 (C) 2 (D) none of these
(iii) =
(A) (B) (C) (D)
(iv) =
(A) (B) (C) (D) none of these
Ans. (i) (B)
= =9x1=9
(ii) (C)
=
=
= =2
(iii) (D)
= = =
(iv) (D)
= =
= = =
5. Prove the following identities, where the angles involved are acute angles
for which the expressions are defined:
(i)
(ii)
(iii)
(iv)
(v) , using the identity
(vi)
(vii)
(viii)
(ix)
(x)
Ans. Proof:
(i) L.H.S.
=
=
= = R.H.S.
(ii) L.H.S.
= =
= = = R.H.S
(iii) L.H.S.
=
= =
(iv) L.H.S.
= =
=
=
= = R.H.S.
(v) L.H.S.
Dividing all terms by
= =
= = R.H.S.
(vi) L.H.S.
=
= =
= = R.H.S.
(vii) L.H.S. =
= = = = R.H.S
(viii) L.H.S.
=
= R.H.S.
(ix) L.H.S.
= =
Dividing all the terms by ,
= =
= = R.H.S.
(x) L.H.S.
= = = R.H.S.
Now, Middle side = =
=
= = = = R.H.S.
6. Use Euclid’s division algorithm to find the HCFof:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Ans. (i) 135 and 225
We have 225 > 135,
So, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Here remainder 90 ≠ 0, we apply the division lemma again to 135 and 90 to
obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45≠ 0, and apply the
division lemma to obtain
90 = 2 × 45 + 0
Since that time the remainder is zero, the process get stops.
The divisor at this stage is 45
Therefore, the HCF of 135 and 225 is 45.
(ii) 196 and 38220
We have 38220 > 196,
So, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since we get the remainder is zero, the process stops.
The divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii) 867 and 255
We have 867 > 255,
So, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Here remainder 102 ≠ 0, we apply the division lemma again to 255 and 102
to obtain
255 = 102 × 2 + 51
Here remainder 51 ≠ 0, we apply the division lemma again to 102 and 51 to
obtain
102 = 51 × 2 + 0
Since we get the remainder is zero, the process stops.
The divisor at this stage is 51,
Therefore, HCF of 867 and 255 is 51.
7. Evaluate:
(i)
(ii)
(iii)
(iv)
(v)
Ans. (i) =
= =1
(ii) =
= =2
(iii)
= =
= =
=
=
(iv) =
= =
(v)
=
=
8. Prove the following identities, where the angles involved are acute
angles for which the expressions are defined:
(i)
(ii)
(iii)
(iv)
(v) , using the identity
(vi)
(vii)
(viii)
(ix)
(x)
Ans. (i) L.H.S.
= = = R.H.S.
(ii) L.H.S.
= = =
= = R.H.S
(iii) L.H.S.
=
(iv) L.H.S.
= =
= = R.H.S.
(v) L.H.S.
Dividing all terms by
= =
= = R.H.S.
(vi) L.H.S.
=
= = = = R.H.S.
(vii) L.H.S. =
= = = = R.H.S
(viii) L.H.S.
=
=
= R.H.S.
(ix) L.H.S.
= =
Dividing all the terms by ,
= =
= = R.H.S.
(x) L.H.S. =
= = = R.H.S.
Now, Middle side =
= =
= = R.H.S.
