Chapter 8 – Introduction to Trigonometry (Ex – 8.1)

Introduction to Trigonometry

Question 1.
In ∆ ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A (ii) sin C, cos C

Solution:
Let us draw a right ∆ ABC.

By using the Pythagoras theorem, we have :
AC² = AB² + BC²
or AC² = (24)² + (7)²
or AC² = 576 + 49
or AC² = 625
or AC = √625 cm
∴ AC = 25 cm
(i) sin A = BC/AC = 7/25, cos A = AB/AC = 24/25
(ii) sin C = AB/AC = 24/25, cos C = BC/AC = 7/25

Question 2.
In adjoining figure, find tan P – cot R.

Solution:
By using the Pythagoras theorem, we have :
PR² = PQ² + QR²
or 13² = 12² + QR²
or QR² = 13² – 12²
or QR² = 169 – 144
or QR² = 25
or QR = √25 cm
∴ QR = 5 cm
∴ tan P = QP/PQ = 5/12 and cot R = QR/PQ = 5/12
Hence, tan P – cot R = 5/12 – 5/12 = 0.

Question 3.
If sin A = 3/4, calculate cos A and tan A.

Solution:
Consider a ∆ ABC in which ∠B = 90°.

For ∠A, we have :
Base (adjacent side) = AB,
Perp. (opposite side) = BC
and Hyp. = AC.

Question 4.
Given 15 cot A =8, find sin A and sec A.

Solution:
Consider a ∆ ABC in which ∠B = 90°.
For ∠A, we have:
Base AB, Perp. = BC and Hyp. = AC.

Question 5.
Given sec θ = 13/12, calculate all other trigonometric ratios.

Solution:
Consider a ∆ ABC in which ∠A = 0° and ∠B = 90° Then, Base = AB, Perp. = BC and Hyp. = AC.

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Solution:
Let us consider two right triangles PQA and RSB in which cos A = cos B (see the figure).

We have:
cos A = QA/PA
and cos B = SB/RB
Thus, it is given that
QA/PA = SB/RB
so, QA/SB = PA/RB = K(say) … (1)
Now, by Pythagoras Theorem,

Therefore, from (1) and (2), we have
QA/SB = PA/RB = PQ/RS
Hence, ∆ PQA ~ ∆ RSB [SSS similarity]
Therefore, ∠A = ∠B [Corresponding angles]

Question 7.
If cot θ = 7/8, evaluate :
(i) {(1+sinθ)(1−sinθ)}/{(1+cosθ)(1−cosθ)}
(ii) cot² θ

Solution:
Consider a ∆ ABC in which ∠A = 0 and ∠B = 90°.

Then, Base = AB, Perp = BC and Hyp = AC
∴ cot θ = Base/Perp = AB/BC = 7/8
Let AB = 7k and BC = 8k

Question 8.
If 3 cot A = 4, check whether (1−tan²A)/(1+tan²A) = cos² A – sin² A or not.

Solution:
Consider a ∆ ABC in which ∠B = 90°.
Foe ∠A, we have:

Base = AB, Perp = BC and Hyp = AC
∴ cot A = Base/Perp = AB/BC = 4/3
Let AB = 4k and BC = 3k [3 cot A = 4 ⇒ cot A = 4/3 ]

Question 9.
In ∆ ABC right angled at B, if tan A = 1/√3, find the value of
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C

Solution:
Consider a ∆ ABC, in which ∠B = 90°

For ∠A, we have :
Base = AB,
Perp. = BC .
and Hyp. = AC
∴ tan A = Perp/Base = BC/AB = 1/√3
Let BC = k and AB = √3k.

(i) sin A cos C + cos A sin C = 1/2 x 1/2 + √3/2 x √3/2
= 1/4 + 3/4 = 4/4 = 1
(ii) cos A cos C – sin A sin C = √3/2 x 1/2 – 1/2 x √3/2 = 0

Question 10
In ∆ PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution:
In ∆ PQR, right ∠d at Q,

PR + QR = 25 cm and PQ = 5 cm.
Let QR = x cm
∴ PR = (25 – x) cm
By Pythagoras theorem, we have :
RP² = RQ² + QP²
So, (25 – x)² = x² + 5²
or 625 – 50x + x² = x² + 25
or – 50x = – 600
or x = −600/−50 = 12
∴ RQ = 12cm
So, RP = (25 – 12)cm = 13 cm
Now, sin P = RQ/RP = 12/13
cos P = PQ/RP = 5/13
and, tan P = RQ/PQ = 12/5

Question 11.
State whether the following are true or false. Justify your answer.

1. The value of tan A is always less than 1.
2. sec A = 125 for some value of angle A.
3. cos A is the abbreviation used for the cosecant of angle A.
4. cot A is the product of cot and A.
5. sin θ = 43 for some angle θ.

Solution:

1. False because sides of a right triangle may have any length, so tan A may have any value.
2. True as sec A is always greater than or equal to 1.
3. False as cos A is the abbreviation used for cosine A.
4. False as cot A is not the product of ‘cot’ and A. ‘cot separated from A has no meaning.
5. False as sin 0 cannot be > 1.