Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Questions and Answers.
BSEB Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1
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 = (24)² + (7)²
= 576 + 49 = 625
So, AC = \(\sqrt{625}\) cm = 25 cm
(i) sin A = \(\frac { BC }{ AC }\) = \(\frac { 7 }{ 25 }\), cos A = \(\frac { AB }{ AC }\) = \(\frac { 24 }{ 25 }\)
(ii) sin C = \(\frac { AB }{ AC }\) = \(\frac { 24 }{ 25 }\), cos C = \(\frac { BC }{ AC }\) = \(\frac { 7 }{ 25 }\)
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 = 169 – 144 = 25
So, QR = \(\sqrt{25}\) cm = 5 cm
∴ tan P = \(\frac { QP }{ PQ }\) = \(\frac { 5 }{ 12 }\) and cot R = \(\frac { QR }{ PQ }\) = \(\frac { 5 }{ 12 }\)
Hence, tan P – cot R = \(\frac { 5 }{ 12 }\) – \(\frac { 5 }{ 12 }\) = 0.
Question 3.
If sin A = \(\frac { 3 }{ 4 }\), 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 = \(\frac { 13 }{ 12 }\), 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 = \(\frac { QA }{ PA }\)
and cos B = \(\frac { SB }{ RB }\)
Thus, it is given that
\(\frac { QA }{ PA }\) = \(\frac { SB }{ RB }\)
so, \(\frac { QA }{ SB }\) = \(\frac { PA }{ RB }\) = K(say) … (1)
Now, by Pythagoras Theorem,
Therefore, from (1) and (2), we have
\(\frac { QA }{ SB }\) = \(\frac { PA }{ RB }\) = \(\frac { PQ }{ RS }\)
Hence, ∆ PQA ~ ∆ RSB [SSS similarity]
Therefore, ∠A = ∠B [Corresponding angles]
Question 7.
If cot θ = \(\frac { 7 }{ 8 }\), evaluate :
(i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
(ii) cot² θ
Solution:
Consider a ∆ ABC in which ∠A = 0 and ∠B = 90°.
Then, Base = AB, Perp = BC and Hyp = AC
∴ cot θ = \(\frac { Base }{ Perp }\) = \(\frac { AB }{ BC }\) = \(\frac { 7 }{ 8 }\)
Let AB = 7k and BC = 8k
Question 8.
If 3 cot A = 4, check whether \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = 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 = \(\frac { Base }{ Perp }\) = \(\frac { AB }{ BC }\) = \(\frac { 4 }{ 3 }\)
Let AB = 4k and BC = 3k [3 cot A = 4 ⇒ cot A = \(\frac { 4 }{ 3 }\) ]
Question 9.
In ∆ ABC right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\), 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 = \(\frac { Perp }{ Base }\) = \(\frac { BC }{ AB }\) = \(\frac{1}{\sqrt{3}}\)
Let BC = k and AB = \(\sqrt{3}\)k.
(i) sin A cos C + cos A sin C = \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) + \(\frac{\sqrt{3}}{2}\) x \(\frac{\sqrt{3}}{2}\)
= \(\frac { 1 }{ 4 }\) x \(\frac { 3 }{ 4 }\) = \(\frac { 4 }{ 4 }\) = 1
(ii) cos A cos C – sin A sin C = \(\frac{\sqrt{3}}{2}\) x \(\frac { 1 }{ 2 }\) – \(\frac { 1 }{ 2 }\) \(\frac{\sqrt{3}}{2}\) = 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 :
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RP² = RQ² + QP²
So, (25 – x)² = x² + 5²
or 625 – 50x + x² = x² + 25
or – 50x = – 600 or x = \(\frac { -600 }{ -50 }\) = 12
∴ RQ = 12cm
So, RP = (25 – 12)cm = 13 cm
Now, sin P = \(\frac { RQ }{ RP }\) = \(\frac { 12 }{ 13 }\)
cos P = \(\frac { PQ }{ RP }\) = \(\frac { 5 }{ 13 }\)
and, tan P = \(\frac { RQ }{ PQ }\) = \(\frac { 12 }{ 5 }\)
Question 11.
State whether the following are true or false. Justify your answer.
- The value of tan A is always less than 1.
- sec A = \(\frac { 12 }{ 5 }\) for some value of angle A.
- cos A is the abbreviation used for the cosecant of angle A.
- cot A is the product of cot and A.
- sin θ = \(\frac { 4 }{ 3 }\) for some angle θ.
Solution:
- False because sides of a right triangle may have any length, so tan A may have any value.
- True as sec A is always greater than or equal to 1.
- False as cos A is the abbreviation used for cosine A.
- False as cot A is not the product of ‘cot’ and A. ‘cot separated from A has no meaning.
- False as sin 0 cannot be > 1.