Gujarat Board GSEB Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

Question 1.
In ΔABC, right angled at B, AB = 24 cm, BC = 7 cm, determine
(i) sin A, cos A
(ii) sin C, cos C
Solution:
We have
AB = 24 cm
BC = 7 cm

(i) In right ΔABC
⇒ AC² = AB² + BC²
⇒ AC² = 24² + 7²
⇒ AC² = 576 + 49

⇒AC² = 625
AC = 25 cm
san A = (frac {BC}{AC})
sin A = (frac {7}{25})
cos A = (frac {AB}{AC}) ⇒ cos A = (frac {24}{25})

(ii) sin C = (frac {AB}{AC}) ⇒ sin C = (frac {24}{25})
cos C = (frac {BC}{AC}) ⇒ cos C = (frac {7}{25})

Question 2.
In Fig. find tan P – cot R.

Soution:
In ΔPQR
PR2 = PQ² + QR²
⇒ 13² = 12² + QR²
⇒ 169 = 144 + QR²
⇒ QR² = 169 – 144
⇒ QR² = 25
⇒ QR = 5 cm
tan P = (frac {QR}{PR}) ⇒ tan P = (frac {5}{12})
⇒ cot R = (frac {QR}{PQ}) ⇒ cot R = (frac {5}{12})
∴ tan P – cot R = (frac {5}{12}) – (frac {5}{12}) = 0

Question 3.
If sin A = (frac {3}{4})ca1cu1ate cos A and tan A.
Solution:
We have

BC = 3k
and AC = 4k
In right ΔABC
AC² = AB² + BC²
(4k)² = AB² + (3k)²
16k² = AB² + 9k²
AB² – 16k² – 9k²
AB² – 7k²
AB = (sqrt{7}) k
∴ cos A = (frac{mathrm{AB}}{mathrm{AC}}) = (frac{sqrt{7} k}{4 k}) = (frac{sqrt{7}}{4})
And tan A = (frac{mathrm{BC}}{mathrm{AB}}) = (frac{3 k}{sqrt{7} k}) = (frac{3}{sqrt{7}})

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

AB = 8k
and BC = 15k

In right ΔABC,
AC² = AB² + BC²
⇒ AC² = (8k)² +(15k)²
⇒ AC² = 64k² + 225k²
⇒ AC² = 289k²
⇒ AC = 17k
∴ sin A = (frac {BC}{AC}) = (frac {15k}{17k}) = (frac {15}{17})
and sec A = (frac {AC}{AB}) = (frac {17k}{8k}) = (frac {17}{8})

Question 5.
Given sec θ = (frac {13}{12})calculate all other trigono metric ratios.
Solution:
We have

AC = 13k and AB = 12k

In right ΔABC,
AC² = AB² + BC²
(13k)² = (12k)² + BC²
BC² = 169k² – 144k²
BC² = 25k²
BC = 5k
sin θ = (frac {BC}{AC}) = (frac {5k}{13k}) = (frac {5}{13})
Cos θ = (frac {AB}{AC}) = (frac {12k}{13k}) = (frac {12}{13})
tan θ = (frac {BC}{AB}) = (frac {5k}{12k}) = (frac {5}{12})
cosec θ = (frac {AC}{BC}) = (frac {13k}{5k}) = (frac {13}{5})
cot θ = (frac {AB}{BC}) = (frac {12k}{5k}) = (frac {12}{5})

Question 6.
1f ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
Let us consider two right triangles LMN and PQR.
cos A = cos B
(frac {PQ}{PR}) = (frac {LM}{LN}) = k Also = (frac {PQ}{LM}) = (frac {PR}{LN}) …….( 1)
(where k is a positive number)
PQ = kPR
LM = kLN

Now, in ΔPQR
PR² = PQ² + QR²
⇒ PR² = (kPR)² + QR²
⇒ QR² = PR² – k²PR²
⇒ QR = (sqrt{mathrm{PR}^{2}-k^{2} mathrm{PR}^{2}})
⇒ QR = PR (sqrt{1-k^{2}}) ……….(2)
In right ΔLMN
LN² = LM² + MN²
⇒ LN² = (kLN)² + MN²
⇒ MN² = LN² – k²LN²
⇒ MN = (sqrt{mathrm{L} mathrm{N}^{2}-k^{2} mathrm{L} mathrm{N}^{2}})
⇒ MN = LN(sqrt{1-k^{2}}) ……….(3)
Dividing eqn. (2) by eqn. (3)
(frac {QR}{MN}) = (frac{mathrm{PR} sqrt{1-k^{2}}}{mathrm{L} mathrm{N} sqrt{1-k^{2}}}) = (frac{Q R}{M N}=frac{P R}{L N}) ………..(4)
From equation (1) and (4),
(frac{Q R}{M N}=frac{P R}{L N}=frac{P Q}{L M})
∴ Δ PQR – ΔLMN (by SSS similarity)
∴ ∠PQR = ∠LMN
(corresponding angle of two similar triangles)
∠A = ∠B

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:
We have

AB = 7k and BC = 8k

In right ΔABC,
AC² = AB² + BC²
⇒ AC² = (7k)² + (8k)² = 49k² + 64k²
⇒ AC² = 113k²
⇒ AC = (sqrt{113 k})
sin θ = (frac{mathrm{BC}}{mathrm{AC}}=frac{8 k}{sqrt{113 k}}=frac{8}{sqrt{113}})
cos θ = (frac{mathrm{AB}}{mathrm{AC}}=frac{7 k}{sqrt{133} k}=frac{7}{sqrt{133}})

(i) (frac{(1+sin theta)(1-sin theta)}{(1+cos theta)(1-cos theta)})

(ii) cot² θ = (cot θ)²
= ((frac {7}{8}))² (cot θ = (frac {AB}{BC}) = (frac {7}{8}))
= (frac {49}{64})

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:
We have

⇒ AB = 4k
⇒ BC = 3k
In right ΔABC,
AC² = AB² + BC²
= (4k)² + (3k)²
= 16k² + 9k²
⇒ AC² = 25k²
⇒ AC = 5k
tan A = (frac{mathrm{BC}}{mathrm{AB}}=frac{3 k}{4 k}=frac{3}{4})
sin A = (frac{mathrm{BC}}{mathrm{AC}}=frac{3 k}{5 k}=frac{3}{5})
cos A = (frac{A B}{A C}=frac{4 k}{5 k}=frac{4}{5})

RHS = cos²A – sin²A
= ((frac {4}{5}))² – ((frac {4}{5}))²
= (frac{16}{25}-frac{9}{25}=frac{7}{25})
LHS = RHS

Question 9.
In AABC, right angled at B, if tan A = (frac{1}{sqrt{3}}) ,find the value of
(j) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
We have
tan A = (frac{1}{sqrt{3}})
⇒ (frac{mathrm{BC}}{mathrm{AB}}=frac{1}{sqrt{3}})
⇒ (frac{mathrm{BC}}{1}=frac{mathrm{AB}}{sqrt{3}}) = k

∴ BC = k and AB = (sqrt{3} k)

in right ΔABC
AC² = AB² + BC²
⇒ AC² = ((sqrt{3} k))² + k²
AC² = 3k² + k²
AC² = 4k²
AC = 2k
Now sin A = (frac{mathrm{BC}}{mathrm{AC}}=frac{k}{2 k}=frac{1}{2})
Cos A = (frac{mathrm{AB}}{mathrm{AC}}=frac{sqrt{3} k}{2 k}=frac{sqrt{3}}{2})
sin C = (frac{mathrm{AB}}{mathrm{AC}}=frac{sqrt{3} k}{2 k}=frac{sqrt{3}}{2})
Cos C = (frac{mathrm{BC}}{mathrm{AC}}=frac{k}{2 k}=frac{1}{2})

(i) sin A cos C + cos A sin C
= (frac{1}{2} times frac{1}{2}+frac{sqrt{3}}{2} times frac{sqrt{3}}{2})
= (frac{1}{4}+frac{3}{4}=frac{1+3}{4}=frac{4}{4}=1)

(ii) cos A cos C – sin A sin C
= (frac{sqrt{3}}{2} times frac{1}{2}-frac{1}{2} times frac{sqrt{3}}{2})
= (frac{sqrt{3}}{4}-frac{sqrt{3}}{4}) = 0

Question 10.
In APQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the value of sin P, cos P and tan P.
Solution:
In right triangle PQR,
PR² = PQ² + QR² ……..(1)
(By Pythagoras theorem)
PR + QR = 25
⇒ PR = 25 – QR ……….(2)

From eqn. (1) and (2)
(25 – QR)² = (5)² + QR²
625 – 2 x 25 x QR + QR²
= 25 + QR²
= 625 – 5O x QR = 5²
– 5O QR = 25 – 625
– 5O QR = – 600
QR = (frac{600}{50}) = 12
Putting in eqn. (2),
PR = 25 – QR = 25 – 12
PR = 13 cm
sin P = (frac{QR}{PR}) = (frac{12}{13})
cos P = (frac{P Q}{P R}=frac{5}{13})
and tan P = (frac{Q R}{P Q}=frac{12}{5})

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1
(ii) sec A = (frac{12}{5}) for some value of A
(iii) cos A is the abbreviation used for the cosecant of angle A
(iv) cot A is the product of cot and A
(v) sin θ = (frac{4}{3}) for some angle.
Solution:
(i) False.

the value of the perpendicular may be longer than the base.

(ii) True

Because Hypotenuse is always greater than the base.

(iii) False
Since cos A is the abbreviation used for the cosine of angle A.

(iv) False
Since cot is meaningless without angle A.

(v) False
Since value of sin O is always less than 1.