# HBSE 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1

Haryana State Board HBSE 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1

Question 1.
In âˆ†ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C
Solution :
(i) In a right’triangle ABC we have âˆ B = 90Â°, AB = 24 cm and BC = 7 cm.
By Pythagoras theorem, we have
AC2 = AB2 + BC2
â‡’ AC2 = 242 + 72
â‡’ AC2 = 576 + 49
â‡’ AC2 = 625
â‡’ AC = $$\sqrt{625}$$ = 25

âˆ´ sin A = $$\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{25}$$
cos A = $$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24}{25}$$
Here, sin A = $$\frac{7}{25}$$, cos A = $$\frac{24}{25}$$

(ii) For trigonometric ratio of âˆ C, We have base = BC = 7 cm
Perpendicular = AB = 24 cm
Hypotenuse = AC = 25 cm
sin C = $$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24}{25}$$
and cos C = $$\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{25}$$
Hence, sin C = $$\frac{24}{25}$$, cos C = $$\frac{7}{25}$$

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

Solution:
We have, PQ = 12 cm, PR = 13 cm.
In right âˆ†PQR, âˆ Q = 90Â°
By Pythagoras theorem, we have PR2 = QR2 + PQ2
â‡’ (13)2 = QR2 + (12)2
â‡’ QR2 = (13)2 – (12)2
â‡’ QR2 = (13 + 12) (13 – 12)
â‡’ QR2 = 25
â‡’ QR = âˆš25 = 5
cot R = $$\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{5}{12}$$
For trigonometric ratio of âˆ P, we have
Base = PQ = 12 cm
Perpendicular = QR = 5 cm
Hypotenuse = PR = 13 cm
âˆ´ tan P = $$\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{5}{12}$$
Now, tan P – cot R = $$\frac{5}{12}-\frac{5}{12}$$ = 0
Hence, tan P – cot R = 0.

Question 3.
If sin A = $$\frac{3}{4}$$ Calculate cos A and tan A.
Solution :
Consider a right triangle ABC in which âˆ B = 90Â°
Then, sin A = $$\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{4}$$
Let BC = 3k and AC = 4k, where k is a positive integer.
By Pythagoras theorem, we have

AC2 = AB2 + BC2
(4k)2 = AB2 + (3k)2
AB2 = (4k)2 – (3k)2
AB2 = 16k2 – 9k2
AB2 = 7k2
AB = âˆš7 k
âˆ´ cos A = $$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{7} k}{4 k}=\frac{\sqrt{7}}{4}$$
and tan A = $$\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}}$$
Hence, cos A = $$\frac{\sqrt{7}}{4}$$, tan A = $$\frac{3}{\sqrt{7}}$$.

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

we have,
15 cot A = 8
cot A = $$\frac{8}{15}$$
Now consider a right 15 triangle ABC, in which âˆ B = 90Â°
cot A = $$\frac{8}{15}$$
â‡’ $$\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{8}{15}$$
Let AB = 8k and BC = 15k,where k is a positive integer.
By Pythagoras theorem, we have
AC2 = AB2 + BC2
AC2 = (8k)2 + (15k)2
AC2 = 64k2 + 225k2
AC = 289k2
AC = $$\sqrt{289 k^2}$$ = 17k.
sin A = $$\frac{\mathrm{BC}}{\mathrm{AC}}:=\frac{15 k}{17 k}=\frac{15}{17}$$
and sec A = $$\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{17 k}{8 k}=\frac{17}{8}$$
Hence, sin A = $$\frac{15}{17}$$ and sec A = $$\frac{17}{8}$$.

Question 5.
Given sec Î¸ = $$\frac{13}{12}$$, calculate all other trigonometric ratios.
Solution:
Consider a right triangle ABC in which âˆ B = 90Â°, âˆ CAB = 9, sec Î¸ = $$\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{13}{12}$$
Let AC = 13k and AB = 12k, where k is a positive integer.

By Pythagoras theorem, we have
AC2 = AB2 + BC2
(13k)2 = (12k)2 + BC2
BC2 = (13k)2 – (12k)2
BC2 = (13k + 12k) (13k – 12k)
BC2 = 25k Ã— k = 25k2
BC = $$\sqrt{25 k^2}$$ = 5k

âˆ´ sin Î¸ = $$\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 k}{13 k}=\frac{5}{13}$$
cos Î¸ = $$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12 k}{13 k}=\frac{12}{13}$$
tan Î¸ = $$\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5 k}{12 k}=\frac{5}{12}$$
cot Î¸ = $$\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12 k}{5 k}=\frac{12}{5}$$
cosec Î¸ = $$\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13 k}{5 k}=\frac{13}{5}$$

Hence, sin Î¸ = $$\frac{5}{13}$$,
cos Î¸ = $$\frac{12}{13}$$,
tan Î¸ = $$\frac{5}{12}$$,
cot Î¸ = $$\frac{12}{5}$$,
cosec Î¸ = $$\frac{13}{5}$$.

Question 6.
If âˆ A and âˆ B are acute angles such that cos A = cos B; then show that âˆ A = âˆ B.
Solution :
Let us consider right triangle ABC in which âˆ C = 90Â° and cos A = cos B.
Draw CD âŠ¥ AB.
In right triangle ADC, âˆ ADC = 90Â°.
cos A = $$\frac{\mathrm{AD}}{\overline{\mathrm{AC}}}$$ ………………….(1)
In right âˆ†BDC,
âˆ BDC = 90Â°
cos B = $$\frac{\mathrm{BD}}{\overline{\mathrm{BC}}}$$ ………………….(2)

According to question,
cos A = cos B
$$\frac{\mathrm{AD}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{BC}}$$
$$\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{BC}}$$ [each = $$\frac{C D}{C D}$$]
âˆ†ADC ~ âˆ†BDC [By SSS similarity criterion]
[corresponding âˆ s of two similar âˆ†s are equal]
Hence Proved.

Question 7.
If cot Î¸ = $$\frac{7}{8}$$, evaluate:
(i) $$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$$
(ii) cot2 Î¸
Solution :

(i) Consider a triangle ABC in which âˆ B = 90Â°,
âˆ ACB = Î¸ and cot Î¸ = $$\frac{7}{8}$$
â‡’ $$\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{7}{8}$$
Let BC = 7k and AB = 8k, where k is a positive integer,
By Pythagoras theorem, we have
AC2 = BC2 + AB2
AC2 = (7k)2 + (8k)2
AC2 = 49k2 + 64k2
AC2 = 113k2
AC = âˆš113 k
âˆ´ sin Î¸ = $$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{8 k}{\sqrt{113} k}=\frac{8}{\sqrt{113}}$$

(ii) cot2 Î¸ = (cot Î¸)2
= $$\left(\frac{\mathrm{BC}}{\mathrm{AB}}\right)^2=\left(\frac{7 k}{8 k}\right)^2$$
= $$\left(\frac{7}{8}\right)^2=\frac{49}{64}$$

Hence, (i) $$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{49}{64}$$
(ii) cot2 Î¸ = $$\frac{49}{64}$$

Question 8.
If 3 cot A = 4, check whether $$\frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}$$ = cos2 A – sin2 A or not.
Solution :
Consider a triangle ABC in which âˆ B = 90Â° and 3 cot A = 4
â‡’ cot A = $$\frac{4}{3}$$
â‡’ $$\frac{A B}{B C}=\frac{4}{3}$$
Let AB = 4k and BC = 3k,where k is a positive integer.

By pythagoras theorem, we have
AC2 = AB2 + BC2
â‡’ AC2 = (4k)2 + (3k)2
â‡’ AC2 = 16k2 + 9k2
â‡’ AC2 = 25k2
â‡’ AC = $$\sqrt{25 k^2}$$
â‡’ AC = 5k

Hence, $$\frac{1-\tan ^2 A}{1+\tan ^2 A}$$ = cos2 A – sin2 A.

Question 9.
In triangle ABC, right angled at B, if tan A = find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
In right triangle ABC, âˆ B = 90Â° and tan A = $$\frac{1}{\sqrt{3}}$$
$$\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{1}{\sqrt{3}}$$
Let BC = k and AB = âˆš3k,where k is positive integer.

By pythagoras theorem, we have
AC2 = AB2 + BC2
AC2 =(âˆš3k)2 + k2
AC2 = 3k2 + k2 = 4k2
AC = $$\sqrt{4 k^2}$$ = 2k
âˆ´ sin A = $$\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{k}{2 k}=\frac{1}{2}$$
and cos A = $$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}$$
Now, for trigonometric ratio of âˆ C, we have
Base = BC = k
Perpendicular = AB = âˆš3 k
Hypotenuse = AC = 2k
sin C = $$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}$$
cosC = $$\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{k}{2 k}=\frac{1}{2}$$

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

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

Hence, (i) sin A cos C + cos A sin C = 1
(ii) cos A cos C – sin A sin C = 0.

Question 10.
In âˆ†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 right âˆ†PQR, âˆ Q = 90Â°, PQ = 5 cm and PR + QR = 25 cm

â‡’ PR = 25 – QR
By Pythagoras theorem, we have
PR2 = PQ2 + QR2
(25 – QR)2 = (5)2 + QR2
â‡’ 625 + QR2 – 50 QR = 25 + QR2
â‡’ QR2 – 50QR – QR2 = 25 – 625
â‡’ – 50 QR = – 600
â‡’ QR = $$\frac{-600}{-50}$$ = 12 cm
and PR = 25 – 12 = 13 cm
âˆ´ sin P = $$\frac{\mathrm{QR}}{\mathrm{PR}}=\frac{12}{13}$$
cos P = $$\frac{\mathrm{PQ}}{\mathrm{PR}}=\frac{5}{13}$$
tan P = $$\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{12}{5}$$
Hence, sin P = $$\frac{12}{13}$$, cos P = $$\frac{5}{13}$$ and tan P = $$\frac{12}{5}$$.

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 âˆ A.
(iii) cos A is the abbreviation used, for the cosecant of âˆ A.
(iv) cot A is the product of cot and A.
(v) sin Î¸ = $$\frac{4}{3}$$ for some angle Î¸.
Solution :
(i) False. Because tan 60Â° = âˆš3 > 1
(ii) True. Because value of sec A is always â‰¥ 1.
(iii) False. Because cos A is abbreviation used for cosine A.
(iv) False. Because cot is meaningless without an angle.
(v) False. Because sin Î¸ > 1.