Exercise - 8.1-Solution-Introduction To Trigonometry - Chapter 8- Class 10

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Exercise 8.1

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:

⇨ In △ ABC

⇨ From Pythagoras Theorem

$Exercise - 8.1-Solution-Introduction To Trigonometry - Chapter 8- Class 10 - Mathematics$

$\large \therefore AC^2=AB^2+BC^2$

$\large \Rightarrow AC^2=24^2+7^2$

$\large \Rightarrow AC^2=576+49$

$\large \Rightarrow AC^2=625$

$\large \Rightarrow AC^2=25^2$

$\large \Rightarrow AC=25 cm$

(i) sin A, cos A

Solution:

$\large sin A=\frac{p}{h}=\frac{BC}{AC}=\frac{7}{25}$

$\large cos A=\frac{b}{h}=\frac{AB}{AC}=\frac{24}{25}$

(ii) sin C, cos C

Solution:

$\large sin C=\frac{p}{h}=\frac{AB}{AC}=\frac{24}{25}$

$\large cos C=\frac{b}{h}=\frac{BC}{AC}=\frac{7}{25}$

2. In Fig. 8.13, find tan P – cot R.

Solution:

⇨ From Pythagoras Theorem

$\large h^2=p^2+b^2$

$\large \therefore PR^2=PQ^2+QR^2$

$\large \therefore 13^2=12^2+QR^2$

$\large \Rightarrow QR^2=13^2-12^2$

$\large \Rightarrow QR^2=169-144$

$\large \Rightarrow QR^2=25$

$\large \Rightarrow QR^2=5^2$

$\large \Rightarrow QR=5cm$

⇨Now,

$\large tanP=\frac{p}{b}=\frac{QR}{PQ}=\frac{5}{12}$

$\large cotR=\frac{b}{p}=\frac{QR}{PQ}=\frac{5}{12}$

$\large =tanP-cotR$

$\large =\frac{5}{12}-\frac{5}{12}$

$\large =0$

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

Solution:

⇨Given, $\large sinA=\frac{3}{4}$

⇨We Know that,

$\large sinA=\frac{3}{4}=\frac{p}{h}$

⇨Therefore, p=3, h=4, b=?

$\large \Rightarrow 4^2=3^2+b^2$

$\large \Rightarrow 4^2-3^2=b^2$

$\large \Rightarrow b^2=16-9$

$\large \Rightarrow b^2=7$

$\large \Rightarrow b=\sqrt{7}$

⇨Now,

$\large cosA=\frac{b}{h}=\frac{\sqrt{7}}{4}$

$\large tanA=\frac{p}{b}=\frac{3}{\sqrt{7}}$

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

Solution:

⇨Given, $\large 15cotA=8$

$\large \Rightarrow cotA=\frac{8}{15}$

⇨We Know That,

$\large cotA=\frac{8}{15}=\frac{b}{p}$

⇨Therefore, b=8, p=15, h=?

⇨From Pythagoras Theorem

$\large h^2=p^2+b^2$

$\large \Rightarrow h^2=15^2+8^2$

$\large \Rightarrow h^2=225+64$

$\large \Rightarrow h^2=289$

$\large \Rightarrow h^2=17^2$

$\large \Rightarrow h=17$

⇨Now,

$\large sinA=\frac{p}{h}=\frac{15}{17}$

$\large secA=\frac{h}{b}=\frac{17}{8}$

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

Solution:

⇨Given, $\large sec\Theta=\frac{13}{12}$

⇨We Know That,

$\large sec\Theta=\frac{13}{12}=\frac{h}{b}$

⇨Therefore, h=13, b=12, p=?

⇨From Pythagoras Theorem

$\large h^2=p^2+b^2$

$\large \Rightarrow 13^2=p^2+12^2$

$\large \Rightarrow 13^2-12^2=p^2$

$\large \Rightarrow p^2=169-144$

$\large \Rightarrow p^2=25$

$\large \Rightarrow p^2=5^2$

$\large \Rightarrow p=5$

⇨Now,

$\large sin\Theta =\frac{p}{h}=\frac{5}{13}$

$\large cos\Theta =\frac{b}{h}=\frac{12}{13}$

$\large tan\Theta =\frac{p}{b}=\frac{5}{12}$

$\large cot\Theta =\frac{b}{p}=\frac{12}{5}$

$\large cosec\Theta =\frac{h}{p}=\frac{13}{5}$

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

Solution:

⇨ For acute angle

⇨ cos A = A

⇨ cos B = B

⇨ Now

⇨ cos A = cos B

⇨ A = B

7. If cot θ = 7/8, evaluate

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

(ii) $\large cot^2\theta$

Solution:

⇨ Given

⇨ cot θ = 7/8 = b/p

⇨ Therefore b = 7, p = 8, h = ?

⇨From Pythagoras Theorem

$\large h^2=p^2+b^2$

$\large \Rightarrow h^{2}=8^{2}+7^{2}$

$\large \Rightarrow h^{2}=68+49$

$\large \Rightarrow h^{2}=113$

$\large \Rightarrow h=\sqrt{113}$

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

$\large = \frac{1^{2}-sin^2\theta}{1^{2}-cos^2\theta}$

$\large = \frac{1-(\frac{p}{h})^2}{1-(\frac{b}{h})^2}$

$\large = \frac{1-(\frac{8}{\sqrt{113}})^2}{1-(\frac{7}{\sqrt{113}})^2}$

$\large = \frac{1-\frac{64}{113}}{1-\frac{49}{113}}$

$\large = \frac{\frac{113-64}{113}}{\frac{113-49}{113}}$

$\large = \frac{113-64}{113}\times \frac{113}{113-49}$

$\large = \frac{113-64}{113-49}$

$\large = \frac{49}{64}$

(ii) $\large cot^2\theta$

$\large =\left ( \frac{b}{p} \right )^{2}$

$\large =\left ( \frac{7}{8} \right )^{2}$

$\large =\frac{49}{64}$

8. If 3 cot A = 4, check whether $\frac{1-tan^{2}A}{1+tan^{2}A}=cos^2A-sin^2A$ or not.

Solution :

⇨ Given

⇨ 3cot A = 4

⇨ cot A = 4/3 = b/p

⇨ Therefore b = 4, p = 3, h = ?

⇨From Pythagoras Theorem

$\large h^2=p^2+b^2$

$\Rightarrow h^2=3^2+4^2$

$\Rightarrow h^2=9+16$

$\Rightarrow h^2=25$

$\Rightarrow h=5$

⇨ L.H.S

$\frac{1-tan^2A}{1+tan^2A}$

$=\frac{1-\left ( \frac{p}{b} \right )^{2}}{1+\left ( \frac{p}{b} \right )^{2}}$

$=\frac{1-\left ( \frac{3}{4} \right )^{2}}{1+\left ( \frac{3}{4} \right )^{2}}$

$=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$

$=\frac{\frac{16-9}{16}}{\frac{16+9}{16}}$

$=\frac{16-9}{16}\times \frac{16}{16+9}$

$=\frac{7}{25}$

⇨ R.H.S

$cos^2A-sin^2A$

$= \left ( \frac{b}{h} \right )^{2}-\left ( \frac{p}{h} \right )^{2}$

$= \left ( \frac{4}{5} \right )^{2}-\left ( \frac{3}{5} \right )^{2}$

$=\frac{16}{25}-\frac{9}{25}$

$=\frac{16-9}{25}$

$=\frac{7}{25}$

⇨Therefore L.H.S = R.H.S

9. In triangle 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:

⇨ Given

⇨ tan A = 1/√3 = p/b

⇨ Therefore p = BC = 1 , b = AB = √3, h = AC = ?

⇨ In △ ABC

Exercise 8.1 9.In triangle 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

⇨ From Pythagoras Theorem

$Exercise - 8.1-Solution-Introduction To Trigonometry - Chapter 8- Class 10 - Mathematics$

$\large \therefore AC^2=AB^2+BC^2$

$\large \Rightarrow AC^2=\left ( \sqrt{3} \right )^2+\left ( 1 \right )^2$

$\large \Rightarrow AC^2=3+1$

$\large \Rightarrow AC^2=4$

$\large \Rightarrow AC^2=2^2$

$\large \Rightarrow AC=2$

(i) sin A cos C + cos A sin C

$\large = \frac{BC}{AC}\times \frac{BC}{AC}+\frac{AB}{AC}\times \frac{AB}{AC}$

$\large = \frac{BC^{2}}{AC^{2}}+\frac{AB^{2}}{AC^{2}}$

$\large = \frac{1^{2}}{2^{2}}+\frac{(\sqrt{3})^{2}}{2^{2}}$

$\large = \frac{1}{4}+\frac{3}{4}$

$\large = \frac{1+3}{4}$

$\large = \frac{4}{4}$

$\large =1$

(ii) cos A cos C – sin A sin C

$\large =\frac{AB}{AC}\times \frac{BC}{AC}-\frac{BC}{AC}\times \frac{AB}{AC}$

$\large =\frac{\sqrt{3}}{2}\times \frac{1}{2}-\frac{1}{2}\times \frac{\sqrt{3}}{2}$

$\large =\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

$\large =0$

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:

⇨ Given

⇨ PQ = 5 cm

⇨ PR + QR = 25 cm

⇨ PR = 25 - QR .....(i)

⇨ In ∆ PQR

⇨ From Pythagoras Theorem

$Exercise - 8.1-Solution-Introduction To Trigonometry - Chapter 8- Class 10 - Mathematics$

$\large \Rightarrow PR^2= QR^2+PQ^2$

$\large \Rightarrow \left ( 25-QR \right )^2= QR^2+5^2$

$\large \Rightarrow 25^2-2\times 25\times QR+QR^2= QR^2+25$

$\large \Rightarrow 625-50QR+QR^2-QR^2-25=0$

$\large \Rightarrow 600=50QR$

$\large \Rightarrow QR=\frac{600}{50}$

$\large \Rightarrow QR=12$

∴Equation (i)

⇨ PR = 25 - 12

⇨ PR = 13 cm

∴ sin P = p/h = QR/PR

⇨ sin P = 12/13

∴ cos P = b/h = PQ/PR

⇨ cos P = 5/13

∴ tan P = p/b = QR/PQ

⇨ tan P = 12/5

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 = 12/5 for some value of angle 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 = 4/3 for some angle θ.

Solution:

i. False, because tan 45° = 1.

ii. False, because sec A ≠ 12/5 for any value of A.

iii. True.

iv. False.

v. False, because sin θ ≠ 4/3 for any value of θ.

Exercise - 8.1-Solution-Introduction To Trigonometry - Chapter 8- Class 10 - Mathematics

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