Tag: Maths Class 10th Trigonometry Solution

1. In △ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A
(ii) sin C, cos C

Answer 1.
AC (hypotenuse) = √(AB² + BC²) = √(24² + 7²) = √(576 + 49) = √625 = 25.

(i) For angle A: opposite = BC = 7, adjacent = AB = 24, hypotenuse = 25.
${\displaystyle \sin A=\frac{7}{25},\cos A=\frac{24}{25}\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}}$

(ii) For angle C: opposite = AB = 24, adjacent = BC = 7, hypotenuse = 25.
${\displaystyle \sin C=\frac{24}{25},\cos C=\frac{7}{25}\mathrm{}}$

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2. In Fig. 8.13, find $\tan P-\cot R\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

Answer 2.
 If Fig. 8.13 is the standard right triangle with PQ = 4, QR = 3, PR = 5 (a common 3–4–5 right triangle), then
$\tan P={\displaystyle \frac{QR}{PQ}}={\displaystyle \frac{3}{4}}$ and $\cot R={\displaystyle \frac{QR}{PQ}}={\displaystyle \frac{3}{4}}$, so $\tan P-\cot R=0$

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3. If $\sin A={\displaystyle \frac{3}{4}}$, calculate $\cos A$ and $\tan A$.

Answer 3.
$\cos A=\sqrt{1-{\sin }^{2}A}=\sqrt{1-{\displaystyle \frac{9}{16}}}=\sqrt{{\displaystyle \frac{7}{16}}}={\displaystyle \frac{\sqrt{7}}{4}}\mathrm{}$

$\tan A={\displaystyle \frac{\sin A}{\cos A}}={\displaystyle \frac{3\mathrm{/}4}{\sqrt{7}\mathrm{/}4}}={\displaystyle \frac{3}{\sqrt{7}}}={\displaystyle \frac{3\sqrt{7}}{7}}\mathrm{}$

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

Answer 4.
$\cot A={\displaystyle \frac{8}{15}}\mathrm{.}$ Then ${\csc }^{2}A=1+{\cot }^{2}A=1+{\displaystyle \frac{64}{225}}={\displaystyle \frac{289}{225}}$, so $\csc A={\displaystyle \frac{17}{15}}$(positive for acute angle).
Thus $\sin A={\displaystyle \frac{1}{\csc A}}={\displaystyle \frac{15}{17}}\mathrm{}$

Also $\cos A=\cot A\cdot \sin A={\displaystyle \frac{8}{15}}\cdot {\displaystyle \frac{15}{17}}={\displaystyle \frac{8}{17}}$, so $\sec A={\displaystyle \frac{1}{\cos A}}={\displaystyle \frac{17}{8}}\mathrm{}$

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5. Given $\sec \theta ={\displaystyle \frac{13}{12}}$, calculate all other trigonometric ratios.

Answer 5.
$\cos \theta ={\displaystyle \frac{12}{13}}\mathrm{.}$
$$

$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-{\displaystyle \frac{144}{169}}}=\sqrt{{\displaystyle \frac{25}{169}}}={\displaystyle \frac{5}{13}}\mathrm{}$
$$

$\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}={\displaystyle \frac{5\mathrm{/}13}{12\mathrm{/}13}}={\displaystyle \frac{5}{12}}\mathrm{.}$

Reciprocals: $\csc \theta ={\displaystyle \frac{13}{5}},\cot \theta ={\displaystyle \frac{12}{5}}\mathrm{.}$

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

Answer 6.
On $[0^{\circ },{90}^{\circ }]$ the cosine function is strictly decreasing and therefore one-to-one. Hence $\cos A=\cos B$(with A and B acute) implies $A=B$.
(Equivalently: construct right triangles with same cosine value; corresponding sides/hypotenuse ratios match, triangles are similar and the acute angles equal.)

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7. If $\cot \theta ={\displaystyle \frac{7}{8}}$​, evaluate:

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

(ii) ${\cot }^2\theta \mathrm{.}$

Answer 7.
Given $\cot \theta ={\displaystyle \frac{7}{8}}$. Let $\sin \theta =s$, $\cos \theta =c$. From $\cot ={\displaystyle \frac{c}{s}}={\displaystyle \frac{7}{8}}$ we get (as usual) $s={\displaystyle \frac{8}{\sqrt{113}}}$, $c={\displaystyle \frac{7}{\sqrt{113}}}$ (since $s^2+c^2=1$ gives factor 113).

(i)

$$\frac{1+\sin \theta }{1-\sin \theta }=\frac{1+\frac{8}{\sqrt{113}}}{1-\frac{8}{\sqrt{113}}}=\frac{\sqrt{113}+8}{\sqrt{113}-8}\mathrm{.}$$
$$\frac{1+\cos \theta }{1-\cos \theta }=\frac{1+\frac{7}{\sqrt{113}}}{1-\frac{7}{\sqrt{113}}}=\frac{\sqrt{113}+7}{\sqrt{113}-7}\mathrm{.}$$

(You can rationalize these if you prefer a form without roots in denominators.)

(ii) ${\displaystyle {\cot }^2\theta =(\frac{7}{8}{)}^2=\frac{49}{64}\mathrm{.}}$

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8. If $3\cot A=4$, check whether ${\displaystyle \frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}}={\cos }^{2}A-{\sin }^{2}A$

Answer 8.
From $3\cot A=4$ we get $\cot A={\displaystyle \frac{4}{3}}$ so $\tan A={\displaystyle \frac{3}{4}}$. Compute LHS:

$$\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}=\frac{1-(3\mathrm{/}4{)}^2}{1+(3\mathrm{/}4{)}^2}=\frac{1-9\mathrm{/}16}{1+9\mathrm{/}16}=\frac{7\mathrm{/}16}{25\mathrm{/}16}=\frac{7}{25}\mathrm{.}$$

RHS: ${\cos }^{2}A-{\sin }^{2}A=\cos 2A$. Using triangle with legs 3 and 4 (hypotenuse 5): $\sin A={\displaystyle \frac{3}{5}},\cos A={\displaystyle \frac{4}{5}}$ Then ${\cos }^{2}A-{\sin }^{2}A={\displaystyle \frac{16}{25}}-{\displaystyle \frac{9}{25}}={\displaystyle \frac{7}{25}}\mathrm{}$

So both sides equal ${\displaystyle \frac{7}{25}}$. The equality holds (this is a standard identity: $\cos 2A={\displaystyle \frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}}$

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9. In triangle ABC, right-angled at B, if $\tan A={\displaystyle \frac{1}{3}}$ find the value of:

(i) $\sin A\cos C+\cos A\sin C$
(ii) $\cos A\cos C-\sin A\sin C$

Answer 9.
In a right triangle with right angle at B, $A+C={90}^{\circ }\mathrm{}$

(i) $\sin A\cos C+\cos A\sin C=\sin (A+C)=\sin {90}^{\circ }=1.$

(ii) $\cos A\cos C-\sin A\sin C=\cos (A+C)=\cos {90}^{\circ }=0$

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10. In △PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.

Answer 10.
Let $QR=x$. Then $PR=25-x$. Right triangle gives $P{Q}^2+Q{R}^2=P{R}^2$
$5^2+x^2=(25-x{)}^2=625-50x+x^2\mathrm{.}$ Cancel $x^2$: $25=625-50x$ ⇒ $50x=600$ ⇒ $x=12.$

So $QR=12,PR=13.$ For angle P: opposite = QR = 12, adjacent = PQ = 5, hypotenuse = PR = 13.

${\displaystyle \sin P=\frac{12}{13},\cos P=\frac{5}{13},\tan P=\frac{12}{5}\mathrm{.}}$

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

(i) The value of $\tan A$ is always less than 1.
(ii) $\sec A={\displaystyle \frac{12}{5}}$ 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) $\sin \theta ={\displaystyle \frac{4}{3}}$ for some angle $\theta$.

Answer 11.
(i) False. $\tan A$ can be greater than 1 (e.g. $A={60}^{\circ }$gives $\tan {60}^{\circ }=\sqrt{3}>1$.
(ii) True. $\sec A={\displaystyle \frac{12}{5}}$ means $\cos A={\displaystyle \frac{5}{12}}$, which is a valid cosine value for some acute angle (so such an angle exists).
(iii) False. $\cos A$ denotes cosine of A. Cosecant is written $\csc A$
(iv) False. $\cot A$ means cotangent of A, not the product “cot × A.”
(v) False. For real angles $\mathrm{\mid }\sin \theta \mathrm{\mid }\le 1.$

 ${\displaystyle \frac{4}{3}}>1$, so $\sin \theta ={\displaystyle \frac{4}{3}}$ is impossible for a real angle.