[3 marks sum]

Question 13 Marks

If $\tan = 0.75,$ find the other trigonometric ratios for $A.$

Answer

$\tan A =0.75=\frac{75}{100}=\frac{3}{4}=\frac{\text { Perpendicular }}{\text { Base }} $
Hypotenuse
$=\sqrt{(\text { Perpendicular })^2+(\text { Base })^2} $
$=\sqrt{3^2+4^2} $
$=\sqrt{9+16} $
$=\sqrt{25} $
$=5 $
$\sin A =\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{3}{5}=0.6 $
$\cos A =\frac{\text { Base }}{\text { Hypotenuse }}=\frac{4}{5}=0.8 $
$\operatorname{cosec} A =\frac{1}{\sin A }=\frac{5}{3}=1.66 $
$\sec A =\frac{1}{\cos A }=\frac{5}{4}=1.25 $
$\operatorname{cost} A =\frac{1}{\tan A }=\frac{4}{3}=1.33 .$

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Question 23 Marks

If $\sin \theta=\frac{8}{17}$, find the other five trigonometric ratios.

Answer

$\sin \theta=\frac{8}{17}=\frac{\text { Perpendicular }}{\text { Hypotenuse }} $
Base
$=\sqrt{(\text { Hypotenuse })^2-(\text { Perpendicular })^2} $
$=\sqrt{17^2-8^2} $
$=\sqrt{225} $
$=15 $
$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{15}{17} $
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{8}{15} $
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{17}{8} $
$\sec \theta=\frac{1}{\cos \theta}=\frac{17}{15} $
$\cot \theta=\frac{1}{\tan \theta}=\frac{15}{8} .$

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Question 33 Marks

If $\cos B=\frac{1}{3}$ and $\angle C=90^{\circ}$, find $\sin A$, and $B$ and $\cot A$.

Answer

$\cos B=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{B C}{A B} $
$ (A B)^2=(A C)^2+(B C)^2$
$ \Rightarrow A C=\sqrt{(A B)^2-(B C)^2} $
$ \Rightarrow A C=\sqrt{3^2-1} $
$ =\sqrt{9-1} $
$=2 \sqrt{2} $
$ \sin A=\frac{B C}{A B}=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{1}{3} $
$ \tan B=\frac{A C}{B C}=\frac{\text { Perpendicular }}{\text { Base }}=2 \sqrt{2} $
$ \cot A=\frac{1}{\tan A}=\frac{\text { BAse }}{\text { Perpendicular }}=\frac{A C}{B C}=2 \sqrt{2} .$

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Question 43 Marks

If $5 \tan \theta=12$, find the value of $\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}$.

Answer

$5 \tan \theta=12 $
$\Rightarrow \tan \theta=\frac{12}{5}=\frac{\text { Perpendicular }}{\text { Base }} $
Hypotenuse
$=\sqrt{(\text { Perpendicular })^2+(\text { Base })^2} $
$=\sqrt{(12)^2+(5)^2} $
$=\sqrt{144+25} $
$=\sqrt{169} $
$=13 $
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{12}{13},$
$ \cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{5}{13} $
$\Rightarrow \frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta} $
$=\frac{2 \times \frac{12}{13}-3 \times \frac{5}{13}}{4 \times \frac{12}{13}-9 \times \frac{5}{13}} $
$=\frac{24-15}{48-45} $
$=\frac{9}{3} $
$=3$

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Question 53 Marks

If $5 \cos \theta=3$, find the value of $\frac{4 \cos \theta-\sin \theta}{2 \cos \theta+\sin \theta}$

Answer

$\text { If } 5 \cos \theta=3 $
$\Rightarrow \cos \theta=\frac{3}{5}=\frac{\text { Base }}{\text { Hypotenuse }} $
Perpendicular
$=\sqrt{(\text { Hypotenuse })^2-(\text { Base })^2} $
$=\sqrt{(5)^2-(3)^2} $
$=\sqrt{25-9} $
$=\sqrt{16} $
$=4 $
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{4}{5} $
$4 \cos \theta-\sin \theta $
$2 \cos \theta+\sin \theta $
$=\frac{4 \times \frac{3}{5}-\frac{4}{5}}{2 \times \frac{3}{5}+\frac{4}{5}} $
$=\frac{\frac{12}{5}-\frac{4}{5}}{\frac{6}{5}+\frac{4}{5}} $
$=\frac{\frac{8}{5}}{\frac{10}{5}} $
$=\frac{4}{5} . $

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Question 63 Marks

If $8 \tan A = 15,$ find $\sin A - \cos A.$

Answer

$8 \tan A =15 $
$\Rightarrow \tan A =\frac{15}{8}=\frac{\text { Perpendicular }}{\text { Base }} $
Hypotenuse
$=\sqrt{(\text { Perpendicular })^2+(\text { Base })^2} $
$=\sqrt{(15)^2+(8) 2} $
$=\sqrt{225+64} $
$=\sqrt{289} $
$=17 $
$\sin A -\cos A =\frac{\text { Perpendicular }}{\text { Hypotenuse }}-\frac{\text { Base }}{\text { Hypotenuse }} $
$=\frac{15}{17}-\frac{8}{17} $
$=\frac{15-8}{17} $
$\sin A -\cos A =\frac{7}{17} .$

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Question 73 Marks

If $4 \sin \theta=3 \cos \theta$, find $\frac{6 \sin \theta-2 \cos \theta}{6 \sin \theta+2 \cos \theta}$

Answer

$4 \sin \theta=3 \cos \theta $
$\Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{3}{4} $
$\Rightarrow \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{3}{4} $
$\Rightarrow \cot \theta=\frac{1}{\tan \theta}=\frac{4}{3} $
$\frac{6 \sin \theta-2 \cos \theta}{6 \sin \theta+2 \cos \theta} $
$=\frac{6 \times \frac{\sin \theta}{\cos \theta}-2 \times \frac{\cos \theta}{\cos \theta}}{6 \times \frac{\sin \theta}{\cos \theta}+2 \times \frac{\cos \theta}{\cos \theta}} $
$=\frac{6 \tan \theta-2}{6 \tan \theta+2} $
$=\frac{6 \times \frac{3}{4}-2}{6 \times \frac{3}{4}+2} $
$=\frac{\frac{9}{2}-2}{\frac{9}{2}+2} $
$=\frac{\frac{9-4}{2}}{\frac{9+4}{2}} $
$=\frac{5}{13} .$

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Question 83 Marks

If $4 \sin\theta = 3 \cos\theta $, find $tan^2\theta + \cot^2\theta$

Answer

$4 \sin \theta=3 \cos \theta $
$ \Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{3}{4}$
$ \Rightarrow \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{3}{4} $
$\Rightarrow \cot \theta=\frac{1}{\tan \theta}=\frac{4}{3}$
$ \tan ^2 \theta+\cot ^2 \theta $
$ =\left(\frac{3}{4}\right)^2+\left(\frac{4}{3}\right)^2$
$ =\frac{9}{16}+\frac{16}{9} $
$ =\frac{81+256}{144} $
$ =\frac{337}{144} .$

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Question 93 Marks

If $24\cos\theta = 7 \sin\theta ,$ find $\sin\theta + \cos\theta .$

Answer

$24 \cos \theta=7 \sin \theta $
$\Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{24}{7} $
$\Rightarrow \tan \theta=\frac{24}{7}=\frac{\text { Perpendicular }}{\text { Base }}$
Hypotenuse
$=\sqrt{(\text { Perpendicular })^2+(\text { Base })^2} $
$=\sqrt{(24)^2+(7)^2} $
$=\sqrt{576+49} $
$=\sqrt{625} $
$=25 $
$\sin \theta+\cos \theta $
$=\frac{\text { Perpendicular }}{\text { Hypotenuse }}+\frac{\text { Base }}{\text { Hypotenuse }} $
$=\frac{24}{25}+\frac{7}{25} $
$=\frac{24+7}{25} $
$=\frac{31}{25} .$

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Question 103 Marks

In the given figure, $\angle Q=90^{\circ}, P S$ is a median om $Q R$ from $P$, and $R T$ divides $P Q$ in the ratio $1: 2$. Find: $\frac{\tan \angle PSQ }{\tan \angle PRQ }$

Answer

As $P S$ is the median on $QR$ from $P$.
$ \therefore Q S=S R$
$\Rightarrow Q R=2 Q S $
and $R T$ divides $P Q$ in the ratio $1: 2$
$ \therefore QT = x$ and $PT =2 x$
$\Rightarrow PQ =3 x$
$\frac{\tan \angle PSQ }{\tan \angle PRQ }$
$=\frac{\frac{ PQ }{ QS }}{\frac{ PQ }{ QR }}$
$=\frac{ PQ }{ QS } \times \frac{ QR }{ PQ }$
$=\frac{2 QS }{ QS }$
$=2 . $

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Question 113 Marks

In the given figure, $AC = 13\ cm, BC = 12 \ cm$ and $\angle B = 90^\circ .$ Without using tables, find the values of: $\sin A \cos A$

Answer

$\triangle A B C$ is a right$-$angled triangle.
$\therefore A C^2=A B^2+B C^2$
$\Rightarrow A B^2$
$=A C^2-B C^2$
$=13^2-12^2$
$=169-144$
$=25$
$\Rightarrow A B=5 \ cm$
$\sin A=\frac{B C}{A C}=\frac{12}{13}$
$\cos A=\frac{A B}{A C}=\frac{5}{13}$
$\sin A \cos A$
$=\frac{12}{13} \times \frac{5}{13}$
$=\frac{60}{169} . $

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Question 123 Marks

In the given figure, $\text{PQR}$ is a triangle, in which $Q S \perp P R, Q S=3\ cm , P S=4\ cm$ and $Q R=12\ cm$, find the value of: $4 \sin ^2 R-\frac{1}{\tan ^2 P}$

Answer

$\sin R=\frac{Q S}{Q R}=\frac{3}{12}$
$4 \sin ^2 R-\frac{1}{\tan ^2 P} $
$ =4 \sin ^2 R-\cot ^2 P $
$ =4 \times\left(\frac{3}{12}\right)^2-\left(\frac{\cos P}{\sin P}\right)^2$
$ =4 \times \frac{9}{144}-\left(\frac{\frac{4}{5}}{\frac{3}{5}}\right)^2 $
$ =\frac{9}{36}-\frac{16}{9} $
$ =-\frac{55}{36} .$

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MATHEMATICS [3 marks sum]

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