5 Marks Questions

Question 15 Marks

Evaluate the following:
If A = 60° and B = 30°, verify that:
$\sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}$

Answer

$\text{A}=60^\circ$ and $\text{B}=30^\circ$
$\sin(\text{A}-\text{B})=\sin30^\circ=\frac12$
$\sin\text{A}\cos\text{B}=\cos\text{A}\sin\text{B}=\sin60^\circ\sin30^\circ$
$=\Big(\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}-\frac12\times\frac12\Big)=\Big(\frac{3}{4}-\frac{1}{4}\Big)$
$=\frac{2}{4}=\frac12$
$\therefore\ \sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}$

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Question 25 Marks

Evaluate the following:
If $\sin(\text{A}+\text{B})=1$ and $\cos(\text{A}-\text{B})=1,0^\circ\leq(\text{A}+\text{B})\leq90^\circ$ and $\text{A}>\text{B}$ then find A and B.

Answer

Here, $\sin(\text{A}+\text{B})=1$
$\Rightarrow\sin(\text{A}+\text{B})=\sin90^\circ$ $[\because\ \sin90^\circ=1]$
$\Rightarrow\text{A}+\text{B}=90^\circ\dots(\text{i})$
Also, $\cos(\text{A}-\text{B})=1$
$\Rightarrow\cos(\text{A}-\text{B})=\cos0^\circ$ $[\because\ \cos0^\circ=1]$
$\Rightarrow\text{A}-\text{B}=0^\circ\dots(\text{ii})$
Solving (i) and (ii), we get:
A = 45° and B = 45°.

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Question 35 Marks

Evaluate the following:
If A = 60° and B = 30°, verify that:
$\cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$

Answer

$\text{A}=60^\circ$ and $\text{B}=30^\circ$
Now, $\text{A}+\text{B}=60^\circ+30^\circ=90^\circ$
Also, $\text{A}-\text{B}=60^\circ-30^\circ=30^\circ$
$\cos(\text{A}+\text{B})=\cos90^\circ=0$
$\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$
$=\cos60^\circ\cos30^\circ-\sin60^\circ\sin30^\circ$
$=\Big(\frac{1}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times\frac12\Big)=\Big(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\Big)=0$
$\therefore\ \cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}$

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Question 45 Marks

Evaluate the following:
If $\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$ and $\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B},$ Find the values of:

$\sin75^\circ$
$\cos15^\circ.$

Answer

Let $\text{A}=45^\circ$ and $\text{B}=30^\circ$
As, $\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$
$\Rightarrow\sin(45^\circ+30^\circ)=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ$
$\Rightarrow\sin(75^\circ)=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac12$
$\Rightarrow\sin75^\circ=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}$
$\therefore\ \sin75^\circ=\frac{\sqrt{3}+1}{2\sqrt{2}}$
As, $\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}$
$\Rightarrow\cos(45^\circ-30^\circ)=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ$
$\Rightarrow\cos(15^\circ)=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac12$
$\Rightarrow\cos15^\circ=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}$
$\therefore\ \cos15^\circ=\frac{\sqrt{3}+1}{2\sqrt{2}}$
Disclaimer: $\cos15^\circ$ can also be calculated by taking A = 60° and B = 45°.

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Question 55 Marks

Evaluate the following:
Using the formula, $\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}},$ Find the value of $\tan60^\circ,$ it being given that $\tan30^\circ=\frac{1}{\sqrt{3}}.$

Answer

$\text{A}=30^\circ$
$\Rightarrow2\text{A}=2\times30^\circ=60^\circ$
By substiting the value of the given T-ratio, we get:
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}$
$\Rightarrow\tan60^\circ=\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$=\frac{2\times\frac{1}{\sqrt{3}}}{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{\frac{2}{\sqrt{3}}}{1-\frac13}$
$=\frac{2}{\sqrt{3}}\times\frac23=\sqrt{3}$
$\therefore\ \tan60^\circ=\sqrt{3}$

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Question 65 Marks

Evaluate the following:
If $3\text{x}=\text{cosec}\theta$ and $\frac{3}{\text{x}}=\cot\theta,$ find the value of $3\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big).$

Answer

$3\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)$
$=\frac93\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)$
$=\frac13\Big(9\text{x}^2-\frac{9}{\text{x}^2}\Big)$
$=\frac13\bigg[\big(3\text{x}\big)^2-\Big(\frac{3}{\text{x}}\Big)^2\bigg]$
$=\frac13\Big[\big(\text{cosec}\theta\big)^2-\big(\cot\theta\big)^2\Big]$
$=\frac13\big(\text{cosec}^2\theta-\cot^2\theta\big)$
$=\frac{1}{3}(1)$
$=\frac13$

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Question 75 Marks

Evaluate the following:
In the adjoining figure, $\triangle\text{ABC}$ is a right-angled triangle in which $\angle\text{B}=90^\circ,\angle\text{A}=30^\circ$ and AC = 20cm.
Find:

Answer

From the given right-angled triangle, we have:
$\frac{\text{BC}}{\text{AC}}=\sin30^\circ$
$\Rightarrow\frac{\text{BC}}{20}=\frac12$
$\Rightarrow\text{BC}=\frac{20}{2}=10\text{cm}$
Also, $\frac{\text{AB}}{\text{AC}}=\cos30^\circ$
$\Rightarrow\frac{\text{AB}}{20}=\frac{\sqrt{3}}{2}$
$\Rightarrow\text{AB}=\Big(20\times\frac{\sqrt{3}}{2}\Big)=10\sqrt{3}\text{cm}$
$\therefore\ \text{BC}=10\text{cm}$ and $\text{AB}=10\sqrt{3}\text{cm}$

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Question 85 Marks

Evaluate the following:
In the adjoining figure, $\triangle\text{ABC}$ is a right-angled at B and $\angle\text{A}=45^\circ,$ If $\text{AC}=3\sqrt{2}\text{cm},$
Find:

Answer

From the given right-angled $\triangle\text{ABC},$ we have:
$\frac{\text{BC}}{\text{AC}}=\sin45^\circ$
$\Rightarrow\frac{\text{BC}}{3\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow\text{BC}=3\text{cm}$
Also, $\frac{\text{AB}}{\text{AC}}=\cos45^\circ$
$\Rightarrow\frac{\text{AB}}{3\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow\text{AB}=3\text{cm}$
$\therefore\ \text{BC}=3\text{cm}$ and $\text{AC}=3\text{cm}$

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Question 95 Marks

Evaluate the following:
If $\tan(\text{A}-\text{B})=\frac{1}{\sqrt{3}}$ and $\tan(\text{A}+\text{B})=\sqrt{3},0^\circ<(\text{A}+\text{B})\leq90^\circ$ and $\text{A}>\text{B}$ then find A and B.

Answer

Here, $\tan(\text{A}-\text{B})=\frac{1}{\sqrt{3}}$
$\Rightarrow\tan(\text{A}-\text{B})=\tan30^\circ$ $\Big[\because\ \tan30^\circ=\frac{1}{\sqrt{3}}\Big]$
$\Rightarrow\text{A}-\text{B}=30^\circ\dots(\text{i})$
Also, $\tan(\text{A}+\text{B})=\sqrt{3}$
$\Rightarrow\tan(\text{A}+\text{B})=\tan60^\circ$ $\Big[\because\ \tan60^\circ=\sqrt{3}\Big]$
$\Rightarrow\text{A}+\text{B}=60^\circ\dots(\text{ii})$
Solving (i) and (ii), we get:
A = 45° and B = 15°.

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Question 105 Marks

Evaluate the following:
If A = 60° and B = 30°, verify that:
$\tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}$

Answer

$\text{A}=60^\circ$ and $\text{B}=30^\circ$
$\tan(\text{A}-\text{B})=\tan30^\circ=\frac{1}{\sqrt{3}}$
$\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}=\frac{\tan60^\circ-\tan30^\circ}{1+\tan60^\circ\tan30^\circ}$
$=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\Big(\sqrt{3}\times\frac{1}{\sqrt{3}}\Big)}=\frac12\times\Big(\frac{3-1}{\sqrt{3}}\Big)=\frac{1}{\sqrt{3}}$
$\therefore\ \tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}$

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Question 115 Marks

Evaluate the following:
Using the formula, $\sin\text{A}=\sqrt{\frac{1-\cos2\text{A}}{2}},$ Find the value of $\sin30^\circ,$ it being given that $\cos60^\circ=\frac{1}{2}.$

Answer

$\text{A}=30^\circ$
$\Rightarrow2\text{A}=2\times30^\circ=60^\circ$
By substiting the value of the given T-ratio, we get:
$\sin\text{A}=\sqrt{\frac{1-\cos2\text{A}}{2}}$
$\Rightarrow\sin30^\circ=\sqrt{\frac{1-\cos60^\circ}{2}}$
$\sqrt{\frac{1-\frac12}{2}}=\sqrt{\frac{\frac12}{2}}$
$=\sqrt{\frac14}=\frac12$
$\therefore\ \sin30^\circ=\frac12$

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