3 Marks Question

Question 13 Marks

Evaluate the following:
If A and B are acute angle such that $\tan\text{A}=\frac13,\tan\text{B}=\frac12$ and $\tan(\text{A}+\text{B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}},$ show that $\text{A}+\text{B}=45^\circ.$

Answer

Given:
$\tan\text{A}=\frac13$ and $\tan\text{B}=\frac12$
$\tan(\text{A}+\text{B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}$
On substituting these values in RHS of the expression, we get:
$\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}=\frac{\Big(\frac13+\frac12\Big)}{\Big(1-\frac13\times\frac12\Big)}$
$=\frac{\Big(\frac56\Big)}{\Big(1-\frac{1}{6}\Big)}=\frac{\Big(\frac56\Big)}{\Big(\frac56\Big)}=1$
$\Rightarrow\tan(\text{A}+\text{B})=1=\tan45^\circ$ $\big[\because\tan45^\circ=1\big]$
$\therefore\ \text{A}+\text{B}=45^\circ$

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Question 23 Marks

Evaluate the following:
Verify the following:
$2\sin45^\circ\cos45^\circ=\sin90^\circ$

Answer

$2\sin45^\circ\cos45^\circ$
$=2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}=1$
Also, $\sin90^\circ=1$
$\therefore\ 2\sin45^\circ\cos45^\circ=\sin90^\circ$

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Question 33 Marks

Evaluate the following:
If A = 30°, verify that:
$\sin2\text{A}=\frac{2\tan\text{A}}{1+\tan^2\text{A}}$

Answer

$\text{A}=30^\circ$
$\Rightarrow2\text{A}=2\times30^\circ=60^\circ$
$\sin2\text{A}=\sin60^\circ=\frac{\sqrt{3}}{2}$
$\frac{2\tan\text{A}}{1+\tan^2\text{A}}=\frac{2\tan30^\circ}{1+\tan^230^\circ}$
$=\frac{2\times\Big(\frac{1}{\sqrt{3}}\Big)}{1+\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{\Big(\frac{2}{\sqrt{3}}\Big)}{1+\frac13}$
$=\frac{\Big(\frac{2}{\sqrt{3}}\Big)}{\frac43}=\Big(\frac{2}{\sqrt{3}}\Big)\times\frac34=\frac{\sqrt{3}}{2}$
$\therefore\ \sin2\text{A}=\frac{2\tan\text{A}}{1+\tan^2\text{A}}$

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Question 43 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 $\text{B}=30^\circ$
$\cos(\text{A}-\text{B})=\cos30^\circ=\frac{\sqrt{3}}{2}$
$\cos\text{A}\cos\text{B}=\sin\text{A}\sin\text{B}\\=\cos60^\circ\cos30^\circ+\sin60^\circ\sin30^\circ$
$=\Big(\frac{1}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times\frac12\Big)=\Big(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\Big)$
$=2\times\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$
$\therefore\ \cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$

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Question 53 Marks

Evaluate the following:
If A = 30°, verify that:
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}$

Answer

$\text{A}=30^\circ$
$\Rightarrow2\text{A}=2\times30^\circ=60^\circ$
$\tan2\text{A}=\tan60^\circ=\sqrt{3}$
$\frac{2\tan\text{A}}{1-\tan^2\text{A}}=\frac{2\tan30^\circ}{1+\tan^230^\circ}$
$=\frac{2\times\Big(\frac{1}{\sqrt{3}}\Big)}{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{\Big(\frac{2}{\sqrt{3}}\Big)}{1-\frac13}$
$=\frac{\Big(\frac{2}{\sqrt{3}}\Big)}{\frac{2}{3}}=\Big(\frac{2}{\sqrt{3}}\Big)\times\frac32=\sqrt{3}$
$\therefore\ \tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}$

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Question 63 Marks

Evaluate the following:
If A = 30°, verify that:
$\cos2\text{A}=\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}$

Answer

$\text{A}=30^\circ$
$\Rightarrow2\text{A}=2\times30^\circ=60^\circ$
$\cos2\text{A}=\cos60^\circ=\frac12$
$\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}=\frac{1-\tan^230^\circ}{1+\tan^230^\circ}$
$=\frac{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}{1+\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{\Big(1-\frac{1}{{3}}\Big)}{\Big(1+\frac13\Big)}$
$=\Big(\frac{2}{3}\Big)\times\frac34=\frac12$
$\therefore\ \cos2\text{A}=\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}$

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Question 73 Marks

Evaluate the following:
If $\sin(\text{A}-\text{B})=\frac12$ and $\cos(\text{A}+\text{B})=\frac12,0^\circ<(\text{A}+\text{B})<90^\circ$ and $\text{A}>\text{B}$ then find A and B.

Answer

Here, $\sin(\text{A}-\text{B})=\frac12$
$\Rightarrow\sin(\text{A}-\text{B})=\sin30^\circ$ $\Big[\because\ \sin30^\circ=\frac12\Big]$
$\Rightarrow\text{A}-\text{B}=30^\circ\dots(\text{i})$
Also, $\cos(\text{A}+\text{B})=\frac12$
$\Rightarrow\cos(\text{A}+\text{B})=\cos60^\circ$ $\Big[\because\ \cos60^\circ=\frac12\Big]$
$\Rightarrow\text{A}+\text{B}=60^\circ\dots(\text{ii})$
Solving (i) and (ii), we get:
A = 45° and B = 15°.

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Question 83 Marks

Evaluate the following:
Using the formula, $\cos\text{A}=\sqrt{\frac{1+\cos2\text{A}}{2}},$ Find the value of $\cos30^\circ,$ it being given that $\cos60^\circ=\frac12.$

Answer

$\text{A}=30^\circ$
$\Rightarrow2\text{A}=2\times30^\circ=60^\circ$
By substiting the value of the given T-ratio, we get:
$\cos\text{A}=\sqrt{\frac{1+\cos2\text{A}}{2}}$
$\Rightarrow\cos30^\circ=\sqrt{\frac{1+\cos60^\circ}{2}}$
$\sqrt{\frac{1+\frac12}{2}}=\sqrt{\frac{\frac32}{2}}$
$=\sqrt{\frac34}=\frac{\sqrt{3}}{2}$
$\therefore\ \cos30^\circ=\frac{\sqrt{3}}{2}$

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Question 93 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 $\text{B}=30^\circ$
Now, $\text{A}+\text{B}=60^\circ+30^\circ=90^\circ$
Also, $\text{A}-\text{B}=60^\circ-30^\circ=30^\circ$
$\sin(\text{A}+\text{B})=\sin90^\circ=1$
$\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$
$=\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ$
$=\Big(\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}+\frac12\times\frac12\Big)=\Big(\frac34+\frac14\Big)=1$
$\therefore\ \sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$

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Question 103 Marks

Evaluate the following:
In the adjoining figure, $\triangle\text{ABC}$ is a right-angled at B and $\angle\text{A}=30^\circ,$ If BC = 6cm.
Find:

Answer

From the given right-angled triangle, we have:
$\frac{\text{BC}}{\text{AB}}=\tan30^\circ$
$\Rightarrow\frac{6}{\text{AB}}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{AB}=6\sqrt{3}\text{cm}$
Also, $\frac{\text{BC}}{\text{AC}}=\sin30^\circ$
$\Rightarrow\frac{6}{\text{AC}}=\frac{1}{2}$
$\Rightarrow\text{AC}=(2\times6)=12\text{cm}$
$\therefore\ \text{AB}=6\sqrt{3}\text{cm}$ and $\text{AC}=12\text{cm}$

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