Exercise 3.10-Additional Questions - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions Solved
Question 1.
Given $a=8, b=9, c=10$, find all the angles.
Solution:
To find $\mathrm{A}$, use the formula
$
\begin{aligned}
a^2 & =b^2+c^2-2 b c \cos \mathrm{A} \\
\cos \mathrm{A} & =\frac{b^2+c^2-a^2}{2 b c}=\frac{81+100-64}{180}=\frac{117}{180} \\
\mathrm{~A} & =49^{\circ} 28^{\prime}
\end{aligned}
$
Similarly $\quad \cos \mathrm{B}=\frac{c^2+a^2-b^2}{2 c a}=\frac{100+64-81}{180}=\frac{83}{180}$
But
$
\begin{aligned}
\mathrm{B} & =58^{\circ} 51^{\prime} \\
\mathrm{A}+\mathrm{B}+\mathrm{C} & =180^{\circ} \\
\therefore \quad \mathrm{C} & =180^{\circ}-\left(49^{\circ} 28^{\prime}+58^{\circ} 51^{\prime}\right) \\
& =71^{\circ} 41^{\prime} \\
\therefore \mathrm{A} & =49^{\circ} 28^{\prime}, \mathrm{B}=58^{\circ} 51^{\prime}, \mathrm{C}=71^{\circ} 41^{\prime}
\end{aligned}
$

Question 2.
Given $\mathrm{a}=31, \mathrm{~b}=42, \mathrm{c}=57$, find all the angles.
Solution:
Since the sides are larger quantities, use half angles formulae
$
\begin{aligned}
& s=\frac{a+b+c}{2}=65 \\
& \tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\left(\frac{23 \times 8}{65 \times 34}\right)^{1 / 2} \\
& \Rightarrow \quad \log \left[\tan \frac{\mathrm{A}}{2}\right]=\frac{1}{2}[\log 23+\log 8-\log 65-\log 34] \\
& =\frac{1}{2}[1.3617+0.9031-1.8129-1.5315] \\
& =\frac{1}{2}[-1.0796]=\frac{1}{2}[-2+0.9204] \\
& =\frac{1}{2}[2+0.9204]=1.4602 \\
& \Rightarrow \quad \frac{\mathrm{A}}{2}=16^{\circ} 6^{\prime} \Rightarrow \mathrm{A}=32^{\circ} 12^{\prime} \\
& \tan \frac{\mathrm{B}}{2}=\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}=\left(\frac{8 \times 34}{65 \times 23}\right)^{1 / 2} \\
& \Rightarrow \quad \log \left[\tan \frac{\mathrm{B}}{2}\right]=\frac{1}{2}[\log 8+\log 34-\log 65-\log 23] \\
& =\frac{1}{2}[-0.7400]=\frac{1}{2}[-2+1.2600]=\frac{1}{2}[2+1.2600]=1.6300 \\
& \Rightarrow \quad \frac{\mathrm{B}}{2}=23^{\circ} 6^{\prime} \Rightarrow \mathrm{B}=46^{\circ} 12^{\prime} \\
& \mathrm{C}=180-(\mathrm{A}+\mathrm{B})=101^{\circ} 36^{\prime} \\
& \therefore \mathrm{A}=32^{\circ} 12^{\prime} ; \mathrm{B}=46^{\circ} 12^{\prime} ; \mathrm{C}=101^{\circ} 36^{\prime} \\
&
\end{aligned}
$

Question 3.
In a triangle $\mathrm{ABC}, \mathrm{A}=35^{\circ} 17^{\prime} ; \mathrm{C}=45^{\circ} 13^{\prime} ; \mathrm{b}=42.1$ Solve the triangle
Solution:
The unknown parts are $B, \mathrm{a}, \mathrm{c}$,
$
\begin{aligned}
& \mathrm{B}=180-(\mathrm{A}+\mathrm{C})=180-\left(35^{\circ} 17^{\prime}+45^{\circ} 13^{\prime}\right) \\
& =99^{\circ} 30^{\prime}
\end{aligned}
$
To find sides, use sine formula

$
\begin{aligned}
& \frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin C} \\
& \Rightarrow \quad a=\frac{b \sin \mathrm{A}}{\sin \mathrm{B}}=\frac{42.1 \times \sin 35^{\circ} 17^{\prime}}{\sin 99^{\circ} 30^{\prime}} \\
& \log a=\log 42.1+\log \sin 36^{\circ} 17^{\prime}-\log \sin 99^{\circ} 30^{\prime} \\
& =1.6243+1.7616-1.9940 \\
& =1.3859-1.9940 \\
& =1.3859-[-1+0.9940]=1.3919 \\
& \Rightarrow \quad a=24.65^{\circ} \\
& \text { Again } \quad c=\frac{b \sin \mathrm{C}}{\sin \mathrm{B}}=\frac{42.1 \times \sin 45^{\circ} 13^{\prime}}{\sin 99^{\circ} 30^{\prime}} \\
& \log \mathrm{c}=\log 42.1+\log \sin 45^{\circ} 31-\log \sin 99^{\circ} 30^{\prime} \\
& =1.6243+1.8511-1.9940 \\
& =1.4754-1.9940 \\
& =1.4754-[-1+0.9940]=1.4814 \\
& \Rightarrow \mathrm{c}=30.3^{\circ} \\
& \text { Thus } \mathrm{B}=99^{\circ} 30^{\prime} ; \mathrm{a}=24.65^{\circ} ; \mathrm{c}=30.3^{\circ} \\
&
\end{aligned}
$
Question 4.
Solve the triangle $\mathrm{ABC}$ if $\mathrm{a}=5, \mathrm{~b}=4$ and $\mathrm{C}=68^{\circ}$.
Solution:
To find c, use $\mathrm{c}^2=\mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab} \cos \mathrm{C}$
$
\begin{aligned}
& c^2=25+16-2 \times 5 \times 4 \cos 68^{\circ} \\
& =41-40 \times 0.3746=26.016 \\
& c=5.1
\end{aligned}
$

To find the other two angles, use sine formula.
$
\begin{aligned}
\Rightarrow \sin \mathrm{B} & =\frac{b \sin \mathrm{C}}{c}=\frac{4 \times \sin 68^{\circ}}{5.1} \\
\log \sin \mathrm{B} & =\log 4+\log \sin 68^{\circ}-\log 5.1 \\
& =0.6021+1.9672-0.7075 \\
& =2.5693-0.7075 \\
& =1.8618 \\
\mathrm{~B} & =46^{\circ} 40^{\prime} \\
\mathrm{A} & =180-(\mathrm{B}+\mathrm{C})=180-\left(114^{\circ} 40^{\prime}\right) \\
& =65^{\circ} 20^{\prime} \\
\therefore \quad \mathrm{B} & =46^{\circ} 40^{\prime} ; \mathrm{A}=65^{\circ} 20^{\prime} ; c=5.1
\end{aligned}
$