The word "trigonometry" is derived from the words "tri" (meaning three), "gon" (meaning sides) and "metry" (meaning measure).Thus trigonometry deals with the measurement of sides and angles of a triangle.
"trigonometry" ဟူသောစကားသည်၊ "tri"သုံး၊ "gon"အနားများ၊ "metry"အတိုင်းအတာ ဟူသောစကားလုံးများမှဆင်းသက်လာသည်။ထို့ကြောင့် "trigonometry" ဟူသောစကားရပ်သည်တြိဂံတစ်ခု၏ထောင့်များနှင့်အနားများ၏အတိုင်းအတာများနှင့်ဆက်စပ်နေသည်။
An arc BC substends an angle of 144° at the centre O of a circle of radius 10 cm. Find the length of arc BC and the area of the sector BOC.
$\theta=144^{\circ}=144 \times \dfrac{\pi}{180}=\dfrac{4 \pi}{5}$ radians and $r=10 \mathrm{~cm}$.
$\begin{aligned}
\text{ The length of arc } B C&=r \theta\\
&=10 \times \dfrac{4 \pi}{5}\\
&=8 \pi \mathrm{~cm}\\
\end{aligned}$
$\begin{aligned}
\text{ The area of sector } B O C&=\dfrac{1}{2} r^{2} \theta\\
&=\dfrac{1}{2} \times 10^{2} \times \dfrac{4 \pi}{5}\\
&=40 \pi \mathrm{~cm}^{2}
\end{aligned}$
Example 4
(a) Using the right triangle $A B C$, find $\sin A$, $\cos A$, $\tan A$.
(b) Using the right triangle $A B^{'} C^{'}$, find $\sin A$, $\cos A$, $\tan A$.
(a) From the right triangle $A B C$,
$\begin{aligned}
\sin A&=\dfrac{\text { opposite side of } \angle A}{\text { hypotenuse }}=\dfrac{B C}{A B}=\frac{3}{5} \\
\cos A&=\dfrac{\text { adjacent side of } \angle A}{\text { hypotenuse }}=\dfrac{A C}{A B}=\frac{4}{5} \\
\tan A&=\dfrac{\text { opposite side of } \angle A}{\text { adjacent side of } \angle A}=\dfrac{B C}{A C}=\dfrac{3}{4}
\end{aligned}$
(b) from the right triangle $A B^{\prime} C^{\prime}$,
$\begin{aligned}
\sin A&=\dfrac{\text { opposite side of } \angle A}{\text { hypotenuse }}=\dfrac{B^{\prime} C^{\prime}}{A B^{\prime}}=\dfrac{6}{10}=\dfrac{3}{5}\\
\cos A&=\dfrac{\text { adjacent side of } \angle A}{\text { hypotenuse }}=\dfrac{A C^{\prime}}{A B^{\prime}}=\dfrac{8}{10}=\frac{4}{5}\\
\tan A&=\dfrac{\text { opposite side of } \angle A}{\text { adjacent side of } \angle A}=\dfrac{B^{\prime} C^{\prime}}{A C^{\prime}}=\dfrac{6}{8}=\dfrac{3}{4}
\end{aligned}$
Example 5
$A B C $ is a right triangle in which $C$ is the right angle. If $a$ = $4$ and $b$ = $3$, find $c$, $\cos A$ and $\sec B$.
In right triangle $A C B $,
By Pythagoras theorem,
$\begin{aligned}
c^{2} &=a^{2}+b^{2} \\
&=4^{2}+3^{2} \\
&=25 \\
c &=5 \\
\cos A &=\dfrac{b}{c}=\dfrac{3}{5} \\
\sec B &=\dfrac{c}{a}=\dfrac{5}{4}
\end{aligned}$
Example 6
Given the right triangle $A B C$ with $\angle C$ = $90^{\circ}$ and $\tan A$ = $\dfrac{5}{12}$, find $\sin A$ and $\cos A$.
$\tan A=\dfrac{5}{12}=\dfrac{B C}{A C}$
Take $ B C $ = $5 k$, $A C$ = $12 k$ where k is constant .
In right triangle $A C B $,
By Pythagoras Theorem,
$\begin{aligned}
A B^{2} &=B C^{2}+A C^{2} \\
&=(5 k)^{2}+(12 k)^{2} \\
&=25 k^{2}+144 k^{2} \\
&=169 k^{2} \\
A B &=13 k \\
\sin A &=\dfrac{B C}{A B}=\dfrac{5 k}{13 k}=\dfrac{5}{13} \\
\cos A &=\dfrac{A C}{A B}=\dfrac{12 k}{13 k}=\dfrac{12}{13}
\end{aligned}$
Example 7
Given the right triangle $A B C$ in which $\angle B$ = $90^{\circ}$, is drawn perpendicular to $C A$
and meets $A B$ produced at $D$. If $B C$ =$6$, $A B$ = $8$, $A C$ = $10$, find $C D$ and $A D$.
In right triangle $A B C, \tan A=\dfrac{B C}{A B}$
In right triangle $A C D, \tan A=\dfrac{C D}{A C}$
$\begin{aligned}
\therefore \dfrac{B C}{A B} &=\dfrac{C D}{A C} \\
C D &=\dfrac{B C}{A B} \times A C \\
&=\dfrac{6}{8} \times 10 \\
&=7.5
\end{aligned}$
In right triangle $A C D, \sec A=\dfrac{A D}{A C}$
In right triangle $A B C, \sec A=\dfrac{A C}{A B}$
$\begin{aligned}
\dfrac{A D}{A C} &=\dfrac{A C}{A B} \\
A D &=\dfrac{A C^{2}}{A B} \\
&=\dfrac{10^{2}}{8}=\dfrac{100}{8}=12.5
\end{aligned}$
Example 8
Prove that $\dfrac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}$ = $\tan\theta$.
Solve the triangle $A B C$ with $\angle B$ = $90^{\circ}$, $a$ = $10$, $b$ = $20$.
By Pythagoras Theorem,
$\begin{aligned}
c^{2} &=b^{2}-a^{2} \\
&=400-100 \\
&=300 \\
c &=10 \sqrt{3} \\
\sin A &=\dfrac{a}{b}\\
&=\dfrac{10}{20}\\
&=\dfrac{1}{2}\\
&=\sin 30^{\circ} \\
\therefore \angle A &=30^{\circ} \\
\angle C &=90^{\circ}-\angle A\\
&=90^{\circ}-30^{\circ}\\
&=60^{\circ}
\end{aligned}$
$\angle A =30^{\circ}$, $\angle C=60^{\circ}$, $c =10 \sqrt{3}$
Example 15
Solve the triangle $A B C$ with $\angle B$ = $90^{\circ}$, $\angle A$ = $30^{\circ}$, $c$ = $6$.
$\begin{aligned}
\angle C &=90^{\circ}-\angle A\\
&=90^{\circ}-30^{\circ}\\
&=60^{\circ} \\
\dfrac{a}{c} &=\tan A \\
\dfrac{a}{6} &=\tan 30^{\circ} \\
a &=6 \tan 30^{\circ} \\
&=6 \times \dfrac{\sqrt{3}}{3}\\
&=2 \sqrt{3} \\
\dfrac{b}{c} &=\sec A \\
\dfrac{b}{6} &=\sec 30^{\circ} \\
b &=6 \sec 30^{\circ} \\
&=6 \times \dfrac{2\sqrt{3}}{3} \\
&=4 \sqrt{3}
\end{aligned}$
$\angle C =60^{\circ}$, $a = 2 \sqrt{3}$, $b = 4 \sqrt{3}$
Example 16
In $ \triangle A B C $, the angles $A$ and $C$ are equal to $30^{\circ}$ and $120^{\circ}$ respectively;
and the side $A C$ = $20$ ft, find the length of the perpendicular from $B$ upon $ A C $ produced.
Draw BD perpendicular to AC produced.
$\begin{aligned}
\theta &=180^{\circ}-120^{\circ}=60^{\circ} \\
\beta&=180^{\circ}-\left(30^{\circ}+120^{\circ}\right)=30^{\circ} \\
\angle A&=\beta=30^{\circ} \\
\therefore A C&=B C=20
\end{aligned}$
In right $\triangle C D B$,
$\begin{aligned}
\dfrac{B D}{B C} &=\sin \theta \\
\dfrac{B D}{20} &=\sin 60^{\circ} \\
B D &=20 \sin 60^{\circ} \\
&=20 \times \dfrac{\sqrt{3}}{2} \\
&=10 \sqrt{3} \mathrm{ft}
\end{aligned}$
Example 17
Solve the triangle $ A B C $, $\angle C$ = $90^{\circ}$ and $\angle B$ = $25^{\circ}43^{'}$ and $c$ = $100$.
Using as much of the information below as necessary.
[ $\cos 25^{\circ}43^{'}$ = $0.9010$, $\tan 25^{\circ}43^{'}$ = $0.4817$ ].
$\begin{aligned}
\angle A &=90^{\circ}-\angle B\\
&=90^{\circ}-25^{\circ} 43^{\prime}\\
&=64^{\circ} 17^{\prime} \\
\dfrac{a}{c} &=\cos B \\
a &=c \cos B \\
&=100 \times \cos 25^{\circ} 43^{\prime} \\
&=100 \times 0.9010 \\
&=90.10 \\
\dfrac{b}{a} &=\tan B \\
b &=a \tan B \\
&=90.10 \times 0.4817 \\
&=43.40
\end{aligned}$
$\angle A=64^{\circ} 17^{\prime}$, $a=90.12$, $b=43.40$
Example 18
A kite is at the end of a $60$ ft string that is taut. It is $30$ ft above the ground. What is the angle of elevation of the kite?
$\begin{aligned}
\sin A&=\dfrac{30}{60}\\
&=\dfrac{1}{2}\\
&=\sin 30^{\circ}\\
\therefore\angle A&=30^{\circ}
\end{aligned}$
The angle of elevation of the kite is $30^{\circ}$.
Example 19
From the top of a house the angle of depression to a point on the ground is $30^{\circ}$.
The point is $45$ ft from the base of the building. How high is the building?
$\cot 60^{\circ}=\dfrac{B C}{A B}$
$\begin{aligned}
B C &=A B \cot 60^{\circ} \\
&=45\times \dfrac{\sqrt{3}}{3} \\
&=15 \sqrt{3} \mathrm{~ft}
\end{aligned}$
The height of the building = $15 \sqrt{3} \mathrm{~ft}$
A central angle $\theta$ substends an arc of $\dfrac{11\pi}{2} \mathrm {~cm}$ on a circle of radius 6 cm. Find the measure of $\theta$ in radians and the area of a sector of a circle which has $\theta$ is its central angle.
The area of a sector of a circle is $143$ $\mathrm{cm}^{2}$ and the length of the arc of a sector is $11 \mathrm {~cm}$. Find the radius of the circle.
$\begin{aligned}
A &=143 \mathrm{~cm}^{2}, s=11 \mathrm{~cm} \\\\
s &=r \theta \\\\
A &=\dfrac{1}{2} r^{2} \theta \\\\
&=\dfrac{1}{2} r(r \theta) \\\\
A &=\dfrac{1}{2} r s \\\\
143 &=\dfrac{1}{2} r \times 11 \\\\
r &=\dfrac{286}{11} \\\\
r &=26 \mathrm{~cm}
\end{aligned}$
A sector cut from a circle of radius $3 \mathrm{~cm}$ has a perimeter of $16 \mathrm{~cm}$ . Find the area of this sector.
$s=3 \mathrm{~cm}$
$\begin{aligned}
\text{ The perimeter of the sector } &=16 \mathrm{~cm}\\\\
r+r+s &=16 \\\\
3+3+s &=16 \\\\
6+s &=16 \\\\
s &=10 \mathrm{~cm}
\end{aligned}$
$\begin{aligned}
\text { The area of the sector }
&=\dfrac{1}{2} r^{2} \theta \\\\
&=\dfrac{1}{2} r(r \theta) \\\\
&=\dfrac{1}{2} r s \\\\
&=\dfrac{1}{2} \times 3 \times 10 \\\\
&=15 \mathrm{~cm}^{2}
\end{aligned}$
A piece of wire of fixed length $L \mathrm {~cm}$, is bent to form the boundary a sector of a circle. The circle has radius $r \mathrm {~cm}$
and the angle of the sector is $\theta=(\dfrac{32}{r}-2)$ radians. Find the wire of fixed length $L$ and show that the area of the sector,
$ A \mathrm{~cm}^{2}$ is given by $A$ = $16r-r^{2}$.
The length of a piece of wire $=L \mathrm{~cm}$
The radius of the circle $=r \mathrm{~cm}$
The length of $\operatorname{arc}=s \mathrm{~cm}$
$\theta=\left(\dfrac{32}{r}-2\right)$ radians
$\begin{aligned}
\text{ The perimeter of the sector } &=r+r+s\\\\
&=\left(2 r+s\right) \mathrm{~cm} \\\\
s &=r \theta \\\\
&=r\left(\dfrac{32}r-2\right) \\\\
&=\left(32-2 r\right)\mathrm{~cm}
\end{aligned}$
$\begin{aligned}
\text{ Since the length of a piece of wire } &=\text{ The perimeter of of the sector }\\\\
L &=2 r+s \\\\
L &=2 r+32-2 r \\\\
&=32 \mathrm{~cm}
\end{aligned}$
$\begin{aligned}
\text{ The area of the sector } =A&=\dfrac{1}{2} r^{2} \theta \\\\
&=\dfrac{1}{2} r^{2}\left(\dfrac{32}{r}-2\right) \\\\
&=\left(16 r-r^{2}\right) \mathrm{cm}^{2}
\end{aligned}$
A race is run at uniform speed on a circular course. In each minute, a runner transverse an arc of a circle which subtends $2\dfrac{6}{7}$ radians at the centre of the course. If each lap is $792$ yards, how long does the runner take to run a mile?
the central angle $=\theta=2 \dfrac{6}{7}=\dfrac{20}{7}$ radians
$\begin{aligned}
\text{ each lap } &=792 \text{ yards }\\\\
\text{ each lap } &= \text{ one complete circumference }\\\\
792 &=2 \pi r \\\\
792 &=2 \times \dfrac{22}{7} \times r \\\\
r &=126 \text { yards }
\end{aligned}$
In every minute, the runner transverses an arc of a circle.
$\begin{aligned}
\text{ The distance travelled by the runner } &= s\\\\
s &=r \theta \\\\
&=126 \times \dfrac{20}{7} \\\\
&=360 \text { yards }
\end{aligned}$
$1$ mile =$1760$ yards
The runner takes $1$ minute in $360$ yards.
$\begin{aligned}
360 \text{ yards } &=1 \text{ minute }\\\\
1760 \text { yards } &=\dfrac{1760 \times 1}{360} \\\\
&=4.8889 \text { min }
\end{aligned}$
The large hand of a clock is $28$ inches long; how many inches does its extremity move in $20$ minutes?
Let the length of the hand of a clock = $r$ = $28\mathrm{ ~in } $
Let the angle taken by the hand in $20\mathrm{ minutes }$ = $\theta$
$\begin{aligned}
\dfrac{\theta}{360^{\circ}} &= \dfrac{20}{60}\\\\
\theta &= \dfrac{20}{60}\times 360\times\dfrac{\pi}{180}\\\\
&=\dfrac{2\pi}{3}\text{ radians }
\end{aligned}$
Let the arc length taken by the hand = $s$
$\begin{aligned}
s&=r\theta\\\\
&=28\times \dfrac{2\pi}{3}\\\\
&=28\times \dfrac{2}{3}\times \dfrac{22}{7}\\\\
&=58.67\mathrm{~in}
\end{aligned}$
The figure shows two sectors in which the arcs $ A B $ and $ C D $ are arcs of concentric circles, centre $ O $. If $\angle A O B =\dfrac{2}{3}$ radians, $ A C =3 \mathrm{~cm}$ and the area of sector $ A O B $ is $12 \mathrm{~cm}^{2}$, calculate the area and the perimeter of $ A B D C $.
the central angle $=\angle A O B=\dfrac{2}{3}$ radians
the radius of small sector $=O A=O B$
the radias of large sector $=O C=O D$
$A C=B D=3 \mathrm{~cm}$
the area of sector $A O B=12 \mathrm{~cm}^{2}$
$\begin{aligned}
\text{ the area of sector } A O B&=\dfrac{1}{2} r^{2} \theta\\
12&=\dfrac{1}{2} \times O A^{2} \times \angle A O B \\
12&=\dfrac{1}{2} \times O A^{2} \times \dfrac{2}{3} \\
O A^{2}&=36 \\
O A&=6 \mathrm{~cm}
\end{aligned}$
$\begin{aligned}
\text{ the length of arc } A B &=r \theta \\
&=O A \times \angle A O B \\
&=6 \times \dfrac{2}{3}\\
&=4 \mathrm{~cm} \\\\
O C &=O A+A C \\
&=6+3=9 \mathrm{~cm}
\end{aligned}$
$\begin{aligned}
\text { the length of arc } C D=&r \theta \\
&= O C \times \angle A O B \\
=& 9 \times \dfrac{2}{3} \\
=& 6 \mathrm{~cm} \\
\text { The area of sector } C O D=& \dfrac{1}{2} r^{2} \theta \\
=& \dfrac{1}{2} \times O C^{2} \times \angle A O B \\
=& \dfrac{1}{2} \times 9 \times 9 \times \dfrac{2}{3} \\
=& 27 \mathrm{~cm}^{2}
\end{aligned}$
$\begin{aligned}
\text{ the area of } A B D C&=\text { the area of sector } C O D-\text { the area of sector } A O B \\
&=27-12\\
&=15 \mathrm{~cm}^{2}
\end{aligned}$
$\begin{aligned}
\text{ The perimeter of } A B C D&=A C+B D+\text{ the length of arc } A B+\text{ the length of arc } C D\\
&=3+3+4+6\\
&=16 \mathrm{~cm}\\
\end{aligned}$
Exercise 10.2
Find the following trigonometric ratios for a right triangle with sides as indicated.
$\sin A$, $\cos A$, $\tan A$, $\cot A$, $sec A$, $\csc A$
$\sin B$, $\cos B$, $\tan B$, $\cot B$, $sec B$, $\csc B$
Given a right $\triangle A B C$ with $\angle C = 90^{\circ}$ and $\cos A$=$\dfrac{5}{13}$ , determine the values of $\tan A$ and $\sin A$ .
$\angle C=90^{\circ}, \cos A=\dfrac{5}{13}$
Let $A C=5 k, A B=13 k$
In right $\triangle A C B$,
$\begin{aligned}
B C^{2} &=A B^{2}-A C^{2} \\
&=(13 k)^{2}-(5 k)^{2} \\
&=169 k^{2}-25 k^{2} \\
&=144 k^{2} \\
B C &=12 k \\
\tan A &=\dfrac{B C}{A C} \\
&=\dfrac{12 k}{5 k}\\
&=\dfrac{12}{5} \\
\sin A &=\dfrac{B C}{A B} \\
&=\dfrac{12 k}{13 k}\\
&=\dfrac{12}{13}
\end{aligned}$
Given a right $\triangle P Q R$ with $\angle R = 90^{\circ}$ , If $P Q$=$13$, $Q R$=$5$, $P R$=$12$, find the values of $\tan P$, $\sec Q$, $\csc Q$, $\sin P$.
$P Q R S$ is a quadrilateral in which $\angle P S R$= $90^{\circ}$. If the diagonal $P R$ is at right angles to $R Q$, and $R P$ =$21$, $R S$ = $16$,
find $\sin \angle P R S$, $\tan \angle R P S$, $\cos \angle R P Q$, and $\csc\angle P Q R$.
In right $\triangle P S R$,
$\begin{aligned}
P S^{2} &=P R^{2}-S R^{2} \\
&=20^{2}-16^{2} \\
&=400-256 \\
&=144 \\
P S &=12
\end{aligned}$
In right $\triangle P R Q$,
$\begin{aligned}
P Q^{2} &=P R^{2}+Q R^{2} \\
&=20^{2}+21^{2} \\
&=400+441 \\
&=841 \\
P Q &=29
\end{aligned}$
$\begin{aligned}
\sin \angle P R S &=\dfrac{P S}{P R} \\
&=\dfrac{12}{20} \\
&=\dfrac{3}{5} \\
\tan \angle R P S &=\dfrac{S R}{P S} \\
&=\dfrac{16}{12} \\
&=\dfrac{4}{3} \\
\cos \angle R P Q &=\dfrac{P R}{P Q} \\
&=\dfrac{20}{29} \\
\csc \angle P Q R &=\dfrac{P Q}{P R} \\
&=\dfrac{29}{20}
\end{aligned}$
Find the value of acute angle $\alpha$ in each of the following equations:
(a) $\cos 2\alpha$ = $\sin 7\alpha$ (b) $\tan 3\alpha$ = $\cot 2\alpha$ (c) $\sec \alpha$ = $\csc 5\alpha$
A ladder is placed along a wall such that it upper end is touching the top of the wall.
The foot of the ladder is $5$ ft away from the wall and the ladder is making an angle of
$60^{\circ}$ with the level of the ground. Find the height of the wall.
$\begin{aligned}
\dfrac{B C}{5} &=\tan 60^{\circ} \\
B C &=5 \tan 60^{\circ} \\
&=5 \sqrt{3}
\end{aligned}$
The height of the wall $=5 \sqrt{3} \mathrm{~ft}$
Given $\triangle A B C $ with $\angle A$ = $30^{\circ}$,$\angle B$ = $135^{\circ}$ and $ A B$ = $100$ .
Find the length of the perpendicular from $C$ to $ A B $ produced.
$\angle A=30^{\circ}, \angle B=135^{\circ}$ and $A B=100, C D=$ ?
Draw CD perpendicular to $A B$ produced.
$\angle C B D=180^{\circ}-135^{\circ}=45^{\circ}$
since $\triangle B D C$ is a $45^{\circ}-45^{\circ}$ right triangle,
Let $B D=D C=x$
In right $\triangle A D C$,
$\begin{aligned}
\dfrac{C D}{A D}&=\tan A\\
C D&=A D \tan 30^{\circ}\\
x&=(100+x) \dfrac{\sqrt{3}}{3}\\
3 x&=100 \sqrt{3}+\sqrt{3} x\\
(3-\sqrt{3}) x &=100 \sqrt{3} \\
x &=\dfrac{100 \sqrt{3}}{3-\sqrt{3}} \times \dfrac{3+\sqrt{3}}{3+\sqrt{3}} \\
&=\dfrac{300 \sqrt{3}+300}{3^{2}-(\sqrt{3})^{2}}\\
x &=\dfrac{300 \sqrt{3}+300}{9-3} \\
&=\dfrac{6(50 \sqrt{3}+50)}{6} \\
x &=50 \sqrt{3}+50 \\
\therefore C D &=50+50 \sqrt{3}
\end{aligned}$
If $ B D $ is perpendicular to the base $ A C $ of a triangle $ A B C $ , find $a$ and $c$, given $\angle A$ = $30^{\circ}$,$\angle C$ = $45^{\circ}$, $ B D $ = $10$.
$B D \perp A C, \angle A=30^{\circ}, \angle C=45^{\circ}$, $B D=10, a, c=$ ?
In right $\triangle A D B$,
$\begin{aligned}
\dfrac{c}{10} &=\csc A \\
c &=10 \csc 30^{\circ} \\
&=10 \times 2 \\
&=20 \\
\end{aligned}$
In right $\triangle C D B$,
$\begin{aligned}
\dfrac{10}{a}&=\sin 45^{\circ}\\
a &=\dfrac{10}{\sin 45^{\circ}} \\
&=\dfrac{10}{\dfrac{\sqrt{2}}{2}} \\
&=\dfrac{20}{\sqrt{2}} \\
&=10 \sqrt{2} \\
\therefore a &=10 \sqrt{2}, c=20
\end{aligned}$
In the triangle $ A B C $, the angle $B$ and $C$ are equal to $45^{\circ}$ and $120^{\circ}$ respectively;
if $a$ = $40$, find the length of the perpendicular from $A$ on $ B C $ produced.
Draw AD perpendicular to BC produced.
$\angle A C D=180^{\circ}-120^{\circ}=60^{\circ}$
since $\triangle A D C$ is a $30^{\circ}-60^{\circ}$ right triangle,
Let $C D=x$
$A D=\sqrt{3} x, A C=2 x$
In right $\triangle A D B$,
$\begin{aligned}
\dfrac{A D}{B D}&=\tan 45^{\circ}\\
\dfrac{\sqrt{3} x}{40+x} &=1 \\
\sqrt{3} x&=40+x \\
(\sqrt{3}-1) x &=40 \\
x &=\dfrac{40}{\sqrt{3}-1} \times \dfrac{\sqrt{3}+1}{\sqrt{3}+1} \\
&=\dfrac{40(\sqrt{3}+1)}{\sqrt{3}^{2}-1^{2}}\\
&=\dfrac{40(\sqrt{3}+1)}{3-1} \\
&=\dfrac{40(\sqrt{3}+1)}{2} \\
x &=20+20 \sqrt{3} \\
\text { AD } &=\sqrt{3} x \\
&=\sqrt{3}(20+20 \sqrt{3}) \\
&=60+20 \sqrt{3}
\end{aligned}$
Solve the isosceles triangle. (Draw the perpendicular from the vertex to the base.)
(a) $b$ = $c$ = $12$, $\angle B$ = $30^{\circ}$.
(b) $a$ = $b$ = $9$, $\angle A$ = $60^{\circ}$.
(c) $b$ = $c$ = $10$, $\angle A$ = $120^{\circ}$.
(a) $b$ = $c$ = $12$, $\angle B$ = $30^{\circ}$
$\begin{aligned}
b=c&=12, \angle B=\angle C=30^{\circ}\\
\angle B A C&=180^{\circ}-(\angle B+\angle C)\\
&=180^{\circ}-\left(30^{\circ}+30^{\circ}\right) \\
&=120^{\circ}\\
\text{ Draw } A D &\perp B C\\
\text{ In right } \triangle& A D B,\\
\dfrac{B D}{A B} &=\cos B \\
\dfrac{B D}{c} &=\cos 30^{\circ} \\
B D &=c \cos 30^{\circ} \\
&=12 \times \dfrac{\sqrt{3}}{2} \\
&=6 \sqrt{3} \\
\text { similarly }, D C&=6 \sqrt{3} \\
B C &=B D+D C \\
&=6 \sqrt{3}+6 \sqrt{3} \\
&=12 \sqrt{3}
\end{aligned}$
(b) $a$ = $b$ = $9$, $\angle A$ = $60^{\circ}$
$\begin{aligned}
a=b &=9, \angle A=\angle B=60^{\circ} \\
\angle C &=180^{\circ}-(\angle A+\angle B) \\
&=180^{\circ}-\left(60^{\circ}+60^{\circ}\right) \\
&=60^{\circ} \\
\triangle&A B C \text{ is an equilateral triangle }.\\
\therefore a &=b=c=9 \\
\therefore A B&=9
\end{aligned}$
(c) $b$ = $c$ = $10$, $\angle A$ = $120^{\circ}$
$\begin{aligned}
b=c&=10, \quad \angle A=120^{\circ}\\
\angle B&=\angle C=\dfrac{180^{\circ}-\angle A}{2}\\
&=\dfrac{180^{\circ}-120^{\circ}}{2}\\
&=30^{\circ}\\
\text{ Draw } AD &\perp B C\\
\text{ In right } \triangle& A D B,\\
\dfrac{B D}{A B}&=\cos A\\
\dfrac{B D}{c}&=\cos 30^{\circ}\\
B D&=c \cos 30^{\circ}\\
B D&=10 \times \dfrac{\sqrt{3}}{2}\\
&=5 \sqrt{3}\\
\text { similarly }, D C&=5 \sqrt{3} \\
B C&=B D+D C\\
&=5 \sqrt{3}+5 \sqrt{3}\\
&=10 \sqrt{3}\\
\end{aligned}$
In $\triangle A B C $, $\angle A$ = $60^{\circ}$,$ A C$ = $12$ and $A B$ = $20$.
Find the length of the perpendicular drawn from $C$ to $ A B $. Also find the area of the triangle $ A B C $.
In $\triangle A B C$,
$\angle A=60^{\circ}, A C=12, A B=20$
Draw CD $\perp A B$.
In right $\triangle A D C$,
$\begin{aligned}
\dfrac{C D}{A C}&=\sin 60^{\circ}\\
C D&=A C \sin 60^{\circ}\\
&=12 \times \dfrac{\sqrt{3}}{2}\\
&=6 \sqrt{3}\\
\end{aligned}$
$\begin{aligned}
\text{ The area of } \triangle A B C &=\dfrac{1}{2} \times A B \times C D \\
&=\dfrac{1}{2} \times 20 \times 6 \sqrt{3} \\
&=60 \sqrt{3} \text { square units }
\end{aligned}$
Find the area of triangle $ A B C $, given $A B$ = $30$, $B C$ = $16$ and $\angle B$ = $30^{\circ}$.
In $\triangle A B C$
$A B=30, B C=16, \angle B=30^{\circ}$
Draw $C D \perp A B$.
In right $\triangle B D C$,
$\begin{aligned}
\dfrac{C D}{B C} &=\sin 30^{\circ} \\
C D &=B C \sin 30^{\circ} \\
&=16 \times \dfrac{1}{2} \\
&=8
\end{aligned}$
$\begin{aligned}
\text{ The area of } \triangle A B C&=\dfrac{1}{2} \times A B \times C D\\
&=\dfrac{1}{2} \times 30 \times 8 \\
&=120 \text { square units }
\end{aligned}$
Solve the triangle $ A B C $, $\angle B$ = $90^{\circ}$,$\angle A$ = $36^{\circ}$ and $c$ = $100$.
Using as much of the information below as necessary.
[ $\tan 36^{\circ}$ = $0.7265$, $\sec 36^{\circ}$ = $1.2361$ ]
$\begin{aligned}
\angle B &=90^{\circ}, \angle A=36^{\circ}, C=100 \\
\angle C &=180^{\circ}-(\angle A+\angle B) \\
&=180^{\circ}-\left(36^{\circ}+90^{\circ}\right) \\
&=54^{\circ} \\
\dfrac{a}{c} &=\tan A \\
a &=c \tan 36^{\circ} \\
&=100 \times 0.7265 \\
&=72.65 \\
\dfrac{b}{c} &=\sec A \\
b &=c \sec 36^{\circ} \\
&=100 \times 1.2361 \\
&=123.61 \\
\therefore \angle C &=54{ }^{\circ}, a=72.65, b=123.61
\end{aligned}$
Solve the triangle $ A B C $, $\angle A$ = $90^{\circ}$,$c$ = $37$ and $a$ = $100$.
Using as much of the information below as necessary.
[ $\sin 21^{\circ}43^{'}$ = $0.3700$, $\cos 21^{\circ}43^{'}$ = $0.9290$ ]
$\begin{aligned}
\angle A=90^{\circ}, c &=37, a=100 \\
\sin C &=\dfrac{c}{a} \\
& =\dfrac{37}{100} \\
&=0.37 \\
&=\sin 21^{\circ} 43^{\prime} \\
\therefore\angle C &=21^{\circ} 43^{\prime} \\
\angle B &=180^{\circ}-(\angle A+\angle C) \\
&=180^{\circ}-\left(90^{\circ}+21^{\circ} 43^{\prime}\right) \\
&=68^{\circ} 17^{\prime} \\
\dfrac{b}{a} &=\cos C \\
b &=a \cos 21^{\circ} 17^{\prime} \\
&=100 \times 0.9290 \\
&=92.9
\end{aligned}$
$ \therefore \angle B =68^{\circ} 17^{\prime}, \angle C=21^{\circ} 43^{\prime}, b=92.9$
Exercise 10.6
A mountain railway runs for $400$ yards at a uniform slope of $30^{\circ}$ with the horizontal.
What is the horizontal distance between its two ends?
$\begin{aligned}
\dfrac{A B}{A C} &=\cos A \\
A B &=A C \cos A \\
&=400 \cos 30^{\circ} \\
&=400 \times \dfrac{\sqrt{3}}{2} \\
&=200 \sqrt{3}
\end{aligned}$
The horizontal distance between its two ends $=200 \sqrt{3}$ yards.
A vertical mast is secured from its top by straight cables $600$ ft long fixed into the ground.
Each cable makes angle of $60^{\circ}$ with the ground. What is the height of the mast?
$\begin{aligned}
\dfrac{B C}{A C} &=\sin A \\
B C &=A C \sin A \\
&=600 \sin 60^{\circ} \\
&=600 \times \dfrac{\sqrt{3}}{2} \\
&=300 \sqrt{3}
\end{aligned}$
The height of the mast $=300 \sqrt{3} \mathrm{~ft}$.
A kite flying at a height of $60$ yards is attached to a string inclined at $45^{\circ}$ to the horizontal.
What is the length of the string? Assume there is no slack in the string.
$\begin{aligned}
\dfrac{A C}{B C} &=\csc A \\
A C &=B C \csc A \\
&=60 \csc 45^{\circ} \\
&=60 \sqrt{3}
\end{aligned}$
The length of the setring $=60 \sqrt{3}$ yards.
An observer, $6$ ft tall, is $20$ yards away from a tower $22$ yards high.
Determine the angle of elevation from his eye to the top of the tower.
$\begin{aligned}
A B&=E D=20 \text{ yards } =20 \times 3=60 \mathrm{~ft}\\
C D&=22 \text{ yards } =22 \times 3=66 \mathrm{~ft}\\
B D&=A E=6 \mathrm{~ft}\\
B C&=C D-B D=66-6=60 \mathrm{~ft} \\
A B&=E D=20 \text { yards }=60 \mathrm{ft} \\
\tan \theta&=\dfrac{B C}{A B}\\
&=\dfrac{60}{60}\\
&=1\\
&=\tan 45^{\circ} \\
\therefore \theta&=45^{\circ}
\end{aligned}$
The angle of elevation $=45^{\circ}$
Two observers are on the opposite sides of a tower. They measure the angles of elevation to the top of the tower as $30^{\circ}$ and $45^{\circ}$ respectively.
If the height of the tower is $40$ yards, find the distance between them.
In right $\triangle A D C$,
$\begin{aligned}
\dfrac{A D}{C D} &=\cot A \\
A D &=C D \cot 30^{\circ} \\
&=40 \sqrt{3}
\end{aligned}$
In right $\triangle B D C$,
$\begin{aligned}
\dfrac{D B}{C D} &=\cot B \\
D B &=C D \cot 45^{\circ} \\
&=40 \times 1 \\
&=40 \\
A B &=A D+D B \\
&=40 \sqrt{3}+40 \\
&=40(\sqrt{3}+1)
\end{aligned}$
The distance between the two observers $=40(\sqrt{3}+1)$ yards.
Find the angle of elevation of the sun when the shadow of a pole $12$ ft high is $4\sqrt{3}$ ft long.
In right $\triangle A B C$,
$\begin{aligned}
\tan A&=\dfrac{B C}{A B}\\
&=\dfrac{12}{4 \sqrt{3}} \\
&=\sqrt{3} \\
&=\tan 60^{\circ} \\
\therefore \angle A &=60^{\circ}\\
\end{aligned}$
The angle of elevation $=60^{\circ}$
Two masts are $60$ ft and $40$ ft high, and the line joining their tops makes an angle of $30^{\circ}$
with the horizontal, find the distance between them.
Let $A E$ and $D C$ be two masts.
$\begin{aligned}
B D&=A E=40 \mathrm{~ft} \\
B C&=C D-B D=60-40=20 \mathrm{~ft} \\
\text { In right }& \triangle A B C, \\
\dfrac{A B}{B C}&=\cot 30^{\circ} \\
A B&=B C \cot 30^{\circ} \\
&=20 \sqrt{3} \\
E D&=A B=20 \sqrt{3}
\end{aligned}$
The distance between two masts $=20 \sqrt{3} \mathrm{~ft}$.
From the foot of a tower the angle of elevation to the top of a $60\mathrm{~ft}$ column is $60^{\circ}$.
The angle of elevation from the top of the tower to the top of the column is $30^{\circ}$. Find the height of the column.
Let $A D$ be the height of the tower.
Let BE be the height of the column.
$\begin{aligned}
B E&=60 \mathrm{~ft}\\
\angle B A E&=60^{\circ}, \angle C D E=30^{\circ}\\
\text{ In right } &\triangle A B E,\\
\dfrac{A B}{B E}&=\cot 60^{\circ}\\
A B&=B E \cot 60^{\circ}\\
&=60 \times \dfrac{\sqrt{3}}{3} \\
&=20 \sqrt{3} \mathrm{~ft}
\end{aligned}$
Since $A B C D$ is a rectangle,
$\begin{aligned}
D C&=A B=20 \sqrt{3} \mathrm{~ft} \\
A D&=B C\\
\text{ In right } &\triangle D C E,\\
\dfrac{C E}{D C} &=\tan 30^{\circ} \\
C E &=D C \tan 30^{\circ}\\
C E &=20 \sqrt{3} \times \dfrac{\sqrt{3}}{3} \\
&=20 \mathrm{~ft} \\
B C &=B E-C E \\
&=60-20\\
&=40 \mathrm{~ft} \\
\therefore A D &=40 \mathrm{~ft}
\end{aligned}$
The height of the tower $=40 \mathrm{~ft}$.
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