Do you like to know the solutions of Chapter 1: Angle and its Measurement – Exercise 1.1 of your Maths Part 1 book? In this article, I will provide detailed solutions for this exercise.
The solution contains a necessary diagrammatical representation of a problem whenever required for better understanding. For a few questions, I will share important points to remember and explain the reason why the arrived answer differs from the answer provided behind the book.
If you face any doubt or query while going through any solution, then feel free to drop a comment below. I will be more than happy to help you out.
Angle and its Measurement Exercise 1.1 Solutions
Q.1 A) Determine which of the following pairs of angles are co-terminal.
i) $210^\circ, -150^\circ$
Solution: Co-terminal angles have same position of initial and terminal rays despite having different measures. It is implicitly assumed in this question (in fact most of other questions) that the angles have been drawn in standard position. That means, their initial arms already coincide. Now, we are left with terminal rays.
To determine whether the terminal rays of two angles coincide, we have to find out difference of their measures. If this difference is an integer multiple of $360^\circ$, then the pair is co-terminal. Mathematically, two directed angles $\measuredangle A$ and $\measuredangle B$ are co-terminal if
\begin{equation*}
\text{m}\measuredangle A – \text{m}\measuredangle B = n*360^\circ, \quad n \in \mathbb{Z},
\end{equation*}
where $\mathbb{Z}$ denotes set of all integers. Applying this, the difference is
\begin{align*}
210^\circ -(-150^\circ) &= 210^\circ + 150^\circ, \\
&= 360^\circ,
\end{align*}
which is a multiple of $360^\circ$.
$\therefore$ The given pair of angles is co-terminal.
ii) $360^\circ, -30^\circ$
Solution:
\begin{align*}
\text{Difference of angles } &= 360^\circ – (-30^\circ), \\
&= 360^\circ +30^\circ, \\
&= 390^\circ,
\end{align*}
which is not an integer multiple of $360^\circ$.
$\therefore$ The given pair of angles is not co-terminal.
iii) $-180^\circ, 540^\circ$
Solution: We can take any angles as the first angle. Here, we take the second one first, so that difference comes positive.
\begin{align*}
\text{Difference of angles } &= 540^\circ – (-180^\circ), \\
&= 540^\circ +180^\circ, \\
&= 720^\circ,\\
&= 2*360^\circ,
\end{align*}
which is an integer multiple of $360^\circ$.
$\therefore$ The given pair of angles is co-terminal.
iv) $-405^\circ, 675^\circ$
Solution:
\begin{align*}
\text{Difference of angles } &= 675^\circ – (-405^\circ), \\
&= 675^\circ +405^\circ, \\
&= 1080^\circ,\\
&= 3*360^\circ,
\end{align*}
which is an integer multiple of $360^\circ$.
$\therefore$ The given pair of angles is co-terminal.
v) $860^\circ, 580^\circ$
Solution:
\begin{align*}
\text{Difference of angles } &= 860^\circ – 580^\circ, \\
&= 280^\circ,
\end{align*}
which is not an integer multiple of $360^\circ$.
$\therefore$ The given pair of angles is not co-terminal.
vi) $900^\circ, -900^\circ$
Solution:
\begin{align*}
\text{Difference of angles } &= 900^\circ – (-900^\circ), \\
&= 1800^\circ,\\
&= 5*360^\circ,
\end{align*}
which is an integer multiple of $360^\circ$.
$\therefore$ The given pair of angles is co-terminal.
Q.1 B) Draw the angles of the following measures and determine their quadrants.
i) $-140^\circ$
Solution:
As per the above figure, the angle lies in quadrant III.
ii) $250^\circ$
Solution:
As per the above figure, the angle lies in quadrant III.
iii) $420^\circ$
Solution:
As per the above figure, the angle lies in quadrant I.
iv) $750^\circ$
Solution:
As per the above figure, the angle lies in quadrant I.
v) $945^\circ$
Solution:
As per the above figure, the angle lies in quadrant III.
vi) $1120^\circ$
Solution:
As per the above figure, the angle lies in quadrant I.
vii) $-80^\circ$
Solution:
As per the above figure, the angle lies in quadrant IV.
viii) $-330^\circ$
Solution:
As per the above figure, the angle lies in quadrant I.
ix) $-500^\circ$
Solution:
As per the above figure, the angle lies in quadrant III.
x) $-820^\circ$
Solution:
As per the above figure, the angle lies in quadrant III.
Q.2 Convert the following angles in to radian.
i) $85^\circ$
Solution: We have $\theta^\circ = \left( \theta \times \frac{\pi}{180}\right)^c$. Using that
\begin{align*}
85^\circ &= \left( 85 \times \frac{\pi}{180}\right)^c, \\
\therefore 85^\circ &= \left( \frac{17\pi}{36}\right)^c
\end{align*}
Hence, the answer is $\left( \frac{17\pi}{36}\right)^c$. [Note: You need not have to simplify further here.]
ii) $250^\circ$
Solution:
\begin{align*}
250^\circ &= \left( 250 \times \frac{\pi}{180}\right)^c, \\
\therefore 250^\circ &= \left( \frac{25\pi}{18}\right)^c
\end{align*}
iii) $-132^\circ$
Solution:
\begin{align*}
-132^\circ &= \left( -132 \times \frac{\pi}{180}\right)^c, \\
\therefore -132^\circ &= -\left( \frac{11\pi}{15}\right)^c
\end{align*}
iv) $65^\circ 30’$
Solution: Here, the angle is in degree minutes. We have to first convert it into complete degree form first.
\begin{align*}
65^\circ 30^\prime &= 65^\circ + 30^\prime \\
&= 65^\circ + \left(\frac{30}{60} \right)^\circ \quad (1^\circ = 60^\prime) \\
&= 65^\circ + \left(\frac{1}{2} \right)^\circ \\
&= \left(\frac{131}{2} \right)^\circ
\end{align*}
Now, we can convert the angle into radian as
\begin{align*}
65^\circ 30^\prime &= \left( \frac{131}{2} \times \frac{\pi}{180}\right)^c, \\
&= \left( \frac{131\pi}{360}\right)^c.
\end{align*}
v) $75^\circ 30’$
Solution:
\begin{align*}
75^\circ 30^\prime &= 75^\circ + 30^\prime \\
&= 75^\circ + \left(\frac{30}{60} \right)^\circ \quad (1^\circ = 60^\prime) \\
&= 75^\circ + \left(\frac{1}{2} \right)^\circ \\
&= \left(\frac{151}{2} \right)^\circ
\end{align*}
Now, we can convert the angle into radian as
\begin{align*}
75^\circ 30^\prime &= \left( \frac{151}{2} \times \frac{\pi}{180}\right)^c, \\
&= \left( \frac{151\pi}{360}\right)^c.
\end{align*}
vi) $40^\circ 48’$
Solution:
\begin{align*}
40^\circ 48^\prime &= 40^\circ + 48^\prime \\
&= 40^\circ + \left(\frac{48}{60} \right)^\circ \quad (1^\circ = 60^\prime) \\
&= 40^\circ + \left(\frac{4}{5} \right)^\circ \\
&= \left(\frac{204}{5} \right)^\circ
\end{align*}
Now, we can convert the angle into radian as
\begin{align*}
40^\circ 48^\prime &= \left( \frac{204}{5} \times \frac{\pi}{180}\right)^c, \\
&= \left( \frac{51\pi}{5\times 45}\right)^c \\
&= \left( \frac{17\pi}{5\times 15}\right)^c \\
&= \left( \frac{17\pi}{75}\right)^c
\end{align*}
In the answer $\left( \frac{51\pi}{225}\right)^c$ behind the book, there is common factor of $3$ present in both numerator and denominator. On removing that, we have the same answer $\left( \frac{17\pi}{75}\right)^c$.
Q.3 Convert the following angles in degree.
i) $\frac{7\pi^c}{12}$
Solution: We know that $\theta^c = \left( \theta \times \frac{180}{\pi}\right)^\circ$.
\begin{align*}
\therefore \frac{7\pi^c}{12} &= \left( \frac{7\pi}{12} \times \frac{180}{\pi}\right)^\circ \\
&= 105^\circ
\end{align*}
ii) $\frac{-5\pi^c}{3}$
Solution:
\begin{align*}
\frac{-5\pi^c}{3} &= \left( \frac{-5\pi}{3} \times \frac{180}{\pi}\right)^\circ \\
&= -300^\circ
\end{align*}
iii) $5^c$
Solution:
\begin{align*}
5^c &= \left( 5 \times \frac{180}{\pi}\right)^\circ \\
&= \left( \frac{900}{\pi}\right)^\circ
\end{align*}
There is no need to simplify further the answer.
iv) $\frac{11\pi^c}{18}$
Solution:
\begin{align*}
\frac{11\pi^c}{18} &= \left( \frac{11\pi}{18} \times \frac{180}{\pi}\right)^\circ \\
&= 110^\circ
\end{align*}
v) $\left(\frac{-1}{4} \right)^c$
Solution:
\begin{align*}
\left(\frac{-1}{4} \right)^c &= \left( \frac{-1}{4} \times \frac{180}{\pi}\right)^\circ \\
&= -\left( \frac{45}{\pi}\right)^\circ
\end{align*}
To match with the answer behind the book, we can simplify the answer and get it in degree minutes as
\begin{align*}
-\left( \frac{45}{\pi}\right)^\circ &= -\left( \frac{45}{22/7}\right)^\circ \\
&= -\frac{315^\circ}{22} \\
&= – 14.318^\circ \\
&= – (14^\circ + 0.318 \times 60^\prime) \\
&= – (14^\circ + 19.08^\prime) \\
&\approx – 14^\circ 19^\prime
\end{align*}
Hence, the answer is $-\left( \frac{45}{\pi}\right)^\circ $ or $- 14^\circ 19’$ approx.
Q.4 Express the following angles in degree, minute and second.
i) $(183.7)^\circ$
Solution:
\begin{align*}
(183.7)^\circ &= 183^\circ + 0.7^\circ \\
&= 183^\circ + 0.7 \times 60^\prime \quad (1^\circ = 60^\prime) \\
&= 183^\circ + 42^\prime \\
&= 183^\circ 42^\prime
\end{align*}
ii) $(245.33)^\circ$
Solution:
\begin{align*}
(245.33)^\circ &= 245^\circ + 0.33^\circ \\
&= 245^\circ + 0.33 \times 60^\prime \quad (1^\circ = 60^\prime) \\
&= 245^\circ + 19.8^\prime \\
&= 245^\circ + 19^\prime + 0.8^\prime \\
&= 245^\circ 19^\prime + 0.8 \times 60^{\prime\prime} \quad (1^\prime = 60^{\prime\prime})\\
&= 245^\circ 19^\prime + 48^{\prime\prime} \\
&= 245^\circ 19^\prime48^{\prime\prime}
\end{align*}
iii) $\left(\frac{1}{5}\right)^c$
Solution: We first convert the given angle from radian to degrees.
\begin{align*}
\left(\frac{1}{5}\right)^c &= \left( \frac{1}{5} \times \frac{180}{\pi} \right)^\circ \quad (\theta^c = \left( \theta \times \frac{180}{\pi}\right)^\circ) \\
&= \left( \frac{36}{\pi} \right)^\circ \\
&= \left( \frac{36}{22/7} \right)^\circ \\
&= \left( \frac{126}{11} \right)^\circ \\
&= 11.45454^\circ
\end{align*}
We next convert the angle in degree, minute and second.
\begin{align*}
11.45454^\circ &= 11^\circ + 0.45454^\circ \\
&= 11^\circ + 0.45454 \times 60^\prime \quad (1^\circ = 60^\prime) \\
&= 11^\circ + 27.2724^\prime \\
&= 11^\circ + 27^\prime + 0.2724^\prime \\
&= 11^\circ 27^\prime + 0.2724 \times 60^{\prime\prime} \quad (1^\prime = 60^{\prime\prime})\\
&= 11^\circ 27^\prime + 16.344^{\prime\prime} \\
&\approx 11^\circ 27^\prime 16^{\prime\prime}
\end{align*}
The answer behind the book differs at second part. We can get this answer when we compute using a calculator and put exact value of $\pi$ to get radian to degree conversion.
Q.5 In $\triangle$ABC, if m$\angle$A = $\frac{7\pi^c}{36}$, m$\angle$B = $120^\circ$, find m$\angle$C in degree and radian.
Solution: Given
\begin{align*}
\text{m}\angle \text{A} &= \frac{7\pi^c}{36} \\
&= \left( \frac{7\pi}{36} \times \frac{180}{\pi} \right)^\circ \quad (\theta^c = \left( \theta \times \frac{180}{\pi}\right)^\circ ) \\
&= 35^\circ \\
\text{m}\angle \text{B} &= 120^\circ
\end{align*}
Since, sum of angles of a triangle = $180^\circ$.
\begin{align*}
\therefore \text{m}\angle \text{A} + \text{m}\angle \text{B} + \text{m}\angle \text{C} &= 180^\circ \\
35^\circ + 120^\circ + \text{m}\angle \text{C} &= 180^\circ \\
\text{m}\angle \text{C} &= 25^\circ \\
&= \left( 25 \times \frac{\pi}{180} \right)^c \quad (\theta^\circ = \left( \theta \times \frac{\pi}{180}\right)^c) \\
&= \left(\frac{5\pi}{36} \right)^c
\end{align*}
Ans: $\text{m}\angle \text{C}$ = $25^\circ$ or $\left(\frac{5\pi}{36} \right)^c$.
Q.6 Two angles of a triangle are $\frac{5\pi^c}{9}$ and $\frac{5\pi^c}{18}$. Find the degree and radian measure of third angle.
Solution: Let $\triangle$ABC be the required triangle with m$\angle$A = $\frac{5\pi^c}{9}$ and m$\angle$B = $\frac{5\pi^c}{18}$.
$\because$ sum of angles of a triangle = 180$^\circ$ or $\pi^c$.
\begin{align*}
\therefore \text{m}\angle \text{A} + \text{m}\angle \text{B} + \text{m}\angle \text{C} &= \pi^c \\
\frac{5\pi^c}{9} + \frac{5\pi^c}{18} + \text{m}\angle \text{C} &= \pi^c \\
\text{m}\angle \text{C} &= \frac{\pi^c}{6} = 30^\circ
\end{align*}
Hence, third angle measure is $30^\circ$ or $\frac{\pi^c}{6}$.
Q.7 In a right angled triangle, the acute angles are in the ratio 4:5. Find the angles of the triangle in degree and radian.
Solution: Let $\triangle$ABC be the right angled triangle with m$\angle$C = $90^\circ$.
Given m$\angle$A : m$\angle$B = 4:5
Let m$\angle$A = $4k$ and m$\angle$B = $5k$, where $k$ is the common factor.
$\because$ sum of angles of a triangle = $180^\circ$.
\begin{align*}
\therefore \text{m}\angle \text{A} + \text{m}\angle \text{B} + \text{m}\angle \text{C} &= 180^\circ \\
4k + 5k + 90^\circ &= 180^\circ \\
9k &= 90^\circ \\
k &= 10^\circ
\end{align*}
$\therefore$ The measure of the angles in degrees are
\begin{align*}
\text{m}\angle \text{A} &= 4*10^\circ = 40^\circ \\
\text{m}\angle \text{B} &= 5*10^\circ = 50^\circ \\
\text{m}\angle \text{C} &= 90^\circ
\end{align*}
The measure of the angles in radians are
\begin{align*}
\text{m}\angle \text{A} &= \left( 40 \times \frac{\pi}{180} \right)^c = \frac{2\pi^c}{9} \\
\text{m}\angle \text{B} &= \left( 50 \times \frac{\pi}{180} \right)^c = \frac{5\pi^c}{18} \\
\text{m}\angle \text{C} &= \left( 90 \times \frac{\pi}{180} \right)^c = \frac{\pi^c}{2}
\end{align*}
Q.8 The sum of two angles is $5\pi^c$ and their difference is $60^\circ$. Find their measures in degree.
Solution: Let $\angle$A and $\angle$B be the two angles.
Given
\begin{align*}
\text{m}\angle \text{A}+ \text{m}\angle \text{B} &= 5\pi^c \\
&= \left(5\pi \times \frac{180}{\pi}\right)^\circ \\
&= 900^\circ \quad \text{- – – – } (1) \\
\text{m}\angle \text{A}- \text{m}\angle \text{B} &= 60^\circ \quad \text{- – – – } (2)
\end{align*}
Adding eq. (1) and (2), we get
\begin{align*}
2 \text{m}\angle \text{A} &= 900^\circ + 60^\circ \\
\text{m}\angle \text{A} &= 480^\circ
\end{align*}
Substituting value of $\angle$A in eq. (1), we get
\begin{align*}
480^\circ + \text{m}\angle \text{B} &= 900^\circ \\
\text{m}\angle \text{B} &= 420^\circ
\end{align*}
Ans: $420^\circ$ and $480^\circ$.
Q.9 The measures of the angles of a triangle are in the ratio of 3:7:8. Find their measures in degree and radian.
Solution: Let the measures of the angles are $3k, 7k$ and $8k$, where $k$ is the common factor.
$\because$ sum of the angles of a triangle = $180^\circ$.
\begin{align*}
\therefore 3k + 7k + 8k &= 180^\circ \\
18k &= 180^\circ \\
k &= 10^\circ
\end{align*}
Hence, the measures of the angles in degrees are $30^\circ, 70^\circ$ and $80^\circ$. The measure in radian are
\begin{align*}
30^\circ &= \left(30 \times \frac{\pi}{180} \right)^c = \frac{\pi^c}{6} \\
70^\circ &= \left(70 \times \frac{\pi}{180} \right)^c = \frac{7\pi^c}{18} \\
80^\circ &= \left(80 \times \frac{\pi}{180} \right)^c = \frac{4\pi^c}{9}
\end{align*}
Q.10 The measures of the angles of a triangle are in A.P. and the greatest is 5 times the smallest (least). Find the angles in degrees and radians.
Solution: Let the measures of the angles of the triangle are $a-d, a$ and $a+d$.
Since, sum of angles of a triangle = 180$^\circ$.
\begin{align*}
\therefore a-d + a + a+d &= 180^\circ \\
3a &= 180^\circ \\
\therefore a &= 60^\circ
\end{align*}
Given
\begin{align*}
a+d &= 5 (a-d) \\
6d &= 4a \\
6d &= 4*60 \\
d&= 40^\circ
\end{align*}
Hence, the measures of the angles are
\begin{align*}
a-d &= 60^\circ – 40^\circ = 20^\circ = \left( 20 \times \frac{\pi}{180}\right)^c = \frac{\pi^c}{9} \\
a &= 60^\circ = \left( 60 \times \frac{\pi}{180}\right)^c = \frac{\pi^c}{3} \\
a+d &= 60^\circ + 40^\circ = 100^\circ = \left( 100 \times \frac{\pi}{180}\right)^c = \frac{5\pi^c}{9}
\end{align*}
Ans: $20^\circ, 60^\circ$ and $100^\circ$ that is $\frac{\pi^c}{9}, \frac{\pi^c}{3}$ and $\frac{5\pi^c}{9}$.
Q.11 In a cyclic quadrilateral two adjacent angles are $40^\circ$ and $\frac{\pi^c}{3}$. Find the angles of the quadrilateral in degrees.
Solution: A cyclic quadrilateral is a quadrilateral whose all vertices lie on a circle.
Let ABCD be the required cyclic quadrilateral with m$\angle$A = $40^\circ$ and m$\angle$B = $\frac{\pi^c}{3}$ = $60^\circ$.
$\because$ the sum of opposite angles of a cyclic quadrilateral is $180^\circ$.
\begin{align*}
\therefore \text{m} \angle \text{A} + \text{m} \angle \text{C} &= 180^\circ \\
40^\circ + \text{m} \angle \text{C} &= 180^\circ \\
\therefore \text{m} \angle \text{C} &= 140^\circ
\end{align*}
Similarly,
\begin{align*}
\therefore \text{m} \angle \text{B} + \text{m} \angle \text{D} &= 180^\circ \\
60^\circ + \text{m} \angle \text{D} &= 180^\circ \\
\therefore \text{m} \angle \text{D} &= 120^\circ
\end{align*}
Hence, the angles of the quadrilateral are $40^\circ, 60^\circ, 140^\circ$ and $120^\circ$.
Q.12 One angle of a quadrilateral has measure $\frac{2\pi^c}{5}$ and the measures of other three angles are in the ratio 2:3:4. Find their measures in degrees and radians.
Solutions: Let ABCD be the given quadrilateral with
\begin{align*}
\text{m} \angle \text{A} &= \frac{2\pi^c}{5} \\
&= \left( \frac{2\pi^c}{5} \times \frac{180}{\pi} \right)^\circ \\
&= 72^\circ \\
\end{align*}
and m$\angle$B:m$\angle$C: m$\angle$D = 2:3:4.
Let m$\angle$B = 2$k$, m$\angle$C = 3$k$ and m$\angle$D = 4$k$, where $k$ is the common multiple.
$\because$ sum of angles of a quadrilateral = 360$^\circ$.
\begin{align*}
\therefore \text{m} \angle \text{A} + \text{m} \angle \text{B} + \text{m} \angle \text{C} +\text{m} \angle \text{D} &= 360^\circ \\
72^\circ + 2k + 3k+4k &= 360^\circ \\
9k &= 288^\circ \\
k &= 32^\circ
\end{align*}
$\therefore$ The measure of other three angles in degrees are
\begin{align*}
\text{m} \angle \text{B} &= 2k = 2*32^\circ = 64^\circ \\
\text{m} \angle \text{C} &= 3k = 3*32^\circ = 96^\circ \\
\text{m} \angle \text{D} &= 4k = 4*32^\circ = 128^\circ
\end{align*}
The measures of angles in radian are
\begin{align*}
\text{m} \angle \text{B} &= \left(64^\circ \times \frac{\pi}{180} \right)^c = \frac{16\pi^c}{45} \\
\text{m} \angle \text{C} &= \left(96^\circ \times \frac{\pi}{180} \right)^c = \frac{8\pi^c}{15} \\
\text{m} \angle \text{D} &= \left(128^\circ \times \frac{\pi}{180} \right)^c = \frac{32\pi^c}{45}
\end{align*}
Q.13 Find the degree and radian measure of exterior and interior angle of a regular
i) Pentagon
Solution:
no. of sides = 5
Since, sum of exterior angles of a polygon = $360^\circ$.
As the pentagon is regular, all exterior angles are equal.
\begin{align*}
\therefore \text{each exterior angle} &= \frac{360^\circ}{5} \\
&= 72^\circ = \left( 72 \times \frac{\pi}{180} \right)^c = \frac{2 \pi^c}{5}
\end{align*}
$\because$ interior angle + exterior angle = 180$^\circ$
\begin{align*}
\therefore \text{each interior angle} &= 180^\circ – 72^\circ \\
&= 108^\circ = \left( 108 \times \frac{\pi}{180} \right)^c = \frac{3 \pi^c}{5}
\end{align*}
Ans: exterior angle = $72^\circ$ or $\frac{2 \pi^c}{5}$ and interior angle = $108^\circ$ or $\frac{3 \pi^c}{5}$.
ii) Hexagon
Solution: no. of sides = 6
Since, sum of exterior angles of a polygon = $360^\circ$.
As the pentagon is regular, all exterior angles are equal.
\begin{align*}
\therefore \text{each exterior angle} &= \frac{360^\circ}{6} \\
&= 60^\circ = \frac{\pi^c}{3}
\end{align*}
$\because$ interior angle + exterior angle = 180$^\circ$
\begin{align*}
\therefore \text{each interior angle} &= 180^\circ – 60^\circ \\
&= 120^\circ = \frac{2 \pi^c}{3}
\end{align*}
Ans: exterior angle = $60^\circ$ or $\frac{\pi^c}{3}$ and interior angle = $120^\circ$ or $\frac{2 \pi^c}{3}$.
iii) Septagon
Solution: no. of sides = 7
Since, sum of exterior angles of a polygon = $360^\circ$.
As the pentagon is regular, all exterior angles are equal.
\begin{align*}
\therefore \text{each exterior angle} &= \frac{360^\circ}{7} \\
&= 51.43^\circ \\
& \text{or} \\
&= \left( \frac{360}{7} \times \frac{\pi}{180} \right)^c \\
&= \frac{2 \pi^c}{7}
\end{align*}
$\because$ interior angle + exterior angle = 180$^\circ$
\begin{align*}
\therefore \text{each interior angle} &= 180^\circ – \frac{360^\circ}{7} \\
&= \frac{900^\circ}{7} \\
&= 128.57^\circ
& \text{or} \\
&= \left( \frac{900}{7} \times \frac{\pi}{180} \right)^c = \frac{5 \pi^c}{7}
\end{align*}
Ans: exterior angle = $51.43^\circ$ or $\frac{2 \pi^c}{7}$ and interior angle = $128.57^\circ$ or $\frac{5 \pi^c}{7}$.
iv) Octagon
Solution: no. of sides = 8
Since, sum of exterior angles of a polygon = $360^\circ$.
As the pentagon is regular, all exterior angles are equal.
\begin{align*}
\therefore \text{each exterior angle} &= \frac{360^\circ}{8} \\
&= 45^\circ = \left( 45 \times \frac{\pi}{180} \right)^c = \frac{ \pi^c}{4}
\end{align*}
$\because$ interior angle + exterior angle = 180$^\circ$
\begin{align*}
\therefore \text{each interior angle} &= 180^\circ – 45^\circ \\
&= 135^\circ = \left( 135 \times \frac{\pi}{180} \right)^c = \frac{3 \pi^c}{4}
\end{align*}
Ans: exterior angle = $45^\circ$ or $\frac{ \pi^c}{4}$ and interior angle = $135^\circ$ or $\frac{3\pi^c}{4}$.
Q.14 Find the angle between hour-hand and minute-hand in a clock at
i) ten past eleven
Solution:
Minute-hand is at mark 2.
Hour-hand is between marks 11 and 12.
Angle between two adjacent marks = $\frac{360^\circ}{12}$ = $30^\circ$
$\therefore$ Angle moved by hour-hand in 1-hour (or 60 minutes) = $30^\circ$
\begin{align*}
\text{Angle moved by hour-hand in 10 minutes } &= \frac{30^\circ}{60} \times 10 \\
&= 5^\circ
\end{align*}
\begin{align*}
\therefore \text{Angle between hour-hand and minute hand } &= \text{Angle between marks 11 and 2} \\
& \quad – \text{Angle moved by hour-hand in 10 minutes} \\
&= 3\times 30^\circ-5^\circ \\
&= 85^\circ
\end{align*}
Ans: $85^\circ$
ii) twenty past seven
Solution:
Minute-hand is at mark 4
Hour-hand is between marks 7 and 8.
Angle between two adjacent marks = $\frac{360^\circ}{12}$ = $30^\circ$
$\therefore$ Angle moved by hour-hand in 1-hour (or 60 minutes) = $30^\circ$
\begin{align*}
\text{Angle moved by hour-hand in 20 minutes } &= \frac{30^\circ}{60} \times 20 \\
&= 10^\circ
\end{align*}
\begin{align*}
\therefore \text{Angle between hour-hand and minute hand } &= \text{Angle between marks 11 and 2} \\
& \quad + \text{Angle moved by hour-hand in 20 minutes} \\
&= 3\times 30^\circ+10^\circ \\
&= 100^\circ
\end{align*}
Ans: $100^\circ$
iii) thirty five past one
Solution:
Minute-hand is at mark 7.
Hour-hand is between marks 1 and 2.
Angle between two adjacent marks = $\frac{360^\circ}{12}$ = $30^\circ$
$\therefore$ Angle moved by hour-hand in 1-hour (or 60 minutes) = $30^\circ$
\begin{align*}
\text{Angle moved by hour-hand in 35 minutes } &= \frac{30^\circ}{60} \times 35 \\
&= \frac{35^\circ}{2} \\
&= 17.5^\circ
\end{align*}
\begin{align*}
\therefore \text{Angle between hour-hand and minute hand } &= \text{Angle between marks 1 and 7} \\
& \quad – \text{Angle moved by hour-hand in 35 minutes} \\
&= 6\times 30^\circ-17.5^\circ \\
&= 162.5^\circ \\
&= 162^\circ 30′
\end{align*}
Ans: $162^\circ 30’$.
iv) quarter to six
Solution:
Minute-hand is at mark 9.
Hour-hand is between marks 5 and 6.
Angle between two adjacent marks = $\frac{360^\circ}{12}$ = $30^\circ$
$\therefore$ Angle moved by hour-hand in 1-hour (or 60 minutes) = $30^\circ$
\begin{align*}
\text{Angle to be moved by hour-hand in 15 minutes } &= \frac{30^\circ}{60} \times 15 \\
&= \frac{15^\circ}{2} \\
&= 7.5^\circ
\end{align*}
\begin{align*}
\therefore \text{Angle between hour-hand and minute hand } &= \text{Angle between marks 6 and 9} \\
& \quad + \text{Angle to be moved by hour-hand in 15 minutes} \\
&= 3\times 30^\circ+7.5^\circ \\
&= 97.5^\circ \\
&= 97^\circ 30′
\end{align*}
Ans: $97^\circ 30’$
v) 2:20
Solution:
Minute-hand is at mark 4
Hour-hand is between marks 2 and 3.
Angle between two adjacent marks = $\frac{360^\circ}{12}$ = $30^\circ$
$\therefore$ Angle moved by hour-hand in 1-hour (or 60 minutes) = $30^\circ$
\begin{align*}
\text{Angle moved by hour-hand in 20 minutes } &= \frac{30^\circ}{60} \times 20 \\
&= 10^\circ
\end{align*}
\begin{align*}
\therefore \text{Angle between hour-hand and minute hand } &= \text{Angle between marks 2 and 4} \\
& \quad – \text{Angle moved by hour-hand in 20 minutes} \\
&= 2\times 30^\circ-10^\circ \\
&= 50^\circ
\end{align*}
Ans: $50^\circ$
vi) 10:10
Solution:
Minute-hand is at mark 2
Hour-hand is between marks 10 and 11.
Angle between two adjacent marks = $\frac{360^\circ}{12}$ = $30^\circ$
$\therefore$ Angle moved by hour-hand in 1-hour (or 60 minutes) = $30^\circ$
\begin{align*}
\text{Angle moved by hour-hand in 10 minutes } &= \frac{30^\circ}{60} \times 10 \\
&= 5^\circ
\end{align*}
\begin{align*}
\therefore \text{Angle between hour-hand and minute hand } &= \text{Angle between marks 10 and 2} \\
& \quad – \text{Angle moved by hour-hand in 10 minutes} \\
&= 4\times 30^\circ-5^\circ \\
&= 115^\circ
\end{align*}
Ans: $115^\circ$.
Summary
In this article, we have seen solutions for all problems of exercise 1.1 of chapter Angle and its measurement. If you face any problem or query related to any problem in this exercise, then you may drop a comment.
Happy learning!
