Exercise 16 B
Question 1
Ans: Using the since, cosine and tangent tables
(a) $\quad 15^{\circ} 27^{\prime}$
$=\sin 15^{\circ} 24^{\prime}+3^{\prime}$ (mean difference of 3)
$=0.26556+84$
$=0.26640=0.2664$
$\cos 15^{\prime} 27^{\prime}=\cos 15^{\circ} 24^{\prime}+3^{\prime}$
$=0.96410-23$ (Mean difference of 3 )
$=0.96387=0.9639$
$\tan 15^{\circ} 27^{\prime}=\tan 15^{\circ} 24^{\prime}+3^{\prime}$
$=0.27545+94=0.27639=0.2764$
(b) $\sin 3748^{\prime}=$ $0.6129$
$\operatorname{Cos} 37^{\circ} 48^{\prime}=0.79015=0.7902$
$\tan 37.48^{\prime}=0.77568=0.7757$
(c) $\sin 55^{\circ} 17^{\prime}=\sin 5555^{\circ} 12^{\prime}+5^{\prime}$
$=0.82115+82$ (Mean dillerence of $5^{\prime}$ )
$=0.82917=0.8219$
$\cos 55^{\circ} 17^{\prime}=\cos 55^{\circ} 12^{\prime}+5^{\prime}$
$=0.57071-120=0.56951=0.5695$
$\tan 55^{\circ} 17^{\prime}=\tan 55^{\circ} 12^{\prime}+5^{\prime}$
$=1.4388+453=1.44334=0.4433 .$
(d) $\sin 83^{\circ} 37^{\prime}=$ $\sin 83^{\circ} 36^{\prime}+1^{\prime}$
$0.99377+3=0.99380=0.9938$
$\operatorname{Cos} 83^{\circ} 37^{\prime}=\operatorname{Cos} 83^{\circ} 36^{\prime}+1$
$0.11147-29=0.1118=0.1112$
$\tan 8337^{\prime}=8.91520=8.9152$
Question 2
Ans: Find the acute angle A, given
(a) $\quad \sin A=0.4919$
$=0.4919=0.49090+$ difference
$=100$
$=\sin 2924^{\prime}+4=\sin 29.28^{\prime}$
$\therefore A=29.281$
(b) $\tan A=2.7775$
$=2.7775=2.77761$ (it is nearest to 2.777 So)
$\begin{aligned} \therefore \tan A &=\tan 70^{\circ} 12^{\prime} \\ \therefore \quad A &=70^{\circ} 12^{\prime} \end{aligned}$
(c) $\tan A \quad 3.412$
=3.41973
$=\tan 73^{\circ} 42'$ (∵ 3.91973 is nearest to 3.412)
$\therefore \quad A=73^{\circ} 42^{\prime}$
(d) $\cos A=0.4651$
$=0.46484+16$
$=\cos 62^{\prime} 18-1^{\prime}=\operatorname{Cos} 62^{\circ} 17^{\prime}$
$\therefore A=62.17^{\prime}$
(e)
$\begin{aligned} \sin A &=0.95190=095150+31 \\ &=\sin 72^{\circ} 6^{\prime}+3^{\prime}=\sin 72^{\circ} 9^{\prime} \\ \therefore A &=72^{\circ} 9^{\prime} \end{aligned}$
(f) $\operatorname{Cos} A=$ $0.57570=.57501+69$
$=\operatorname{Cos} 54^{\circ} 54^{\prime}-3^{\prime}=$
$\operatorname{Cos} 54^{\prime} 51$
$\therefore A=54.5^{\prime}$
Question 3
Ans: Using tables , find the value of $(2sin\theta- Cos \theta)$
(i) when $\theta=35^{\circ}$
(a) $2 \sin \theta-\cos \theta=2 \sin 35^{\circ}-\cos 35^{\circ}$
$=2(0.57358)-0.81915$ (from table)
$=1.14716-0.81915$
$=0.32801=0.3280$
(b) When tan $\theta=0.2679$
tan 19.56
$\begin{aligned} \therefore & 2 \sin \theta-\cos \theta \\=& 2 \sin 1456^{\prime}-\cos 1456^{\prime} \\=& 2(0.25769)-0.9662 e \\ & 0.51538-0.96622=-0.45084 \end{aligned}$
Question 4
Ans: State for any acute angle $\theta$
(1) Whether sin $\theta$ increase or decrease as increase
(i) We know that sin $\theta$=0 and sin 90= 1
ஃ It is clear that sin$\theta$ increase as $\theta$ increase
(2) Whether cos $\theta$ increase or decrease as $\theta$ decrease
(ii) We know that cos $\theta$ and cos 90 = 0
ஃ it is clear that as $\theta$ decrease , cos $\theta$ increase
Question 5
Ans: $\begin{aligned} \sin x^{\circ} &=0.67 \\ &=0.67043 \\=& \sin 42^{\circ} .6^{\prime}-2^{\prime} \\=& \sin 42^{\circ} 4^{\prime} \end{aligned}$
(a)
$\begin{aligned} \cos x^{\circ} &=\cos 42^{\circ} 4^{\prime} \\ &=0.74314-77 \\ &=0.742 .37 . \\ &=0.7423 \end{aligned}$
(b) $\cos x^{\circ}+\tan x^{\circ}$
$=\cos 42^{\circ} 4^{\prime}+42^{\circ} 4^{\prime}$
$=0.7423+(0.90040+214)$
$=0.7423+0.90254$
$=0.7423+0.90 .25$
$=1.6448$
Question 6
Ans: $\sin A=0.1822$
$\sin A=0.18224$
$A=\sin 10^{\circ} 30$
$A=10^{\circ} 30$
Question 7
Ans: Given,
Rectangle ABCD , AC is its diagonal
AB = 23CM
$\angle C A B=35^{\circ}$
Let $B C=x$
In Right $\triangle A B C$
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$\tan \theta=\frac{B C}{A B}$
$\tan 35^{\circ}=\frac{x}{23}$
$0.70021=\frac{x}{23}$
$\begin{aligned} x &=23 \times 0.70021 \\ &=16.10483 \\ &=16.1048 \\ &=16.11 \\ BC &=16.11 \mathrm{~cm} \end{aligned}$
Question 9
Ans: Given,
$\begin{aligned} B C &=12 \mathrm{~cm} \\ A B &=4 \mathrm{~cm} ; \\ \angle A E B &=50^{\circ}, \\ \angle B &=50^{\circ} \text { and } \\ \angle C &=30^{\circ} \end{aligned}$
(i) In right angle $\triangle A E B$;
$\operatorname{Cos} 50^{\circ}=\frac{B E}{A B}$
$.6428=\frac{8 E}{4}$
$B E=: 6428 \times 4 .$
$B E=2.5712 \mathrm{~cm}$
(ii) $\begin{aligned} \operatorname{Sin} 50^{\circ} &=\frac{A E}{A B} \\ \cdot 7660 &=\frac{A E}{4} \\ \cdot 7660 \times 4 &=A E \\ 3.0640 &=A E \\ A E &=3.064 \mathrm{~cm} \end{aligned}$
In $\triangle A E C$
$\begin{aligned}\sin 30^{\circ} &=\frac{A E}{A C} \\.5000 &=\frac{3.064}{A C}\end{aligned}$
$A C=\frac{3.0640}{.5000}$
$A C=\frac{3064}{500}$
$A C=6.128 \mathrm{~cm} .$
Question 10
Ans: Given,
From $\triangle A B C$
$\angle B=90^{\circ}$
$\angle C=30^{\circ}$
$\begin{aligned} \therefore \angle A &=180^{\circ}-\left(90^{\circ}+30^{\circ}\right) \\ \angle A &=180^{\circ}-920^{\circ} \\ \angle A &=60^{\circ} \end{aligned}$
(i) In right angle $\triangle A B C$,
$\tan 30^{\circ}=\frac{A B}{B C} .$
$\frac{1}{\sqrt{3}}=\frac{12}{B C}$
$B C=12 \sqrt{3} \mathrm{~cm}$.
(ii) In right angle $\triangle B D A$
$\cos 60^{\circ}=\frac{A D}{A B} .$
$\frac{1}{2}=\frac{A D}{12}$
$\frac{12}{2}=A D$
$6=A D$
$A D=6 \mathrm{~cm} .$
(iii) In right angle $\triangle A B C$
$\sin 30^{\circ}=\frac{A B}{A C}$
$\frac{1}{2}=\frac{12}{A C}$
$A C=12 \times 2$
$A C=24 \mathrm{~cm} .$
Question 11
Ans: Radius of the circle with center C is 15cm
(IMAGE TO BE ADDED)
ex $A C=13 C=15 \mathrm{~cm}$
$\angle A C B=131^{\circ}$
From C, Draw CL $\perp A B$ Now in $D A B C$, $\angle C=131^{\circ} \mathrm{C}$ $A C=B C$
$\begin{aligned} \therefore \angle A=\angle B &=\frac{180^{\circ}-131^{\circ}}{2} \\ &=\frac{49^{\circ}}{2} \\ &=24. 5^{\circ}=24^{\circ} 30^{\prime} \end{aligned}$
(i) Now in right triangle ACL, LA = $20^{\circ} 36^{\circ}$
$\begin{aligned} \therefore \quad \cos o=\frac{A L}{A C} &=\cos 24^{\circ} 30^{\circ} \\ &=\frac{A L}{15} \end{aligned}$
$0.90996=\frac{A L}{15}$ $A L=15 \times 0.90996$
$=\quad A L=13.6494$
and $\begin{aligned} A B &=2 A L=2 \times 13.6494 \\ &=27.2988=27.3 \mathrm{~cm} \end{aligned}$
(ii) Sinθ $=\frac{C L}{A C}$ So, $\sin 24^{\circ} 30^{\circ}=\frac{C L}{15}$
$=0.41469=\frac{C L}{15}$ $C L=15 \times 0.41469$
$\Rightarrow C L=6.22035=6.22 \mathrm{CM}$
Hence , the distance of AB from the center C= 6.22cm
Question 12
Ans: (IMAGE TO BE ADDED)
Given,
$A P=20 \mathrm{~km}$
$A B=80 \mathrm{~km} .$
$A B$ making an angle of $30^{\circ}$
$\begin{aligned}\angle B A D &=90^{\circ}-30^{\circ} \\&=60^{\circ}\end{aligned}$
(i) In right angle $\triangle A D B$,
$\sin 60^{\circ}=\frac{B D}{A B}$
$\frac{\sqrt{3}}{2}=\frac{B D}{80}$
$B D=\frac{80}{2} \sqrt{3}$
$B D=40 \sqrt{3}$
$B D=40 \times 1.732 .$
$B D=69.280 \mathrm{~km}$
$\begin{aligned} \therefore B C &=B D+D C . \\ &=69.280420 \\ &=89.280 \mathrm{~km} \end{aligned}$
(ii) In right angle $\triangle A D B$
$\operatorname{Cos} 60^{\circ}=\frac{A D}{A B}$
$\frac{1}{2}=\frac{A D}{80}$
$A D=\frac{80}{2}$
$A D=40 \mathrm{~km}$
So, $A D=P C=40 \mathrm{~km}$.
Hence the horizontal distance of point C from point P is 40km
Question 13
Ans: BCDE is a rectangle in which ED = 3.88cm
BC = 3.88CM
A is a point such that AD = 10cm and A lie
On CB on producing AE is joined
Let angle AEB = $\theta$
(Image to be added)
(i) In right triangle ACD
$\begin{aligned} & \sin 0=\frac{C D}{A D} \\ & \sin 23^{\circ} 35^{\circ}=\frac{C D}{10} \\ \therefore & 0.40008=\frac{C D}{10} \\ \text { so, } C D &=4.00 \mathrm{⊥} \mathrm{CM} \end{aligned}$
(ii) $\cos \theta=\frac{A C}{A D}=$ $\cos 23^{\circ} 35^{\prime}=\frac{A C}{L O}$
$\begin{aligned} \therefore \quad 0.91648=\frac{A C}{10}=A C &=9.1648 \\ &=9.165 \end{aligned}$
$A C=2.165 \mathrm{~cm}$
(iii) Now $A B=A C-B C=9.165-3.880=5.285$
and $E B=C D=4.00 \mathrm{~L}$
$\therefore \tan \theta=\frac{A B}{E B}=\frac{5.285}{4.001}$
$=\frac{5.285}{4001}=1.32092$
$=1.31745+347$
$=\tan 52^{\circ} .48^{\circ}+5^{\circ}$
$=\tan 52^{\circ} 53^{\prime}$
$\theta=52^{\circ} 53^{\circ}$
$\angle A E B=52^{\circ} 53^{'}$
Question 14
Ans: In right angle triangle, ABC,
$\begin{aligned} \angle B &=90^{\circ} \\ B C &=3 \mathrm{~cm}, \\ A B &=4 \mathrm{~cm} . \end{aligned}$
$\begin{aligned} \therefore A C^{2} &=B C^{2}+A B^{2} . \\ A C^{2} &=(3)^{2}+(4)^{2} \\ A C &=\sqrt{9+16} \\ A C &=\sqrt{25} \\ A C &=5 \mathrm{Cm} . \end{aligned}$
From $\triangle A B C$ and $\triangle D B C$,
$\begin{aligned}&\angle A B C=\angle B D C . \\&\angle C=\angle C\end{aligned}$
$\therefore \triangle A B C \sim \triangle D B C .$ (By AA)
$\triangle A B C \sim \triangle A B D .$
$\frac{A C}{B C}=\frac{A B}{B D}=\frac{B C}{C D} .$
$\frac{A B}{B C}=\frac{B D}{C D} .$ (By alternate)
$\frac{B D}{C D}=\frac{4}{3}$
$\frac{C D}{B D}=\frac{3}{4} .$
(i) $\therefore \tan \angle D B C=\frac{C D}{B D}=\frac{3}{4}$.
(ii) In right angle$\triangle A B D$
sin $\angle D E A=\frac{A D}{A B}=\frac{A B}{A C}=\frac{4}{5}$
Question 15
Ans: Let BC be the building and
AB be the flag pole on the building
So BC = x and
AB= y
Angle of elevation $\angle B D C=63^{\circ}$
and angle ADC= $63^{\circ}+3^{\circ}$
$=66^{\circ}$
From right angle $\triangle B C D$
$\tan 63^{\circ}=\frac{B C}{D C}$
$1.9626=\frac{x}{50}$
$x=1.9626 \times 50$
$x=98.1300$
$x=98 \mathrm{~m}$
From right angle $\triangle A C D$
$\begin{aligned} \tan 66^{\circ} &=\frac{A B+B C}{D C} \\ 2.2460 &=\frac{x+y}{50} \\ 2.2460 &=\frac{98+y}{50} \\ 2-2460 \times 50 &=98+y \\ 112 \cdot 3000 &=98+y \\ 112 \cdot 3000-98 &=y \\ 14 \cdot 3000 &=y \\ y &=14 \mathrm{~m} \end{aligned}$
Question 16
Ans: (IMAGE TO BE ADDED)
TR is the tree which was broken from Q and its top T touched the ground at S. So that SR= 25m
and angle QSR = 30
In the figure TQ= QS
$\tan \theta=\frac{Q R}{S R}=\tan 30^{\circ}=\frac{Q R}{25}$
$=\frac{1}{\sqrt{3}}=\frac{Q R}{25}=Q R=\frac{25}{\sqrt{3}} .$
$=Q \cdot R=\frac{25}{1.732}=14.43$
and $\cos \theta=\frac{S R}{S Q} \Rightarrow \cos 30^{\circ}=\frac{25}{S Q}$
$=\frac{\sqrt{3}}{2}=\frac{25}{SQ}$
$=5Q=\frac{25 \times 2}{\sqrt{3}}=\frac{50}{\sqrt{3}}
Height of tree = TQ +QR
$=Q S+Q R=\frac{25}{\sqrt{3}}+\frac{50}{\sqrt{3}}=\frac{75}{\sqrt{3}}$
$=\frac{75 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{75 \sqrt{3}}{3}$
$=25 \sqrt{3} \mathrm{~m}$
$=25(1.732)$
$=43.3 \mathrm{~m}$
$=43 \mathrm{~m}$
Question 17
Ans: In equilateral triangle ABC with each side 6cm
If D is a point on BC such that BD = 2cm
E is the mid point of BC
DE = DE - BD = 3-1 =2cm
(if E is mid point of BC )
(IMAGE TO BE ADDED)
(i) if $E$ is mid point of $B C$
$\text { so } A E \perp B C$
and $A D=\frac{\sqrt{3}}{2}$ side $=\frac{\sqrt{3}}{2} \times 6=3 \sqrt{3} \mathrm{~cm}$
(ii) In right triangle ADE
$\begin{aligned} \tan \angle A D C &=\tan \angle A D E=\frac{A E}{D E}=\frac{3 \sqrt{3}}{2} \mathrm{CM} \\ &=\frac{3(1.732)}{2}=3 \times 0.866= 2.598 \end{aligned}$
(iii) $\begin{aligned} \tan \angle A D C &=2.59156 \\ &=68^{\circ} 54^{\circ}=69^{\circ} \end{aligned}$
(iv) $\tan \angle D A E=\frac{D E}{A E}$
$=\frac{2}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{3 \times \sqrt{3} \times \sqrt{3}}$
$\begin{aligned}=\frac{2 \sqrt{3}}{9} &=\frac{2(1.732)}{9}=\frac{3.464}{9} \\ &=0.3 .85 \end{aligned}$
$\begin{aligned}=0.38587 &=\tan 21^{\circ} 6^{\prime} \\ &=\tan 21^{\circ} . \end{aligned}$
So $\angle D A E=21^{\circ}$
But $\angle B A C=\angle B A E-\angle D A E$
$=30^{\circ}-21^{\circ}=9^{\circ}$(If AE also bisects angle A)
Question 18
Ans: K be the kite which is 75 m above the ground and its string makes angle of 60 with the ground
(IMAGE TO BE ADDED)
so In $\triangle K B T$
$\begin{aligned}&K T=75 \mathrm{~m} \\&\angle B=60^{\circ} \\&\angle T=90^{\circ} \\&\text { Let } K B=x \mathrm{~m}\end{aligned}$
$\begin{aligned} \therefore \quad & \sin \theta=\frac{K T}{K B} \\ & \sin 60^{\circ}=\frac{75}{x} \\ & \frac{\sqrt{3}}{2}=\frac{75}{x} \end{aligned}$
$x=\frac{75$x=\frac{75 \times 2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{150 \sqrt{3}}{3}=50 \sqrt{3}$ \times 2}{\sqrt{3}}$
$=50(1.732)=86.6=87$
So length of string of the kite = 87m
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