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Questions:

Generic 5

10 x 5 = 50

As shown in the figure, two trees are standing on flat ground. The angle of elevation of the top of both the trees from a point X on the ground is 40° . If the horizontal distance between X and the smaller tree is 8 m and the distance of the top of the two trees is 20 m, calculate
(i) the distance between the point X and the top of the smaller tree.
(ii) the horizontal distance between the two trees.
(cos 40° = 0.7660)

An Aeroplane sets of from G on bearing of 24° towards H, a point 250 km away, at H it changes course and heads towards J deviates further by 55° and a distance of 180 km away.
How for is H to the east of G?

\(\left( \begin{matrix} sin24°=0.4067\quad sin11°=0.1908 \\ cos24°=0.9135\quad cos11°=0.9816 \end{matrix} \right) \)

if cot \(\theta \) + tan\(\theta \) = x and sec\(\theta \) - cos\(\theta \) = y, then prove that \(\begin{equation} \left(x^{2} y\right)^{\frac{2}{3}}-\left(x y^{2}\right)^{\frac{2}{3}}=1 \end{equation}\)

prove that \({ \left( \frac { 1+sin\theta -cos\theta }{ 1+sin\theta +cos\theta } \right) }^{ 2 }=\frac { 1-cos\theta }{ 1+cos\theta } \)

Two ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45° respectively. If the distance between the ships is 200\(\left( \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } } \right) \) metres, find the height of the lighthouse.

A bird is flying from A towards B at an angle of 35°, a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48° and distance 32 km away.
How far is B to the North of A? (sin 55° = 0.8192, cos 55° = 0.5736,sin 42° = 0.6691.cos 42° = 0.7431)

If x sin³\(\theta \) + ycos³\(\theta \) = sin\(\theta \) cos\(\theta \) and x sin\(\theta \) = ycos\(\theta \), then prove that x²+ y²= 1.

Three villagers A, B and C can see each other across a valley. The horizontal distance between A and B is 8 km and the horizontal distance between B and C is 12 km. The angle of depression of B from A is 20° and the angle of elevation of C from B is 30° . Calculate : the vertical height between A and B.(tan20° = 0.3640,(\(\sqrt { 3 } \) = 1.732)

The angle of elevation of the top of a cell phone tower from the foot of a high apartment is 60° and the angle of depression of the foot of the tower from the top of the apartment is 30° . If the height of the apartment is 50 m, find the height of the cell phone tower. According to radiations control norms, the minimum height of a cell phone tower should be 120 m. State if the height of the above mentioned cell phone tower meets the radiation norms.

10.

A flag pole of height ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30°. Find (i) the height of the pole (ii) radius of the (\( \sqrt { 3 } \) =1.732)

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Answers Key:

Generic 5

10 x 5 = 50

Let AB be the height of the biggest tree and CD be the highest of the smaller tree and x is the point on the ground.
(i) In the right angled ΔXCD, cos 40°= \(\frac{C X}{X D}\)
\(X D=\frac{8}{0.7660}=10.44 \mathrm{~m}\)
Therefore, the distance between X and top of the smaller tree = XD = 10.44 m
(ii) In the right triangle XAB,
cos40° =\(\frac { AX }{ BX } =\frac { AC+CX }{ BD+DX } \)
0.7660 = \(\frac { AC+8 }{ 20+10.44 } \) gives AC = 23.32 - 8 = 15.32m
Therefore the horizontal distance between two trees = AC = 15.32.

In right triangle GOH,
sin24° = \(\frac { OG }{ GH } \)
0.4067 = \(\frac { OH }{ 250 } \) ; OH = 101.68
Distance of H to the east of G = 101.68 km.

We have \(\cot \theta+\tan \theta=x\)
\(\sec \theta-\cos \theta=y \)
Taking \(\cot \theta+\tan \theta=x \)
\(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=x \)
\(\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}=x \frac{1}{\sin \theta \cos \theta}=x \)
\(\sec \theta-\cos \theta =y \)
\(\frac{1}{\cos \theta}-\cos \theta =y \)
\(\frac{1-\cos ^{2} \theta}{\cos \theta} =y \)
\(\frac{\sin ^{2} \theta}{\cos \theta} =y \)
Now
\(=\left[\left(\frac{1}{\cos \theta \sin \theta}\right)^{2}\left(\frac{\sin ^{2} \theta}{\cos \theta}\right)\right]^{\frac{2}{3}}-\left[\left(\frac{1}{\cos \theta \sin \theta}\right)\left(\frac{\sin ^{4} \theta}{\cos ^{2} \theta}\right)\right]^{\frac{2}{3}}\)
\(=\left[\frac{1}{\cos ^{2} \theta \sin ^{2} \theta} \times \frac{\sin ^{2} \theta}{\cos \theta}\right]^{\frac{3}{3}}-\left[\frac{1}{\cos \theta \sin \theta} \times \frac{\sin ^{4} \theta}{\cos ^{2} \theta}\right]^{\frac{3}{3}}\)
\(=\left(\frac{1}{\cos ^{3} \theta}\right)^{\frac{2}{3}}-\left(\frac{\sin ^{3} \theta}{\cos ^{3} \theta}\right)^{\frac{2}{3}} \)
\(=\frac{1}{\cos ^{2} \theta}-\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \)
\(=\sec ^{2} \theta-\tan ^{2} \theta \)
= 1 = RHS

\( \text { LHS }=\left(\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\right)^{2} \)
\(=\left[\frac{(1+\sin \theta)-\cos \theta}{(1+\sin \theta)+\cos \theta} \times \frac{(1+\sin \theta)-\cos \theta}{(1+\sin \theta)-\cos \theta}\right)^{2} \)
\(=\left[\frac{((1+\sin \theta)-\cos \theta)^{2}}{(1+\sin \theta)^{2}-(\cos \theta)^{2}}\right]^{2} \)
\(=\left[\frac{1+\sin ^{2} \theta+2 \sin \theta+\cos ^{2} \theta-2 \cos \theta(1+\sin \theta)}{1+\sin ^{2} \theta+2 \sin \theta-\cos ^{2} \theta}\right]^{2} \)
\( =\left[\frac{\sin ^{2} \theta+2 \sin \theta+1+\cos ^{2} \theta-2 \cos \theta-2 \sin \theta \cos \theta}{1+\left(1-\cos ^{2} \theta\right)+2 \sin \theta-\cos ^{2} \theta}\right]^{2} \)
\(=\left[\frac{\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+2 \sin \theta+1-2 \cos \theta(1+\sin \theta)}{1+1-2 \cos ^{2} \theta+2 \sin \theta}\right]^{2} \)
\(=\left[\frac{1+1+2 \sin \theta-2 \cos \theta(1+\sin \theta)}{2-2 \cos ^{2} \theta+2 \sin \theta}\right]^{2} \)
\(=\left[\frac{2+2 \sin \theta-2 \cos \theta(1+\sin \theta)}{2-2 \cos ^{2} \theta+2 \sin \theta}\right]^{2} \)
\(=\left(\frac{2(1+\sin \theta)-2 \cos \theta(1+\sin \theta)}{2\left(1-\cos ^{2} \theta\right)+2 \sin \theta}\right)^{2} \)
\(=\left(\frac{(1+\sin \theta) 2(1-\cos \theta)}{2 \sin ^{2} \theta+2 \sin \theta}\right)^{2}\)
\( =\left(\frac{2(1+\sin \theta)(1-\cos \theta)}{2 \sin \theta(1+\sin \theta)}\right)^{2} \)
\(=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2} \)
\(=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta} \)
\(=\frac{(1-\cos \theta)(1-\cos \theta)}{1-\cos { }^{2} \theta} \)
\(=\frac{(1-\cos \theta)(1-\cos \theta)}{(1+\cos \theta)(1-\cos \theta)} \)
\(=\frac{1-\cos \theta}{1+\cos \theta} \)
= RHS

Let C and D are two ships.
Let AB be the height of the light house.
\(\mathrm{CD}=200\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right) m\)
In the right ABD
\(\tan 60^{\circ}=\frac{A B}{B D} \)
\(\sqrt{3}=\frac{A B}{B D} \)
\(\mathrm{BD}=\frac{A B}{\sqrt{3}} \)
In the right triangle ABC
\(\tan 45^{\circ} =\frac{A B}{B C} \)
\(1 =\frac{A B}{B C} \)
\(B C =A B \)
\((1)+(2) \Rightarrow B D +B C=\frac{A B}{\sqrt{3}}+A B \)
\(\mathrm{CD}=A B\left(\frac{1}{\sqrt{3}}+1\right) \quad[\because \mathrm{CB}+\mathrm{BD}=\mathrm{CD}] \)
\(\frac{C D}{\left(\frac{1}{\sqrt{3}}+1\right)}=\mathrm{AB} \)
\(A B=\frac{200\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}{\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}\)
AB = 200 m
Height of the light house is 200 m.

Let A be the initial position of the bird.
B be the position after travelling 30 km at an angle of 35^o from A.
Let C be the position after travelling 32 km at an angle of 48^o from B.
o
o
[complementary angle]
In AOB
\(\sin 55^{\circ}=\frac{O B}{A B} \)
\(0.8192=\frac{O B}{30} \)
OB = 30 x 0.8192 = 24.58 km

We have x sin³\(\theta \) + ycos³\(\theta \) = sin\(\theta \) cos\(\theta \)
\( \Rightarrow \ (x \sin \theta)\left(\sin ^{2} \theta\right)+(y \cos \theta) \cos ^{2} \theta \)
\(= \sin \theta \cos \theta \)
\(\Rightarrow \ x \sin \theta\left(\sin ^{2} \theta\right)+(x \sin \theta) \cos ^{2} \theta \)
\(= \sin \theta \cos \theta \quad[\because x \sin \theta=y \cos \theta] \mathrm{S} \)
\(\Rightarrow \ x \sin \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\sin \theta \cos \theta \)
\(\Rightarrow x \sin \theta=\sin \theta \cos \theta \)
\( \mathrm{x}=\cos \theta\) ...(1)
\(\text { Now, } x \sin \theta=y \cos \theta \)
\( \Rightarrow \cos \theta \sin \theta=y \cos \theta \)
\(\Rightarrow [\because x=\cos \theta \text { from (1) }] \)
y = sin\(\theta \) ...(2)
From (1) and (2)
\( x^{2}+y^{2}=\cos ^{2} \theta+\sin ^{2} \theta \)
= 1
x²+ y²= 1.

In the right \(\triangle\)ADB
\( \tan 20^{\circ} =\frac{A D}{D B} \)
\(0.3640 =\frac{A D}{8}\)
AD = 8 x 0.3640 = 2.91 km
Vertical height between A and B = 2.91 km

Let AB be the cell phone tower.
CD be the apartment.
\( \angle X C B=\angle C B D=30^{\circ} \)
\(\angle A D B=60^{\circ} \)
ln right triangle ABD
\(\tan 60^{\circ}=\frac{A B}{B D}\)
\( \sqrt{3}=\frac{A B}{B D} \)
\(B D=\frac{A B}{\sqrt{3}} \) ...(1)
In the right triangle \(\triangle\)CDB
\( \tan 30^{\circ} =\frac{C D}{B D} \)
\(\frac{1}{\sqrt{3}} =\frac{50}{B D} \)
\(B D =50 \sqrt{3} \)
From (1) and (2)
\( \frac{A B}{\sqrt{3}}=50 \sqrt{3} \)
\(A B=50 \times \sqrt{3} \times \sqrt{3}=50 \times 3=150 \mathrm{~m} \)
Height of the cell phone tower = 150 m.
Since height of the tower > 120 m, yes, the tower meets the radiation norms.

10.

Let BD be the radius of the dome AD is the flag pole of height 'h' m.
(i) From the right triangle \(\triangle\)ABE
\( \tan 45^{\circ} =\frac{A B}{B E} \)
\(1 =\frac{r+h}{r+7} \)
r + 7 = r + h
r + h - r = 7
h = 7m
Height of the flag pole = 7 m
(ii) From the right triangle \(\triangle\)BDC
\( \tan 30^{\circ} =\frac{B D}{B C}=\frac{r}{r+7+5} \)
\(\frac{1}{\sqrt{3}} =\frac{r}{r+12} \)
\(r+12 =\sqrt{3} r \)
\(12 =\sqrt{3} r-r \)
\(12 =r(\sqrt{3}-1) \)
\(r =\frac{12}{1.732-1}=\frac{12}{0.732}=16.39 \mathrm{~m} \)
Height of pole = 7 m
Radius of dome = 16.39 m.