# Chapter 8 Introduction to Trigonometry and its Equation NCERT Exemplar Solutions Exercise 8.4 Class 10 Maths

 Chapter Name NCERT Maths Exemplar Solutions for Chapter 8 Introduction to Trigonometry and its Equation Exercise 8.4 Book Name NCERT Exemplar for Class 10 Maths Other Exercises Exercise 8.1Exercise 8.2Exercise 8.3 Related Study NCERT Solutions for Class 10 Maths

### Exercise 8.4 Solutions

1. If cosecÎ¸ + cotÎ¸ = p, then prove that cosÎ¸ = (p2 – 1)/(p2 + 1) .

Solution

As,
cosecÎ¸ + cot Î¸ = p

2. Prove that √(sec2 Î¸ + cosec2 Î¸) = tan Î¸ + cot Î¸.

Solution

Taking,
L.H.S = √(sec2 Î¸ + cosec2 Î¸)
Since,

3. The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.

Solution

Let us take PR = h meter, the height of the tower.
The distance between the observer and tower is QR = (20 +x) m,
where
QR = QS + SR
= 20 + x
∠PQR = 30°
∠ PSR = Î¸
In ∆PQR,
tan 30° = h/(20 +x )
⇒ 1/√3 = h/(20 + x)
On cross multiplication,
We get,
20 + x = √3h
⇒ x = √3h - 20 ...eq. (i)
In ∆PSR,
tan Î¸ = h/x
As, angle of elevation increases by 15°  when the observer moves 20 m towards the tower.
So,
Î¸ = 30° + 15° = 45°
Therefore,
tan 45° = h/x
⇒ 1 = h/x
⇒ h = x
Putting x = h in eq. (i),
h = √3h - 20
⇒ √3h - h = 20
⇒ h(√3 - 1) = 20
On rationalizing we get,
h = 10(√3 + 1)
The required height of the tower is 10(√3 + 1) meter.

4. If  1 + sin2 Î¸ = 3sin Î¸ cosÎ¸,  then prove that tan Î¸ = 1 or 1/2.

Solution

We have,
1 + sin2 Î¸ = 3sin Î¸ cosÎ¸
on dividing L.H.S and R.H.S equations with sin2 Î¸,
We get,

cosec2 Î¸ + 1 = 3 cot Î¸
Also,
cosec2 Î¸ - cot2 Î¸ = 1
⇒ cosec2 Î¸ = cot2 Î¸ +1
⇒ cot2 Î¸ +1+1 = 3 cot Î¸
⇒ cot2 Î¸ +2 = 3 cot Î¸
⇒ cot2 Î¸ –3 cot Î¸ +2 = 0
We split the middle term and then solve the equation,
cot2 Î¸ – cot Î¸ –2 cot Î¸ +2 = 0
⇒ cot Î¸(cot Î¸ -1)–2(cot Î¸ +1) = 0
⇒ (cot Î¸ – 1)(cot Î¸ – 2) = 0
⇒ cot Î¸ = 1, 2
We have,
tan Î¸ = 1/cotÎ¸
tan Î¸ = 1, 1/2
Proved !!

5. Given that sinÎ¸ + 2cosÎ¸ = 1, then prove that 2sinÎ¸ - cosÎ¸ = 2.

Solution

We have,
sinÎ¸ + 2 cosÎ¸ = 1
Squaring on both sides,
(sinÎ¸ + 2cosÎ¸)2 = 1
⇒ sin2 Î¸ + 4cos2 Î¸ + 4 sinÎ¸ cosÎ¸ = 1
Also,
sin2 Î¸ = 1 - cos2 Î¸
and
cos2 Î¸ = 1 - sin2 Î¸
⇒ (1 - cos2 Î¸) + 4(1 - sin2 Î¸) + 4sin Î¸ cosÎ¸ = 1
⇒ 1 – cos2 Î¸ + 4 – 4 sin2 Î¸ + 4sin Î¸cos Î¸ = 1
⇒ –4 sin2 Î¸ – cos2 Î¸ + 4sin Î¸cos Î¸ = –4
⇒ 4 sin2 Î¸ + cos2 Î¸ – 4sin Î¸cos Î¸ = 4
Also,
a2 + b2 - 2ab = (a - b)2
So,
(2sin Î¸ – cos Î¸)2 = 4
⇒ 2sin Î¸ – cos Î¸ = 2
Proved

6. The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st.
Solution
Let BC = s;
PC = t
Let height of the tower be AB = h.
∠ABC = Î¸ and
∠APC = 90° – Î¸
(∵ the angle of elevation of the top of the tower from two points P and B are complementary)

In ABC,
tan Î¸ = AC/BC
= h/s  ...(i)
In APC,
tan(90 - Î¸) = AC/PC
= h/t
cotÎ¸ = h/t  ...(ii)
So, multiplying (i) and (ii),
tanÎ¸ × cot Î¸ = h/t × h/s
⇒ 1 = h2 /st
⇒ h2  = st
⇒ h = √st
Therefore, the height of the tower is √st.

7. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.
Solution
Let SQ = h be the tower.
∠SPQ = 30° and
∠SRQ = 60°
As given in the question, the length of shadow is 50 m long hen angle of elevation of the sun
is 30° than when it was 60°. So,
PR = 50 m and
RQ = x m

In SRQ,
tan 60° = h/x
⇒ √3 = h/x
⇒ x = h/√3
In SPQ,
tan 30° = h/(50 + x)
1/√3 = h(50 + x)
⇒ 50 + x = √3h
Putting value of x,
50 + h/√3 = √3h
⇒ 50√3 + h = 3h
⇒ 50√3 = 3h - h
⇒ 3h - h = 50√3
⇒ 2h = 50√3
⇒ h = 25√3
So, the required height is 25√3 m.

8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are Î± and Î², respectively. Prove that the height of the tower is  (h tanÎ±)/(tanÎ² - tanÎ±).
Solution
We have, a vertical flag staff of height h is surmounted on a vertical tower of height let (H),
so that,
FP = h and
FO = H.
The angle of elevation of the bottom and top of the flag staff on the plane is ∠PRO = Î± and
∠FRO = Î² respectively.
In ∆PRO, we have,
tanÎ± = PO/RO
= h/x
x = h/tan Î±  ...(i)
In FRO,
tanÎ² = FO/RO
⇒ tanÎ² = (FP + PO)/RO
⇒ tanÎ² = (H + h)/x
⇒ x = (H+h)/tanÎ²  ...(ii)
From (i) and (ii),
(H + h)/tanÎ² = h/tanÎ±
On solving,
H = (h tanÎ±)/(tanÎ± + tanÎ²)
Hence, proved

9. If tanÎ¸ + secÎ¸ = l, then prove that secÎ¸ = (l2 + 1)/2l .
Solution
Given:
tanÎ¸ + secÎ¸ =  l  ...eq.(i)
Multiplying and dividing by (sec Î¸ – tan Î¸) on numerator and denominator of L.H.S,
So,
secÎ¸ - tanÎ¸ = 1/l   ...eq. (ii)
Adding eq. (i) and eq.(ii), we get
(tanÎ¸ + secÎ¸) + (secÎ¸ - tanÎ¸) = l + 1/
Adding eq. (i) and eq. (ii), we get
(tanÎ¸ + secÎ¸) + (secÎ¸ - tanÎ¸) = l + 1/
So,
secÎ¸ =  (l2 + 1)/2l .
Hence, proved.

10. If sinÎ¸ + cosÎ¸ = p and secÎ¸ + cosecÎ¸ = q, then prove that q(p2 - 1) = 2p.
Solution
Given that,
sinÎ¸ + cosÎ¸ = p ...(i)
and
secÎ¸ + cosecÎ¸ = q

Since, sinÎ¸ + cosÎ¸ = p
On squaring both sides, we get
(sinÎ¸ + cosÎ¸)2 = p2
(sin2 Î¸ + cos2 Î¸) + 2sinÎ¸ cosÎ¸ = p2
[as, (a + b)2 = a2 + 2ab + b2 ]
1 + 2sinÎ¸ cosÎ¸ = p2
⇒ 1 + 2(p/q) = p2
⇒ q + 2p = p2q
⇒ 2p = p2q - q
⇒ q(p2 - 1) = 2p

11. If  a sinÎ¸ + bcosÎ¸ = c, the prove that a cosÎ¸ - b sinÎ¸ = √(a2 + b2 - c2)
Solution
Given that,
a sinÎ¸ + b cosÎ¸ = c
On squaring both sides,
(a + sinÎ¸ + cosÎ¸ b)2 = c2
⇒ a2 sin2 Î¸ + b2 cos2 Î¸ + 2ab sinÎ¸ cosÎ¸ = c2
⇒ a2 (1 - cos2 Î¸) + b2 ( 1 - sin2 Î¸) + 2ab sinÎ¸ . cosÎ¸ = c2
⇒ a2 - a2 cos2 Î¸ + b2 - b2 sin2 Î¸ + 2absinÎ¸ . cosÎ¸ = c2
⇒ a2 + b2 - c2 = a2 cos2 Î¸ + b2 sin2 Î¸ - 2absinÎ¸.cosÎ¸
⇒ a2 + b2 - c2  = a2 cos2 Î¸ + b2 sin2 Î¸ - 2absinÎ¸.cosÎ¸
⇒ a2 + b2 - c2  = (acosÎ¸ - bsinÎ¸)2
[as, a2 + b2 - 1ab = (a- b)2]
⇒ a cosÎ¸ - bsinÎ¸ = √(a2 + b2 - c2 )

12. Prove that (1 + secÎ¸ - tanÎ¸)/(1 + secÎ¸ + tanÎ¸) = (1 - sinÎ¸)/cosÎ¸.
Solution
Taking LHS,
On multiplying by (sec Î¸ - tanÎ¸) on numerator and denominator,

13. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.
Solution
Let us take the distance between the two towers = AB
= x m
and height of the other tower = PA
= h m
Given,
height of the tower = QB
= 30 m,
∠QAB = 60° and
∠PBA = 30°

Now, in Î”QAB,
tan 60 = QB/AB = 30/x
⇒ √3 = 30/x
⇒ x = 30/√3
⇒ x = 10√3
In PBA,
tan 30 = PA/AB
1/√3 = h/x
⇒ 1/√3 = h/10√3
⇒ h = 10

14. From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are Î± and Î² (Î² > Î±).  Find the distance between the two objects.
Solution
Let the distance between the two objects is x m and CD = y m
Given,
∠BAX = Î± = ∠ABD
and
∠CAX = Î² = ∠ACD  [Alternate angle]
and
height of tower, AD = h m
In ACD,
⇒ tanÎ² = h/y
⇒ y = h/tanÎ²  ...(i)
In ABD,
⇒ tanÎ± = AD/(BC + CD)
⇒ tanÎ± = h/(x + y)
⇒ y = (h/tanÎ±) - x  ...(ii)
From (i) and (ii),
(h/tanÎ±) - x = h/tanÎ²
⇒ x = (h/tanÎ±) - (h/tanÎ²)
⇒ x = h(cotÎ± - cotÎ²)
Which is the required distance between the two objects.

15. A ladder rests against a vertical wall at an inclination Î± to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle Î² to the horizontal. Show that p/q = (cos Î² - cos Î±)/(sinÎ± - sinÎ²)
Solution
Let,
OQ = x and
OA = y
Given,
BQ = q,
SA = P
and
AB = SQ = Length of ladder
Also,
∠BAO = Î± and ∠QSO = Î²

cos Î± = OA/AB
⇒ cos Î± = y/AB
⇒ y = AB cos Î± = OA  ...(i)
also,
sinÎ± = OB/AB
⇒ OB = AB sinÎ±
In QSO,
cos Î² = OS/SQ
⇒ OS = SQ  cos Î² = AB cos Î²
and,
sin Î² = OQ/SQ
⇒ OQ = SQ sin Î² = AB sinÎ²
Now,
SA = OS - AO
P = Ab(cos Î² - cos Î± )
Also,
BQ = BO - QO
⇒ q = AB(sin Î±  - sin Î²)
So,
p/q  = (cosÎ² - cos Î±)/(sin Î± - sinÎ²)

16. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.
Solution
Let the height the vertical tower be,
OT = H m
and
OP = AB = xm
Given that,
AP = 10 m
and
∠TPO = 60°,
∠TAB = 45°

Now, in Î”TPO,
tan 60° = OT/OP
⇒ tan 60° = H/x
⇒ x = h/√3
In TAB,
tan 45° = TB/AB
⇒ 1 = (H - 10)/x
⇒ x = H - 10
Putting value of x,
h/√3 = H - 10
On solving,
H = 5(√3 + 3)m

17. A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be Î± and  Î², respectively. Prove that the height of the other house is h(1 + tan Î± cot Î²) metres.
Solution
Let the height of the other house,
OQ = H m and
OB = MW = x m
Given,
height of the first house = WB = h = MO
and ∠QWM = Î±,
∠OWM = Î² = ∠WOB  [Alternate angle]
Now, In ∆WOB,
tan Î² = WB/OB
⇒ tan Î² = h/x
⇒ x = h/tanÎ²  ...(i)
In QWM,
tan Î± = QM/WM
⇒ tan Î± = (OQ - MO)/WM
⇒ tan Î± = (H - h)/x
⇒ x = (H - h)/tanÎ±  ...(ii)
From (i) and (ii),
(H - h)/tanÎ± = h/tan Î²
On solving,
H = h(1 + tanÎ± × 1/tanÎ²)

18. The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground.
Solution
Let the height of the balloon from above the ground is H m
and
OP = W2 R
= W2 Q
= x m
Given, height of lower window from the ground = W2 P
= 2m = OR
Height of upper window from the lower window = W1 W2
= 4m = QR
BQ = OB - (QR + RO)
= H - (4 + 2)
= H - 6
∠BW1Q = 30° and
∠BW2R = 60°

In BW2 R,
tan 60° = BR/W2 R
= (BQ + QR)/x
⇒ √3 = [(H - 6) + 4]/x
⇒ x = (H - 2)/√3  ...(i)
In BW1 R,
tan 30° = BR/W1 R
⇒ 1/√3 = (H - 6)/x
⇒ x = √3(H - 6)  ...(ii)
From (i) and (ii),
√3(H - 6) = (H - 2)/√3
On solving,
H = 8m
The required height is 8m.
Therefore, the required height of the balloon from the ground is 8m.