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Solution Of Triangles

Class 11th Mathematics RS Aggarwal Solution
Exercise 18a
  1. a(b cos C – c cos B) = (b2 – c2) In any ΔABC, prove that
  2. ac cos B – bc cos A = (a2 – b2) In any ΔABC, prove that
  3. {c-bcosa}/{b-cosa} = frac {cosb}/{cosc} In any ΔABC, prove that
  4. 2(bc cos A + ca cos B + ab cos C) = (a2 + b2 + c2) In any ΔABC, prove that…
  5. 4 ( bccos^{2} {a}/{2} + cacos^{2} frac {b}/{2} + abcos^{2} frac {c}/{2} ) =…
  6. a sin A – b sin B = c sin (A – B) In any ΔABC, prove that
  7. a2 sin ( B – C ) = (b2 – c2) sin A In any ΔABC, prove that
  8. { sin (a-b) }/{ sin (a+b) } = frac { ( a^{2} - b^{2} ) }/{ c^{2} } In any ΔABC,…
  9. { (b-c) }/{a} cos frac {a}/{2} = sin frac { (b-c) }/{2} In any ΔABC, prove…
  10. { (a+b) }/{c} sin frac {c}/{2} = cos frac { (a-b) }/{2} In any ΔABC, prove…
  11. { (b+c) }/{a} c. cos frac { (b+c) }/{2} = cos frac { (b-c) }/{2} In any ΔABC,…
  12. a2(cos2B – cos2C) + b2(cos2C – cos2A) + c2(cos2A – cos2B) = 0 In any ΔABC,…
  13. {cosa}/{a} + frac {cosb}/{b} + frac {cosc}/{c} = frac { ( a^{2} + b^{2} +…
  14. { (cos^{2}b-cos^{2}c) }/{b+c} + frac { (cos^{2}c-cos^{2}a) }/{c+a} + frac {…
  15. {cos2a}/{ a^{2} } - frac {cos2b}/{ b^{2} } = ( frac {1}/{ a^{2} } - frac {1}/{…
  16. (c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C In any…
  17. If in a ∆ABC, ∠C = 900, then prove that sin (a-b) = { ( a^{2} - b^{2} ) }/{…
  18. In a ∆ABC, if {cosa}/{a} = frac {cosb}/{b} , show that the triangle is…
  19. In a ∆ABC, if sin^{2}a+sin^{2}b = sin^{2}c , show that the triangle is…
  20. Solve the triangle in which a = 2 cm, b = 1 cm and c = root {3} cm.…
  21. In a ΔABC, if a = 3 cm, b = 5 cm and c = 7 cm, find cos A, cos B, cos C.…
  22. If the angles of a triangle are in the ratio 1 : 2 : 3, prove that its…
Exercise 18b
  1. Two boats leave a port at the same time. One travels 60 km in the direction N…
  2. A town B is 12 km south and 18 km west of a town A. Show that the bearing of B…
  3. At the foot of a mountain, the angle of elevation of its summit is 450. After…
Exercise 18a
Question 1.

In any ΔABC, prove that

a(b cos C – c cos B) = (b2 – c2)
Answer:

Left hand side,


a(b cos C – c cos B)


= ab cos C – ac cos B


[As, & ]





b2 – c2


Right hand side. [Proved]
Question 2.

In any ΔABC, prove that

ac cos B – bc cos A = (a2 – b2)
Answer:

Left hand side,


ac cos B – bc cos A


[As, & ]





a2 – b2


Right hand side. [Proved]
Question 3.

In any ΔABC, prove that
Answer:

Need to prove:


Left hand side






[Multiplying the numerator and denominator by ]



Right hand side. [Proved]
Question 4.

In any ΔABC, prove that

2(bc cos A + ca cos B + ab cos C) = (a2 + b2 + c2)
Answer:

Need to prove: 2(bc cos A + ca cos B + ab cos C) = (a2 + b2 + c2)


Left hand side


2(bc cos A + ca cos B + ab cos C)



b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2


a2 + b2 + c2


Right hand side. [Proved]
Question 5.

In any ΔABC, prove that
Answer:

Need to prove:


Right hand side



, where s is half of perimeter of triangle.


4(s(s – a) + s(s – b) + s(s – c))


4(3s2 – s(a + b + c))


We know, 2s = a +b +c


So,




3(a + b + c)2 – 2(a + b + c)2


(a + b + c)2


Right hand side. [Proved]
Question 6.

In any ΔABC, prove that

a sin A – b sin B = c sin (A – B)
Answer:

Need to prove: a sin A – b sin B = c sin (A – B)


Left hand side,


a sin A – b sin B


(b cosC + c cosB) sinA – (c cosA + a cosC) sinB


b cosC sinA + c cosB sinA – c cosA sinB – a cosC sinB


c(sinA cosB – cosA sinB) + cosC(b sinA – a sinB)


We know that, where R is the circumradius.


Therefore,


c(sinA cosB – cosA sinB) + cosC(2R sinB sinA – 2R sinA sinB)


c(si nA cosB – cosA sinB)


c sin (A – B)


Right hand side. [Proved]
Question 7.

In any ΔABC, prove that

a2 sin ( B – C ) = (b2 – c2) sin A
Answer:

Need to prove: a2 sin ( B – C ) = (b2 – c2) sin A


We know that, where R is the circumradius.


Therefore,


a = 2R sinA ---- (a)


Similarly, b = 2R sinB and c = 2R sinC


From Right hand side,


(b2 – c2) sin A


{(2R sinB)2 – (2R sinC)2} sinA


4R2( sin2B – sin2C )sinA


We know, sin2B – sin2C = sin(B + C)sin(B – C)


So,


4R2(sin(B + C)sin(B – C))sinA


4R2(sin( – A)sin(B – C))sinA [ As, A + B + C = ]


4R2(sinAsin(B – C))sinA [ As, sin( – ) = sin ]


4R2sin2A sin(B – C)


a2sin(B – C) [From (a)]


Left hand side. [Proved]
Question 8.

In any ΔABC, prove that
Answer:

Need to prove:


We know that, where R is the circumradius.


Therefore,


a = 2R sinA ---- (a)


Similarly, b = 2R sinB and c = 2R sinC


From Right hand side,









Left hand side. [Proved]
Question 9.

In any ΔABC, prove that
Answer:

Need to prove:


We know that, where R is the circumradius.


Therefore,


a = 2R sinA ---- (a)


Similarly, b = 2R sinB and c = 2R sinC


From Left hand side,








Right hand side. [Proved]
Question 10.

In any ΔABC, prove that
Answer:

Need to prove:


We know that, where R is the circumradius.


Therefore,


a = 2R sinA ---- (a)


Similarly, b = 2R sinB and c = 2R sinC


Now,


Therefore,


⇒


⇒


⇒


⇒ [Proved]
Question 11.

In any ΔABC, prove that
Answer:

Need to prove:


We know that, where R is the circumradius.


Therefore,


a = 2R sinA ---- (a)


Similarly, b = 2R sinB and c = 2R sinC


Now,


⇒


⇒


⇒


⇒


⇒


⇒


⇒


⇒ [Proved]
Question 12.

In any ΔABC, prove that

a2(cos2B – cos2C) + b2(cos2C – cos2A) + c2(cos2A – cos2B) = 0
Answer:

Need to prove: a2(cos2B – cos2C) + b2(cos2C – cos2A) + c2(cos2A – cos2B) = 0


From left hand side,


a2(cos2B – cos2C) + b2(cos2C – cos2A) + c2(cos2A – cos2B)


a2((1 - sin2B) – (1 - sin2C)) + b2((1 - sin2C) – (1 - sin2A)) + c2((1 - sin2A) – (1 - sin2B))


a2( - sin2B + sin2C) + b2( - sin2C + sin2A) + c2( - sin2A + sin2B)


We know that, where R is the circumradius.


Therefore,


a = 2R sinA ---- (a)


Similarly, b = 2R sinB and c = 2R sinC


So,


4R2[ sin2A( - sin2B + sin2C) + sin2B( - sin2C + sin2A) + sin2C( - sin2A + sin2B)


4R2[ - sin2Asin2B + sin2Asin2C – sin2Bsin2C + sin2Asin2B – sin2Asin2C + sin2Bsin2C ]


4R2 [0]


0 [Proved]
Question 13.

In any ΔABC, prove that
Answer:

Need to prove:


Left hand side






Right hand side. [Proved]
Question 14.

In any ΔABC, prove that
Answer:

Need to prove:


We know that, where R is the circumradius.


Therefore,


a = 2R sinA ---- (a)


Similarly, b = 2R sinB and c = 2R sinC


From left hand side,





Now,





0 [Proved]
Question 15.

In any ΔABC, prove that
Answer:

Need to prove:


Left hand side,





We know that, where R is the circumradius.


Therefore,



Hence,


[Proved]
Question 16.

In any ΔABC, prove that

(c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C
Answer:

Need to prove: (c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C


We know,


------ (a)


Similarly, and


Therefore,


[from (a)]


Similarly,


and


Hence we can conclude comparing above equations,


(c2 – a2 + b2) tanA = (a2 – b2 + c2) tanB = (b2 – c2 + a2) tanC


[Proved]
Question 17.

If in a ∆ABC, ∠C = 900, then prove that .
Answer:

Given: ∠C = 900


Need to prove:


Here, ∠C = 900 ; sinC = 1


So, it is a Right-angled triangle.


And also, a2 + b2 = c2


Now,



We know that, where R is the circumradius.


Therefore,


[As, sinC = 1]





Therefore,


⇒


⇒ [Proved]
Question 18.

In a ∆ABC, if , show that the triangle is isosceles.
Answer:

Given:


Need to prove: ∆ABC is isosceles.



⇒


⇒ [Squaring both sides]


⇒


We know,


Therefore,


So,


⇒


⇒


That means a and b sides are of same length. Therefore, the triangle is isosceles. [Proved]
Question 19.

In a ∆ABC, if , show that the triangle is right-angled.
Answer:

Given:


Need to prove: The triangle is right-angled


sin2A + sin2B = sin2C


We know,


So,





This is one of the properties of right angled triangle. And it is satisfied here. Hence, the triangle is right angled. [Proved]
Question 20.

Solve the triangle in which a = 2 cm, b = 1 cm and c = cm.
Answer:

Given: a = 2 cm, b = 1 cm and c = cm


Perimeter = a + b + c = 3 + cm


Area =


=


=


=


= cm2 [Proved]
Question 21.

In a ΔABC, if a = 3 cm, b = 5 cm and c = 7 cm, find cos A, cos B, cos C.
Answer:

Given: a = 3 cm, b = 5 cm and c = 7 cm


Need to find: cos A, cos B, cos C
Question 22.

If the angles of a triangle are in the ratio 1 : 2 : 3, prove that its corresponding sides are in the ratio .
Answer:

Given: Angles of a triangle are in the ratio 1 : 2 : 3


Need to prove: Its corresponding sides are in the ratio


Let the angles are x , 2x , 3x


Therefore, x + 2x + 3x = 1800


6x = 1800


x = 300


So, the angles are 300 , 600 , 900


So, the ratio of the corresponding sides are:


= sin300 : sin600 : sin900


=


= [Proved]
Exercise 18b
Question 1.

Two boats leave a port at the same time. One travels 60 km in the direction N 500 E while the other travels 50 km in the direction S 700 E. What is the distance between the boats?
Answer:


Both the boats starts from A and boat 1 reaches at B and boat 2 reaches at C.


Here, AB = 60Km and AC = 50Km


So, the net distance between ta boats is:


|


=


=


= 55.67Km
Question 2.

A town B is 12 km south and 18 km west of a town A. Show that the bearing of B from A is S 560 20’ W. Also, find the distance of B from A.
Answer:


Distance from A to B is =


Let, bearing from A to B is .


So,




Question 3.

At the foot of a mountain, the angle of elevation of its summit is 450. After ascending 1 km towards the mountain up an incline of 300, the elevation changes to 600 (as shown in the given figure). Find the height of the mountain. [Given : ]
Answer:

After ascending 1 km towards the mountain up an incline of 300, the elevation changes to 600


So, according to the figure given, AB = AF x sin300 = (1 x 0.5) = 0.5 Km.


At point A the elevation changes to 600.


In this figure, ABF ACS


Comparing these triangles, we get AB = AC = 0.5Km


Now, CS = AC x tan600 = (0.5 x 1.73) = 0.865Km


Therefore, the total height of the mountain is = CS + DC


= CS + BA


= (0.865 + 0.5) Km


= 1.365 Km