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1) cos A = b² + c² - a² /2bc = 16 + 9 - 4/ 2 . 3 . 4 = 7/8
2) a / sinA = b/ sinB ⇒ 2/ (2/3) = 3/ sinB ⇒ sin B = 1 ⇒ B = 90 degree or π/2
3) given b : c = √3 : √2 ...........(i)
b / sinB = c / sinC = k ⇒ b = ksinB and c = ksinC
⇒ ksinB : ksinC = √3 : √2 ⇒ sinB : sinC = √3 : √2 ..........from(i)
⇒ sinB : sinC = √3/2 : √2/2 = √3/2 : 1/√2
⇒ B = 60 degree and C = 30 degree
and now A , B and C in A.P. ===== A = 90 degree
4) a = 13 , b = 14 , c = 15 s = a+b+c/2 = 13+14+15/2 = 21
area = √s (s-a)(s-b)(s-c) = √ 21 x (21-13) x (21-14) x (21 -15)
⇒ √21 x 7 x 6 x 8 = 84 sq units
5) B = 120 C= 30 ⇒ A= 30 a = 2
a/sinA = b/sinB = c/sinC
2/sin30 = b/sin120 = c/sin30 ⇒ b=2√3 and c = 2
area = 1/2 x ac x sinB = 1/2 (2)(2) sin120 = √3 sq units
6) a / bc + cosA/a = a/bc + b² + c² - a² /2abc = 2a² + b² + c² - a² / 2bc
= a² + b² + c² / 2abc
similarly b/ca + cosB/b = a² + b² + c² /2abc
and c/ab + cosC/c = a² + b² + c² /2abc
hence proved
7) given c² = a² + b² cosC = a² + b² -c² /2ab = 0
C = 90 degree
therefore area = Δ = 1/2 ab sin C = 1/2 ab ......(i)
now area =Δ = √s(s-a)(s-b)(s-c)
Δ² = s(s-a)(s-b)(s-c) ..........(ii)
A/Q 4s(s-a)(s-b)(s-c) = 4Δ² from (ii)
= 4 (1/2 ab)² = a²b² from (i)
hope u'll get it ....................doubts , ask?
2) a / sinA = b/ sinB ⇒ 2/ (2/3) = 3/ sinB ⇒ sin B = 1 ⇒ B = 90 degree or π/2
3) given b : c = √3 : √2 ...........(i)
b / sinB = c / sinC = k ⇒ b = ksinB and c = ksinC
⇒ ksinB : ksinC = √3 : √2 ⇒ sinB : sinC = √3 : √2 ..........from(i)
⇒ sinB : sinC = √3/2 : √2/2 = √3/2 : 1/√2
⇒ B = 60 degree and C = 30 degree
and now A , B and C in A.P. ===== A = 90 degree
4) a = 13 , b = 14 , c = 15 s = a+b+c/2 = 13+14+15/2 = 21
area = √s (s-a)(s-b)(s-c) = √ 21 x (21-13) x (21-14) x (21 -15)
⇒ √21 x 7 x 6 x 8 = 84 sq units
5) B = 120 C= 30 ⇒ A= 30 a = 2
a/sinA = b/sinB = c/sinC
2/sin30 = b/sin120 = c/sin30 ⇒ b=2√3 and c = 2
area = 1/2 x ac x sinB = 1/2 (2)(2) sin120 = √3 sq units
6) a / bc + cosA/a = a/bc + b² + c² - a² /2abc = 2a² + b² + c² - a² / 2bc
= a² + b² + c² / 2abc
similarly b/ca + cosB/b = a² + b² + c² /2abc
and c/ab + cosC/c = a² + b² + c² /2abc
hence proved
7) given c² = a² + b² cosC = a² + b² -c² /2ab = 0
C = 90 degree
therefore area = Δ = 1/2 ab sin C = 1/2 ab ......(i)
now area =Δ = √s(s-a)(s-b)(s-c)
Δ² = s(s-a)(s-b)(s-c) ..........(ii)
A/Q 4s(s-a)(s-b)(s-c) = 4Δ² from (ii)
= 4 (1/2 ab)² = a²b² from (i)
hope u'll get it ....................doubts , ask?
parisakura98pari:
do u have any doubt in this ques?
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