If the ratio of three sides of a triangle is a:b:c = 7:8:9 then show that cosA:cosB:cosC=14:11:6
Answers
Step-by-step explanation:
Given :-
The ratio of three sides of a triangle is a:b:c = 7:8:9
To find :-
cos A : cos B : cos C = 14:11:6
Solution :-
Given that
The ratio of three sides of a triangle
= 7:8:9
a:b:c = 7:8:9
Let they be 7X , 8X and 9X units
a = 7X
b = 8X
c = 9X
We know that
cos A = (b²+c²-a²)/(2bc)
=> cos A = [(8X)²+(9X)²-(7X)²]/(2×8X×9X)
=> cos A = (64X²+81X²-49X²)/(144X²)
=> cos A = (145X²-49X²)/(144X²)
=> cos A = (96X²)/(144X²)
=> cos A = 96/144
=> cos A = (2×48)/(3×48)
=> cos A = 2/3 ------------------(1)
We know that
cos B = (c²+a²-b²)/(2ac)
=> cos B = [(9X)²+(7X)²-(8X)²]/(2×7X×9X)
=> cos B = (81X²+49X²-64X²)/(126X²)
=> cos B = (130X²-64X²)/(126X²)
=> cos B = (66X²/(126X²)
=> cos B = 66/126
=> cos B = (11×6)/(21×6)
=> cos B = 11/21 ---------------(2)
We know that
cos C = (a²+b²-c²)/(2ab)
=> cos C = [(7X)²+(8X)²-(9X)²]/(2×7X×8X)
=> cos C = (49X²+64X²-81X²)/(112X²)
=> cos C = (113X²-81X²)/(112X²)
=> cos C = (32X²)/(112X²)
=> cos C = 32/112
=> cos C = (2×16)/(7×16)
=> cos C = 2/7 ----------------(3)
We have,
cos A = 2/3
cos B = 11/21
cos C = 2/7
Now,
The ratio of cos A , cos B and cos C
=cos A : cos B : cos C
= (2/3):(11/21):(2/7)
LCM of 3 , 21 and 7 = 21
= [(2/3)×(7/7)] : (11/21) : [ (2/7)×(3/3)]
= (14/21) : (11/21) : (6/21)
= 14 : 11 : 6
Answer :-
cos A : cos B : cos C = 14 : 11 : 6
Used formulae:-
→ cos A = (b²+c²-a²)/(2bc)
→ cos B = (c²+a²-b²)/(2ac)
→ cos C = (a²+b²-c²)/(2ab)
Concept:
The study of the correlation between a right-angled triangle's sides and angles is the focus of one of the most significant branches of mathematics in history: trigonometry. Hipparchus, a Greek mathematician, introduced this idea.
cos A = (b²+c²-a²)/(2bc)
cos B = (c²+a²-b²)/(2ac)
cos C = (a²+b²-c²)/(2ab)
Given :-
The ratio of three sides of a triangle is a:b:c = 7:8:9
Find :-
cos A : cos B : cos C = 14:11:6
Solution :-
The ratio of three sides of a triangle
= 7:8:9
a:b:c = 7:8:9
Let they be 7X , 8X and 9X units
a = 7X
b = 8X
c = 9
We know that
cos A = (b²+c²-a²)/(2bc)
=> cos A = [(8X)²+(9X)²-(7X)²]/(2×8X×9X)
=> cos A = (64X²+81X²-49X²)/(144X²)
=> cos A = (145X²-49X²)/(144X²)
=> cos A = (96X²)/(144X²)
=> cos A = 96/144
=> cos A = (2×48)/(3×48)
=> cos A = 2/3 ------------------(1)
We know that
cos B = (c²+a²-b²)/(2ac)
=> cos B = [(9X)²+(7X)²-(8X)²]/(2×7X×9X)
=> cos B = (81X²+49X²-64X²)/(126X²)
=> cos B = (130X²-64X²)/(126X²)
=> cos B = (66X²/(126X²)
=> cos B = 66/126
=> cos B = (11×6)/(21×6)
=> cos B = 11/21 ---------------(2)
We know that
cos C = (a²+b²-c²)/(2ab)
=> cos C = [(7X)²+(8X)²-(9X)²]/(2×7X×8X)
=> cos C = (49X²+64X²-81X²)/(112X²)
=> cos C = (113X²-81X²)/(112X²)
=> cos C = (32X²)/(112X²)
=> cos C = 32/112
=> cos C = (2×16)/(7×16)
=> cos C = 2/7 ----------------(3)
We have,
cos A = 2/3
cos B = 11/21
cos C = 2/7
Now,
The ratio of cos A , cos B and cos C
=cos A : cos B : cos C
= (2/3):(11/21):(2/7)
LCM of 3 , 21 and 7 = 21
= [(2/3)×(7/7)] : (11/21) : [ (2/7)×(3/3)]
= (14/21) : (11/21) : (6/21)
= 14 : 11 : 6
Hence proved
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