If we let h=CM, we have
x^2 + h^2 = 32
(7-x)^2 + h^2 = 25
eliminating h, we get
x^2 - 32 = (x-7)^2 - 25
x = 4
h = 4
So we have 3-4-5 and a 4-4-4√2 right triangles
Given: △ABC, CM⊥ AB
BC = 5, AB = 7
CA = 4sqrt(2)
Find: CM
3 answers
The angles and the sides of a triangle can be found using the sine rule or cosine rule or trigonometric relationships
The length of the side CM is 4
The reason the above value is correct is presented as follows:
The given parameters are presented in the attached drawing created with MS Visio
By cosine rule, we have'
a² = b² + c² - 2·b·c·cos(A)
In ΔABC, a represents the side BC, b represents the side AC, and c represents the side AB
a = BC = 5
b = AC = 4·√2
c = AB = 7
The cosine rule equation, therefore gives;
b² = a² + c² - 2·a·c·cos(B)
Therefore;
(4·√(2))² = 5² + 7² - 2×5×7×cos(B)
cos(B) = ((4·√(2))² - (5² + 7²))/(-2×5×7) = 0.6
B = arcos(0.6) ≈ 53.13°
By trigonometric ratios, we have;
sin(B) = \dfrac{CM}{BC}
∴ CM = BC × sin(B)
The length of CM = 5 × sin(53.13°) = 5 × 0.8 = 4
The length of the side CM = 4
The length of the side CM is 4
The reason the above value is correct is presented as follows:
The given parameters are presented in the attached drawing created with MS Visio
By cosine rule, we have'
a² = b² + c² - 2·b·c·cos(A)
In ΔABC, a represents the side BC, b represents the side AC, and c represents the side AB
a = BC = 5
b = AC = 4·√2
c = AB = 7
The cosine rule equation, therefore gives;
b² = a² + c² - 2·a·c·cos(B)
Therefore;
(4·√(2))² = 5² + 7² - 2×5×7×cos(B)
cos(B) = ((4·√(2))² - (5² + 7²))/(-2×5×7) = 0.6
B = arcos(0.6) ≈ 53.13°
By trigonometric ratios, we have;
sin(B) = \dfrac{CM}{BC}
∴ CM = BC × sin(B)
The length of CM = 5 × sin(53.13°) = 5 × 0.8 = 4
The length of the side CM = 4
4
1 answer