Pls help Given: △ABC, CM⊥ AB, BC = 5, AB = 7 CA = 4 sqrt(2) Find: CM
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Hi there,
θ = 180º + the angle of the right-angled triangle.
For finding the angle we know that the opposite side measures 6 units and the adjacent side measures 8 units. So, the hypotenuse is 10 units.
If we want to find the angle of the right-angled triangle we have to use the following equation.
sin(the angle of the right-angled triangle) =
⇒ the angle of the right-angled triangle = ≈ 36,87º
So,
θ = 180º + the angle of the right-angled triangle
θ ≈ 180º + 36,87º
θ ≈ 216,87º
sin(θ) = sin(216,87º)
sin(θ) =
sin(θ) =
If you want to do it using properties:
θ = 180º + |the angle of the right-angled triangle|
⇒ sin(θ) = sin(180º + |the angle of the right-angled triangle|)
Using properties:
⇒ sin(θ) = sin(180º)*cos( |the angle of the right-angled triangle|) + cos(180º)*sin(|the angle of the right-angled triangle|)
Sin (180) = 0
⇒ sin(θ) = cos(180º)*sin(|the angle of the right-angled triangle|)
sin(the angle of the right-angled triangle) = -
And cos(180º) = -1
⇒ sin(θ) = -1*
⇒ sin(θ) =
⇒ sin(θ) =