Question

question 9 of 26
given the triangle below, which of the following equations correctly
represents the relationship between a, b, and c?

triangle (right-angled at c, vertices a, b, c; side bc = a, ac = b, ab = c; \triangle not drawn to scale\)

a. ( c^2 = a^2 + b^2 - 2ac cos(90^circ) )

b. ( c^2 = b^2 + a^2 - 2ab cos(90^circ) )

c. ( b^2 = a^2 + c^2 - 2ab cos(90^circ) )

d partially visible

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines states that for any triangle with sides \( x \), \( y \), \( z \) and the angle \( \theta \) opposite side \( z \), \( z^{2}=x^{2}+y^{2}-2xy\cos(\theta) \).

Step2: Identify sides and angle in the right triangle

In the given right triangle \( \triangle ABC \) with right angle at \( C \), the side opposite angle \( B \) (wait, no, angle at \( C \) is \( 90^{\circ} \), and side \( c \) is opposite angle \( C \)? Wait, no, in the triangle, angle \( C \) is the right angle, so side \( c \) is the hypotenuse, and sides \( a \) and \( b \) are the legs. Wait, let's label correctly: in triangle \( ABC \), right - angled at \( C \), so angle \( C = 90^{\circ} \), side \( a \) is \( BC \), side \( b \) is \( AC \), side \( c \) is \( AB \). So for angle \( C \) (which is \( 90^{\circ} \)), the side opposite is \( c \), and the other two sides are \( a \) and \( b \). Applying the Law of Cosines: \( c^{2}=a^{2}+b^{2}-2ab\cos(C) \). Since \( C = 90^{\circ} \), this becomes \( c^{2}=a^{2}+b^{2}-2ab\cos(90^{\circ}) \) (which is option B). Let's check the other options:

• Option A: The formula has \( 2ac\cos(90^{\circ}) \), but according to Law of Cosines, the coefficient should be \( 2ab \) (product of the two sides adjacent to angle \( C \)), not \( 2ac \).
• Option C: The formula is for side \( b \), but side \( b \) is adjacent to angle \( C \), and the side opposite angle \( C \) is \( c \), so the Law of Cosines for side \( b \) would be \( b^{2}=a^{2}+c^{2}-2ac\cos(B) \), not with \( \cos(90^{\circ}) \) in that form.

Answer:

B. \( c^{2}=b^{2}+a^{2}-2ab\cos(90^{\circ}) \)