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In triangle ABC, cos A=sqrt(7/10)  and cos B=sqrt(3/10) Find cos C

 May 9, 2024
 #1
avatar+1071 
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We can find cos(C) using the Law of Cosines. Here's how:

 

Law of Cosines:

 

The Law of Cosines relates the sides and angles of a triangle. For triangle ABC, it states:

 

c^2 = a^2 + b^2 - 2ab * cos(C)

 

where:

 

c is the side opposite angle C (in this case, side AB)

 

a is the side adjacent to angle A (in this case, side BC)

 

b is the side adjacent to angle B (in this case, side AC)

 

Given Information:

 

We are given:

 

cos(A) = sqrt(7/10)

 

cos(B) = sqrt(3/10)

 

We need to find cos(C).

 

Sides are Unknown:

 

We don't have information about the side lengths (a, b, or c). However, we can express them using cos(A) and cos(B) for further calculations.

 

Expressing Sides:

 

Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can find sin^2(A) and sin^2(B):

 

sin^2(A) = 1 - cos^2(A) = 1 - (sqrt(7/10))^2 = 3/10

 

sin^2(B) = 1 - cos^2(B) = 1 - (sqrt(3/10))^2 = 7/10

 

Relating Sides to Cosines:

 

Since we are looking for cos(C), let's focus on side c (AB). Using cosine definitions in right triangles:

 

cos(A) = (adjacent side) / (hypotenuse) = a / c

 

cos(B) = (adjacent side) / (hypotenuse) = b / c

 

Therefore:

 

a = c * cos(A) = c * sqrt(7/10)

 

b = c * cos(B) = c * sqrt(3/10)

 

Applying the Law of Cosines:

 

Substitute the expressions for a and b in the Law of Cosines equation:

 

c^2 = (c * sqrt(7/10))^2 + (c * sqrt(3/10))^2 - 2 * c * sqrt(7/10) * c * sqrt(3/10) * cos(C)

 

Simplifying the equation:

 

c^2 = (7/10)c^2 + (3/10)c^2 - (6/10)c^2 * cos(C)

 

Combining like terms:

 

c^2 = (7 + 3 - 6) / 10 * c^2 - (6/10)c^2 * cos(C)

 

4c^2 / 10 = -(6/10)c^2 * cos(C)

 

Solving for cos(C):

 

Isolate cos(C):

 

cos(C) = -(4c^2 / 10) / -(6/10)c^2

 

cos(C) = 4/6 (Since both c^2 terms are negative, the negative signs cancel out)

 

cos(C) = 2/3

 

Therefore, cos(C) = 2/3.

 May 9, 2024
 #2
avatar+9664 
0

Recall the compound angle formula: 

\(\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B\\ \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B\)

We then have:

\(\quad \cos C\\ = \cos(180^\circ - A - B)\\ = -\cos(A + B)\\ =-\cos A \cos B + \sin A \sin B\\ =-\sqrt{\dfrac7{10}} \sqrt{\dfrac{3}{10}} + \sqrt{1 - \left(\sqrt{\dfrac7{10}}\right)^2} \cdot \sqrt{1 - \left(\sqrt{\dfrac3{10}}\right)^2}\\ =-\sqrt{\dfrac7{10}} \sqrt{\dfrac{3}{10}}+\sqrt{\dfrac7{10}} \sqrt{\dfrac{3}{10}}\\ = 0\)

 

This shows that \(\triangle ABC\) is a right triangle with \(\angle C = 90^\circ\).

 May 9, 2024

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