Triangle Side Length Calculation using Law of Cosines
This problem requires finding the length of side AC in a triangle ABC, given the lengths of sides AB and BC and the included angle B. The Law of Cosines is the appropriate formula to use.
Given Information:
- Length of side AB = 3m
- Length of side BC = 5m
- Angle B = $60^\circ$
Applying the Law of Cosines:
The Law of Cosines states that for any triangle with sides a, b, c and angle B opposite side b: $b^2 = a^2 + c^2 - 2ac \cos(B)$
In this triangle, let side AC be b, side BC be a (5m), and side AB be c (3m).
Calculation Steps:
- Substitute the given values into the Law of Cosines formula:
$AC^2 = BC^2 + AB^2 - 2(BC)(AB) \cos(B)$
$AC^2 = 5^2 + 3^2 - 2(5)(3) \cos(60^\circ)$
- Simplify the equation:
$AC^2 = 25 + 9 - 2(15) \cos(60^\circ)$
$AC^2 = 34 - 30 \cos(60^\circ)$
- Use the value of $\cos(60^\circ) = \frac{1}{2}$:
$AC^2 = 34 - 30 \left(\frac{1}{2}\right)$
$AC^2 = 34 - 15$
- Calculate the final value:
$AC^2 = 19$
$AC = \sqrt{19}\text{ m}$
Result:
The length of side AC is $\sqrt{19}$ meters.
Similar Questions
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Important Questions from Triangles
What is \(AB + BC\) equal to?
In the given figure, if \(RS = 3\sqrt{3} \text{ cm}\) and \(RPS\) is an equilateral triangle, then find the value of \(QR\)
- In a right angled triangle, with angle at \(A\) being \(90^\circ\), the side \(AB\) is of length 3cm and \(BC\) is 12 cm. What is the length of side \(AC\)?
For a pair of similar triangles shown in the figure, the angles made at $B$ and $R$ are same in both the triangles $BAC$ and $RAQ$. If the ratio of the areas of the two triangles is 2:1, and if the area of $BAC$ is $100\text{ cm}^2$ then what is the area of $RAQ$ in sq.cm.?
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