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$In /_\ABC,AB=5cm,BC=6cm and /_B=60^(@). Find the length of AC.$

In $\triangle A B C, A B=5 \mathrm{~cm}, B C=6 \mathrm{~cm}$ and $\angle B=60^{\circ}$ . $\newline$ Find the length of $A C$ .

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Find derivatives using the quotient rule II

Full solution

Q. In

\triangle A B C, A B=5 \mathrm{~cm}, B C=6 \mathrm{~cm}

and

\angle B=60^{\circ}

.

\newline

Find the length of

A C

.

Identify Elements and Rule: Identify the known elements of the triangle and the rule to apply. $\newline$ In triangle $ABC$ , we know side $AB$ ( $5\,\text{cm}$ ), side $BC$ ( $6\,\text{cm}$ ), and angle $B$ ( $60$ degrees). To find the length of side $AC$ , we can use the Law of Cosines, which relates the lengths of the sides of a triangle to the cosine of one of its angles.
Write Law of Cosines: Write down the Law of Cosines formula. $\newline$ The Law of Cosines states that for any triangle $ABC$ with sides $a$ , $b$ , and $c$ opposite angles $A$ , $B$ , and $C$ respectively, the following is true: $c^2 = a^2 + b^2 - 2ab\cos(C)$ . In our case, we want to find $c$ ( $AC$ ), where $a$ $0$ ( $a$ $1$ ), $a$ $2$ ( $a$ $3$ ), and angle $C$ is $a$ $5$ degrees.
Substitute Known Values: Substitute the known values into the Law of Cosines formula. $\newline$ Using the values we have, we get $AC^2 = 5^2 + 6^2 - 2\times5\times6\cos(60 \text{ degrees})$ . We know that $\cos(60 \text{ degrees})$ is $0.5$ .
Calculate AC Length: Calculate the length of AC. $\newline$ $AC^2 = 25 + 36 - 2 \times 5 \times 6 \times 0.5$ $\newline$ $AC^2 = 25 + 36 - 30$ $\newline$ $AC^2 = 61 - 30$ $\newline$ $AC^2 = 31$ $\newline$ $AC = \sqrt{31}$
Check for Errors: Check for any mathematical errors. $\newline$ We have correctly applied the Law of Cosines and performed the arithmetic operations. There are no mathematical errors in the calculations.

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