Find the length of the indicated side, to the nearest tenth of a metre.

Steps to Solve

1. Identify the sides and the angle In triangle ( RKT ), we know:

• Side ( RK = 3.6 , m )
• Side ( RT = 4.3 , m )
• Angle ( R = 36^\circ )

We want to find the length of side ( KT ), denoted as ( x ).

2. Use the Law of Cosines The Law of Cosines states: $$ c^2 = a^2 + b^2 - 2ab \cos(C) $$ Where ( C ) is the angle opposite side ( c ). In this case: $$ KT^2 = (RK)^2 + (RT)^2 - 2(RK)(RT) \cos(36^\circ) $$

Substituting the known values: $$ x^2 = (3.6)^2 + (4.3)^2 - 2(3.6)(4.3) \cos(36^\circ) $$

3. Calculate the squares and cosine First, calculate the squares:

• ( (3.6)^2 = 12.96 )
• ( (4.3)^2 = 18.49 )

Now, find ( \cos(36^\circ) ): $$ \cos(36^\circ) \approx 0.8090 $$

Substituting these values into the equation: $$ x^2 = 12.96 + 18.49 - 2(3.6)(4.3)(0.8090) $$

4. Perform the calculations Calculate ( 2(3.6)(4.3)(0.8090) ): $$ 2(3.6)(4.3)(0.8090) \approx 2(3.6)(4.3)(0.8090) \approx 25.579 $$

Now, substituting back in: $$ x^2 = 12.96 + 18.49 - 25.579 $$

5. Final calculation for ( x^2 ) Now perform the addition and subtraction: $$ x^2 \approx 31.45 - 25.579 $$ $$ x^2 \approx 5.871 $$

6. Find ( x ) Now, take the square root to find ( x ): $$ x \approx \sqrt{5.871} \approx 2.42 $$

Finally, round to the nearest tenth: $$ x \approx 2.4 , m $$