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로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.

For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.

Problem 18 In triangleABC, suppose that AB= 5, AC= 6, and cos(∠BAC) =− 3 5 . Compute the area of triangleABC. [3 points] 더보기

The area of triangle 𝐴 𝐵 𝐶 ABC can be found using the formula:

Area

1 2 ⋅ 𝐴 𝐵 ⋅ 𝐴 𝐶 ⋅ sin ⁡ ( ∠ 𝐵 𝐴 𝐶 ) Area= 2 1 ​

⋅AB⋅AC⋅sin(∠BAC)

Given 𝐴 𝐵

5 AB=5, 𝐴 𝐶

6 AC=6, and cos ⁡ ( ∠ 𝐵 𝐴 𝐶 )

− 3 5 cos(∠BAC)=− 5 3 ​

. Since ∠ 𝐵 𝐴 𝐶 ∠BAC is an interior angle of a triangle, 0 ∘ < ∠ 𝐵 𝐴 𝐶 < 180 ∘ 0 ∘ <∠BAC<180 ∘ , so sin ⁡ ( ∠ 𝐵 𝐴 𝐶 )

0 sin(∠BAC)>0.

sin ⁡ 2 ( ∠ 𝐵 𝐴 𝐶 )

1 − cos ⁡ 2 ( ∠ 𝐵 𝐴 𝐶 )

1 − ( − 3 5 ) 2

1 − 9 25

16 25 sin 2 (∠BAC)=1−cos 2 (∠BAC)=1−(− 5 3 ​

) 2 =1− 25 9 ​

= 25 16 ​

sin ⁡ ( ∠ 𝐵 𝐴 𝐶 )

4 5 sin(∠BAC)= 5 4 ​

Now compute the area:

Area

1 2 ⋅ 5 ⋅ 6 ⋅ 4 5

1 2 ⋅ 30 ⋅ 4 5

15 ⋅ 4 5

12 Area= 2 1 ​

⋅5⋅6⋅ 5 4 ​

= 2 1 ​

⋅30⋅ 5 4 ​

=15⋅ 5 4 ​

=12

\boxed{12}

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