ada segitiga lancip ABC diketahui panjang sisi AC=4 cm,

Nilai cos C adalah \sqrt{7}/4.

Pembahasan

Untuk menyelesaikan soal ini, kita akan menggunakan aturan cosinus dalam segitiga.

Diketahui segitiga lancip ABC dengan:
Panjang sisi AC = b = 4 cm
Panjang sisi AB = c = 5 cm
Nilai cos B = 4/5

Kita perlu mencari nilai cos C.

Aturan cosinus menyatakan:
$a^2 = b^2 + c^2 - 2bc \cos A$
$b^2 = a^2 + c^2 - 2ac \cos B$
$c^2 = a^2 + b^2 - 2ab \cos C$

Kita memiliki b, c, dan cos B. Kita bisa mencari panjang sisi a menggunakan aturan cosinus untuk $b^2$:
$b^2 = a^2 + c^2 - 2ac \cos B$
$4^2 = a^2 + 5^2 - 2(a)(5)(4/5)$
$16 = a^2 + 25 - 8a$
$a^2 - 8a + 25 - 16 = 0$
$a^2 - 8a + 9 = 0$

Kita dapat menyelesaikan persamaan kuadrat ini untuk mencari nilai a menggunakan rumus kuadrat: $a = [-b ± \sqrt{b^2-4ac}] / 2a$
$a = [8 ± \sqrt{(-8)^2 - 4(1)(9)}] / 2(1)$
$a = [8 ± \sqrt{64 - 36}] / 2$
$a = [8 ± \sqrt{28}] / 2$
$a = [8 ± 2\sqrt{7}] / 2$
$a = 4 ± \sqrt{7}$

Jadi, ada dua kemungkinan nilai untuk sisi a: $a_1 = 4 + \sqrt{7}$ atau $a_2 = 4 - \sqrt{7}$.

Sekarang, kita akan mencari cos C menggunakan aturan cosinus untuk $c^2$:
$c^2 = a^2 + b^2 - 2ab \cos C$
$5^2 = a^2 + 4^2 - 2(a)(4) \cos C$
$25 = a^2 + 16 - 8a \cos C$
$8a \cos C = a^2 + 16 - 25$
$8a \cos C = a^2 - 9$
$\cos C = (a^2 - 9) / 8a$

Kasus 1: $a = 4 + \sqrt{7}$
$a^2 = (4 + \sqrt{7})^2 = 16 + 8\sqrt{7} + 7 = 23 + 8\sqrt{7}$
$\cos C = ((23 + 8\sqrt{7}) - 9) / (8(4 + \sqrt{7}))$
$\cos C = (14 + 8\sqrt{7}) / (32 + 8\sqrt{7})$
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The problem states that triangle ABC is an acute triangle. We need to ensure that all angles are less than 90 degrees.

Let's reconsider the problem. We are given AC=b=4, AB=c=5, and cos B = 4/5. We need to find cos C.
We can also use the Law of Sines: $a/\sin A = b/\sin B = c/\sin C$.
First, we need to find sin B. Since cos B = 4/5 and the triangle is acute, sin B must be positive.
$\sin^2 B + \cos^2 B = 1$
$\sin^2 B + (4/5)^2 = 1$
$\sin^2 B + 16/25 = 1$
$\sin^2 B = 1 - 16/25 = 9/25$
$\sin B = 3/5$ (since B is an angle in an acute triangle, sin B > 0).

Now, we can use the Law of Sines: $b/\sin B = c/\sin C$.
$4 / (3/5) = 5 / \sin C$
$4 * (5/3) = 5 / \sin C$
$20/3 = 5 / \sin C$
$\sin C = 5 * (3/20)$
$\sin C = 15/20 = 3/4$

Now that we have sin C, we can find cos C. Since the triangle is acute, angle C must be less than 90 degrees, so cos C must be positive.
$\sin^2 C + \cos^2 C = 1$
$(3/4)^2 + \cos^2 C = 1$
$9/16 + \cos^2 C = 1$
$\cos^2 C = 1 - 9/16 = 7/16$
$\cos C = \sqrt{7/16}$
$\cos C = \sqrt{7} / 4$

Let's verify if the triangle is indeed acute with these values.
We have cos B = 4/5 = 0.8 (acute)
We have sin C = 3/4, so cos C = sqrt(7)/4 ≈ 0.661 (acute)
We need to find cos A.
First, let's find side a.
Using the law of cosines: $b^2 = a^2 + c^2 - 2ac \cos B$
$4^2 = a^2 + 5^2 - 2(a)(5)(4/5)$
$16 = a^2 + 25 - 8a$
$a^2 - 8a + 9 = 0$
$a = (8 ± \sqrt{64 - 36}) / 2 = (8 ± \sqrt{28}) / 2 = 4 ± \sqrt{7}$.

Let's assume $a = 4 + \sqrt{7}$.
Now find cos A using $a^2 = b^2 + c^2 - 2bc \cos A$
$(4 + \sqrt{7})^2 = 4^2 + 5^2 - 2(4)(5) \cos A$
$16 + 8\sqrt{7} + 7 = 16 + 25 - 40 \cos A$
$23 + 8\sqrt{7} = 41 - 40 \cos A$
$40 \cos A = 41 - 23 - 8\sqrt{7}$
$40 \cos A = 18 - 8\sqrt{7}$
$\cos A = (18 - 8\sqrt{7}) / 40 = (9 - 4\sqrt{7}) / 20$
$4\sqrt{7} ≈ 4 * 2.646 = 10.584$
$\cos A ≈ (9 - 10.584) / 20 = -1.584 / 20 = -0.0792$. This means angle A is obtuse, which contradicts the condition that the triangle is acute.

Let's assume $a = 4 - \sqrt{7}$.
$a^2 = (4 - \sqrt{7})^2 = 16 - 8\sqrt{7} + 7 = 23 - 8\sqrt{7}$
$a^2 = b^2 + c^2 - 2bc \cos A$
$23 - 8\sqrt{7} = 4^2 + 5^2 - 2(4)(5) \cos A$
$23 - 8\sqrt{7} = 16 + 25 - 40 \cos A$
$23 - 8\sqrt{7} = 41 - 40 \cos A$
$40 \cos A = 41 - 23 + 8\sqrt{7}$
$40 \cos A = 18 + 8\sqrt{7}$
$\cos A = (18 + 8\sqrt{7}) / 40 = (9 + 4\sqrt{7}) / 20$
$4\sqrt{7} ≈ 10.584$
$\cos A ≈ (9 + 10.584) / 20 = 19.584 / 20 = 0.9792$ (acute)

So, the side length must be $a = 4 - \sqrt{7}$.
With $a = 4 - \sqrt{7}$, $b=4$, $c=5$, we have:
cos B = 4/5 (given)
cos C = sqrt(7)/4
cos A = (9 + 4\sqrt{7}) / 20

All cosines are positive, so all angles are acute.

Therefore, the value of cos C is $\sqrt{7} / 4$.