Question

use the law of cosines. find length x.
x =
(round the final answer to one decimal place as needed. round all intermediate values to two decimal places as needed.)

Explanation:

Step1: Recall Law of Cosines

The Law of Cosines for a triangle with sides \(a\), \(b\), \(c\) and the angle \(C\) opposite side \(c\) is \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). In triangle \(ABC\), we have \(AC = 19\), \(BC=13\), \(\angle C = 70^{\circ}\), and we need to find \(AB=x\). So, using the Law of Cosines, \(x^{2}=19^{2}+13^{2}-2\times19\times13\times\cos(70^{\circ})\).

Step2: Calculate each term

First, calculate \(19^{2}=361\), \(13^{2} = 169\). Then, calculate \(2\times19\times13=494\). Now, find \(\cos(70^{\circ})\approx0.3420\) (rounded to four decimal places for accuracy, but we can use calculator value). Then, \(2\times19\times13\times\cos(70^{\circ})=494\times0.3420\approx168.95\) (rounded to two decimal places).

Step3: Compute \(x^{2}\)

Now, \(x^{2}=361 + 169-168.95=361+169 = 530\); \(530 - 168.95 = 361.05\).

Step4: Find \(x\)

Take the square root of \(361.05\): \(x=\sqrt{361.05}\approx19.0\) (Wait, let's recalculate more accurately. Wait, maybe my intermediate calculation was wrong. Let's do it again. \(\cos(70^{\circ})\approx0.3420\), so \(2\times19\times13\times\cos(70^{\circ})=494\times0.3420 = 494\times0.342\). Let's calculate \(494\times0.3 = 148.2\), \(494\times0.04 = 19.76\), \(494\times0.002=0.988\). So total is \(148.2 + 19.76=167.96+0.988 = 168.948\approx168.95\). Then \(19^{2}+13^{2}=361 + 169 = 530\). \(530-168.95 = 361.05\). \(\sqrt{361.05}\approx19.0\)? Wait, no, wait \(19^{2}=361\), so \(\sqrt{361.05}\) is very close to 19.0. Wait, maybe I made a mistake in the angle. Wait, no, let's use more precise \(\cos(70^{\circ})\). Let's use calculator: \(\cos(70^{\circ})\approx0.3420201433\). Then \(2\times19\times13\times0.3420201433=494\times0.3420201433\). \(494\times0.3420201433 = 494\times0.3 + 494\times0.04 + 494\times0.0020201433=148.2+19.76 + 0.998=148.2+19.76=167.96+0.998 = 168.958\). Then \(19^{2}+13^{2}=361 + 169 = 530\). \(530-168.958 = 361.042\). Then \(\sqrt{361.042}\approx19.0\)? Wait, that can't be. Wait, maybe the triangle sides: \(AC = 19\), \(BC = 13\), angle at C is 70 degrees. Wait, maybe I mixed up the sides. Wait, Law of Cosines: \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\), where \(a = 19\), \(b = 13\), \(C = 70^{\circ}\), so \(c=x\). So \(x^{2}=19^{2}+13^{2}-2\times19\times13\times\cos(70^{\circ})\). Let's compute \(19^{2}=361\), \(13^{2}=169\), \(2\times19\times13 = 494\), \(\cos(70^{\circ})\approx0.3420\), so \(494\times0.3420 = 168.948\). Then \(361+169 = 530\), \(530 - 168.948 = 361.052\). Then \(\sqrt{361.052}\approx19.0\). Wait, but let's check with another approach. Wait, maybe my calculation of \(\cos(70^{\circ})\) is wrong? No, \(\cos(70^{\circ})\) is approximately 0.3420. Wait, maybe the problem is that I made a mistake in the formula. Wait, no, the formula is correct. Wait, let's use a calculator for more precision. Let's compute \(x^{2}=19^{2}+13^{2}-2\times19\times13\times\cos(70^{\circ})\). \(19^2 = 361\), \(13^2 = 169\), sum is 530. \(2\times19\times13 = 494\). \(\cos(70^\circ) \approx 0.3420201433\). So \(494\times0.3420201433 = 494\times0.3420201433\). Let's calculate 494*0.3 = 148.2, 494*0.04 = 19.76, 494*0.0020201433 = 494*0.002 = 0.988, 494*0.0000201433≈0.00995. So total is 148.2+19.76=167.96+0.988=168.948+0.00995≈168.958. Then 530 - 168.958 = 361.042. Then square root of 361.042 is approximately 19.0 (since 19^2=361, so 361.042 is very close to 361, so square root is very close to 19.0). Wait, but maybe I messed up the sides. Wait, the triangle: side AC is 19, side BC is 13, angle at C is 70 degrees, so side AB is x. So yes, Law of Cosines applies here. So the calculation gives x≈19.0. But let's check with a calculator. Let's use a calculator for the Law of Cosines: \(x = \sqrt{19^2 + 13^2 - 2\times19\times13\times\cos(70^\circ)}\). Let's compute each term:

19^2 = 361

13^2 = 169

2*19*13 = 494

cos(70°) ≈ 0.3420201433

494*0.3420201433 ≈ 494*0.3420201433 ≈ 168.95795

Then 361 + 169 = 530

530 - 168.95795 = 361.04205

Square root of 361.04205 is √361.04205 ≈ 19.0 (since 19^2 = 361, 19.0^2 = 361.0, so 361.042 is very close to 361, so square root is approximately 19.0 when rounded to one decimal place. Wait, but maybe my initial assumption is wrong. Wait, let's check with another method. Wait, maybe the angle is not 70 degrees? No, the problem says 70 degrees. Wait, maybe I made a mistake in the formula. Wait, Law of Cosines: \(c^2 = a^2 + b^2 - 2ab\cos(C)\), where C is the angle between sides a and b. So in triangle ABC, angle at C is between AC and BC, so sides AC=19, BC=13, angle C=70°, so AB is opposite angle C? No, wait, no: in triangle ABC, side opposite angle A is BC, side opposite angle B is AC, side opposite angle C is AB. So Law of Cosines: \(AB^2 = AC^2 + BC^2 - 2\times AC\times BC\times\cos(\angle C)\). Yes, that's correct. So AB is x, so \(x^2 = 19^2 + 13^2 - 2\times19\times13\times\cos(70^\circ)\), which is what I did. So the calculation is correct. So x≈19.0 when rounded to one decimal place.

Answer:

19.0