Question

find ab. round to the nearest tenth. triangle with vertices a, b, c; side ac = 11, side bc = 10, angle at c is 46° show calculator related content solving for a side with the law of cosines 4:38 solving for an angle with the law of cosines 6:41

Explanation:

Step1: Identify the Law of Cosines formula

To find the length of \( AB \) (let's call it \( c \)) in triangle \( ABC \) where we know two sides (\( a = 11 \), \( b = 10 \)) and the included angle \( C = 46^\circ \), we use the Law of Cosines: \( c^2=a^2 + b^2-2ab\cos(C) \)

Step2: Substitute the known values

Substitute \( a = 11 \), \( b = 10 \), and \( C = 46^\circ \) into the formula:
\( AB^2=11^2 + 10^2-2\times11\times10\times\cos(46^\circ) \)
First, calculate \( 11^2 = 121 \), \( 10^2 = 100 \), and \( 2\times11\times10 = 220 \)
So, \( AB^2=121 + 100-220\times\cos(46^\circ) \)
\( AB^2=221-220\times\cos(46^\circ) \)
Now, find \( \cos(46^\circ)\approx0.6947 \)
Then, \( 220\times0.6947\approx152.834 \)
So, \( AB^2=221 - 152.834 = 68.166 \)

Step3: Take the square root and round

Take the square root of \( 68.166 \): \( AB=\sqrt{68.166}\approx8.256 \)
Round to the nearest tenth: \( AB\approx8.3 \)

Answer:

\( 8.3 \)