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find the length of a. triangle image with vertices a, b, c; side ab = 3…

Question

find the length of a.

triangle image with vertices a, b, c; side ab = 32, side ac = 21, angle at a is 40°; side bc is labeled a

a = blank box

round to the nearest tenths place.

use the law of cosine to answer this question.

$a^2 = b^2 + c^2 - 2bc \cos a$
$b^2 = a^2 + c^2 - 2ac \cos b$
$c^2 = a^2 + b^2 - 2ab \cos c$

Explanation:

Step1: Identify values for law of cosines

In triangle \(ABC\), we have \(b = 21\), \(c = 32\), and \(\angle A=40^\circ\). The law of cosines formula for \(a^2\) is \(a^{2}=b^{2}+c^{2}-2bc\cos A\).

Step2: Substitute the values

Substitute \(b = 21\), \(c = 32\), and \(\cos(40^\circ)\approx0.7660\) into the formula:

$$ LATEXBLOCK0 $$

Step3: Calculate the product and subtraction

First, calculate \(1344\times0.7660\approx1344\times0.766 = 1029.504\). Then:

$$ LATEXBLOCK1 $$

Step4: Take the square root

Take the square root of \(435.496\) to find \(a\):

$$ a=\sqrt{435.496}\approx20.9 $$

Answer:

\(20.9\)