Question

1. solve for b: a triangle with vertex b having an angle of 47°, side from b to a is 19, side from b to c is 23.

Explanation:

Step1: Identify the Law to Use

We have a triangle with two sides (19, 23) and the included angle (47°) known, so we use the Law of Cosines. The Law of Cosines formula for side \( b \) (opposite angle \( B \)) is \( b^{2}=a^{2}+c^{2}-2ac\cos B \), where \( a = 23 \), \( c = 19 \), and \( B = 47^{\circ} \).

Step2: Substitute the Values

Substitute \( a = 23 \), \( c = 19 \), and \( \cos(47^{\circ})\approx0.6820 \) into the formula:

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Step3: Calculate the Products and Subtract

First, calculate \( 874\times0.6820\approx596.068 \). Then:

$$ b^{2}=890 - 596.068 = 293.932 $$

Step4: Take the Square Root

Take the square root of \( 293.932 \) to find \( b \):

$$ b=\sqrt{293.932}\approx17.14 $$

Answer:

\( b\approx17.14 \)