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1. find the length of the indicated side, to the nearest tenth of a cen…

Question

  1. find the length of the indicated side, to the nearest tenth of a centimetre.5

Explanation:

Step1: Find angle C

The sum of angles in a triangle is 180°. So, $\angle C=180^{\circ}-45^{\circ}-74^{\circ}=61^{\circ}$.

Step2: Use the sine - rule

The sine - rule states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Here, we know $c = 25$ cm, $\angle B = 45^{\circ}$, and $\angle C=61^{\circ}$. We want to find $b$. So, $\frac{b}{\sin B}=\frac{c}{\sin C}$, which can be rewritten as $b=\frac{c\sin B}{\sin C}$.

Step3: Substitute the values

Substitute $c = 25$, $\sin B=\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.707$, and $\sin C=\sin61^{\circ}\approx0.875$ into the formula. Then $b=\frac{25\times0.707}{0.875}$.
$b=\frac{17.675}{0.875}\approx20.2$ cm.

Answer:

$20.2$ cm