Question

question in △hij, j = 1.7 cm, h = 8.9 cm and ∠i = 106°. find the length of i, to the nearest 10th of a centimeter. score: 0/3 penalty: none answer attempt 1 out of 2 submit answer watch video show examples law of cosines (sas)

Explanation:

Step1: Recall the Law of Cosines formula

The Law of Cosines for finding side $i$ in $\triangle HIJ$ is $i^{2}=h^{2}+j^{2}-2hj\cos I$.

Step2: Substitute given values

We have $h = 8.9$, $j = 1.7$, and $I=106^{\circ}$. So $i^{2}=(8.9)^{2}+(1.7)^{2}-2\times8.9\times1.7\times\cos(106^{\circ})$.
First, calculate $(8.9)^{2}=79.21$, $(1.7)^{2}=2.89$. And $\cos(106^{\circ})\approx - 0.2756$. Then $2\times8.9\times1.7\times\cos(106^{\circ})\approx2\times8.9\times1.7\times(- 0.2756)\approx - 8.37$.
So $i^{2}=79.21 + 2.89-(-8.37)=79.21+2.89 + 8.37=90.47$.

Step3: Find the value of $i$

Take the square - root of $i^{2}$. $i=\sqrt{90.47}\approx9.5$.

Answer:

$9.5$ cm