In ΔABC, find the length, to the nearest tenth of an inch, of side a if b=7 in., c=10 in., and the angle formed by these sides measures 72 degrees.(1 point)

Question

a=
in.

Answer 1

To find the length of side $a$ in triangle $\mathrm{\Delta }ABC$ using the Law of Cosines, we can apply the formula:

$$a^2=b^2+c^2-2bc\cdot \cos (A)$$

In this case, we are given:

• $b=7$ inches
• $c=10$ inches
• $A={72}^{\circ }$

First, we need to calculate $\cos ({72}^{\circ })$. You may use a calculator for this:

$$\cos ({72}^{\circ })\approx 0.3090$$

Now, substitute the values into the formula:

$$a^2=7^2+{10}^2-2\cdot 7\cdot 10\cdot \cos ({72}^{\circ })$$

Calculating the squares first:

$$7^2=49$$

$${10}^2=100$$

Now plug in these values:

$$a^2=49+100-2\cdot 7\cdot 10\cdot 0.3090$$

Calculating $2\cdot 7\cdot 10$:

$$2\cdot 7\cdot 10=140$$

Now, we multiply this by $\cos ({72}^{\circ })$:

$$140\cdot 0.3090\approx 43.26$$

Now substitute this into the equation for $a^2$:

$$a^2=49+100-43.26$$

Now calculate:

$$a^2=149-43.26\approx 105.74$$

Now, find $a$ by taking the square root:

$$a\approx \sqrt{105.74}\approx 10.3\text{ inches}$$

Thus, the length of side $a$ to the nearest tenth of an inch is:

$$\boxed{{\displaystyle 10.3}}\text{ in.}$$