Question

Find the value of b. Round your answer to the nearest tenth.

Answer 1

Answer:

Side CA = 7.8

Step-by-step explanation:

Given:

Acute angled $\triangle ABC$.

$\angle B =40^\circ$

AB = 10

BC = 12

We can use cosine rule here to find the side AC = b

Formula for cosine rule:

$cos B = \dfrac{a^{2}+c^{2}-b^{2}}{2ac}$

Where  

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

$cos 40 = \dfrac{12^{2}+10^{2}-b^{2}}{2\times 12\times 10}\\\Rightarrow cos 40 = \dfrac{144+100-b^{2}}{240}\\\Rightarrow 0.77 = \dfrac{244-b^{2}}{240}\\\Rightarrow 244-b^{2} = 0.77 \times 240\\\Rightarrow 244-b^{2} = 183.85\\\Rightarrow 244-183.85 = b^{2}\\\Rightarrow b^2 = 60.15\\\Rightarrow b = 7.76$

To the nearest tenth b = 7.8