Question

In triangle ABC, A=25, c=55 and AB=60. What are the approximate measures of the remaining side lengths of the triangle?

Answer 1

Answer:

$a\approx 31$

$b\approx 72$

Step-by-step explanation:

Please find the attachment.

We have been given that in triangle ABC, A=25, C=55 and AB=60. We are asked to find the approximate measures of the remaining side lengths of the triangle.

We will use Law of Sines to solve for side lengths of given triangle.

$\frac{\text{sin}(A)}{a}=\frac{\text{sin}(B)}{b}=\frac{\text{sin}(C)}{c}$, where a, b and c are opposite sides corresponding to angles A, b and C respectively.

Upon substituting our given values, we will get:

$\frac{\text{sin}(25)}{a}=\frac{\text{sin}(55)}{60}$

$a=\frac{60\text{sin}(25)}{\text{sin}(55)}$

$a=\frac{60*0.422618261741}{0.819152044289}$

$a=\frac{25.35709570446}{0.819152044289}$

$a=30.9552980807967304$

$a\approx 31$

Therefore, the measure of side 'a' is approximately 31 units.

We can find measure of angle B using angle sum property as:

$m\angle A+m\angle B+m\angle C=180$

$25+m\angle B+55=180$

$m\angle B+80=180$

$m\angle B=100$

$\frac{\text{sin}(100)}{b}=\frac{\text{sin}(55)}{60}$

$b=\frac{60\text{sin}(100)}{\text{sin}(55)}$

$b=\frac{60*0.984807753012}{0.819152044289}$

$b=\frac{59.08846518072}{0.819152044289}$

$b=72.1336967815383509$

$b\approx 72$

Therefore, the measure of side 'b' is approximately 72 units.