Question

Develop a formula to find the missing parts of the following triangle if the given parts are:
1. Angle B; side c
2. Angle A; side b
3. Angle B;Angle A

Answer 1

1.Given $\angle B$, side $c$

Find: $a-? b-? \angle A-?$

Solution:

Find the side $a$:

$cos \angle B=\frac{a}{c}$

$a=c⋅cos\angle B$

Find the side $b$:

$a^2+b^2=c^2$

$c^2⋅cos^2\angle B +b^2=c^2$

$b^2=c^2-c^2⋅cos^2\angle B$

$b^2=c^2⋅(1-cos^2\angle B)$

$b=c\sqrt{\smash[b]{1-cos^2\angle B}}$

Find $\angle A$:

$\angle A=180\degree-\angle C-\angle B=180\degree-90\degree-\angle B=$

$=90\degree-\angle B$

2.Given $\angle A$, side $b$

Find: $a-?$$c-?$$\angle B-?$

Solution:

Find the side $c$:

$cos \angle A=\frac{b}{c}$

$c⋅cos\angle A=b$

$c=\frac{b}{cos\angle A}$

Find the side $a$:

$a^2+b^2=c^2$

$a^2+b^2=\frac{b^2}{cos^2\angle A}$

$a^2=\frac{b^2}{cos^2\angle A}-b^2$

$a^2=\frac{b^2-b^2⋅cos^2\angle A}{cos^2\angle A}$

$a^2=\frac{b^2}{cos^2\angle A}⋅(1-cos^2\angle A)$

$a=\frac{b}{cos\angle A}⋅\sqrt{\smash[b]{1-cos^2\angle A}}$

Find $\angle B$:

$\angle B=180\degree-\angle A-\angle C=180\angle-\angle A- 90\degree=$

$=90\degree-\angle A$

3.Given $\angle A, \angle B$

Find: $a-?$ $b-?$ $c-?$

Solution:

Let side $c=x$, then find the side $b$ :

$cos \angle A=\frac{b}{c}$

$b=c⋅cos\angle A$

$b=x⋅cos\angle A$

Find the side $a$ :

$a^2+b^2=c^2$

$a^2+x^2⋅cos^2\angle A=x^2$

$a^2=x^2-x^2⋅cos^2\angle A$

$a^2=x^2(1-cos^2\angle A)$

$a=x⋅\sqrt{\smash[b]{1-cos^2\angle A}}$