Answer:
[tex]\angle B = 60^o[/tex]
[tex]b =17.3[/tex]
[tex]c = 20[/tex]
Step-by-step explanation:
Given
[tex]a= 15[/tex]
[tex]\angle A = 30^o[/tex]
[tex]\angle C = 90^o[/tex]
See attachment for illustration
Required
Solve the triangle
First, we calculate the measure of B
[tex]\angle A + \angle B + \angle C = 180^o[/tex] --- angles in a triangle
[tex]30^o + \angle B + 90^o = 180^o[/tex]
Collect like terms
[tex]\angle B = 180^o-90^o-30^o[/tex]
[tex]\angle B = 60^o[/tex]
Solve for (c) using sine function
[tex]\sin(30) = \frac{a}{c}[/tex]
Make c the subject
[tex]c = \frac{a}{\sin(30)}[/tex]
Substitute known values
[tex]c = \frac{10}{0.5}[/tex]
[tex]c = 20[/tex]
Solve for (b) using Pythagoras
[tex]c^2 = a^2 + b^2[/tex]
This gives:
[tex]20^2 = 10^2 + b^2[/tex]
[tex]400 = 100 + b^2[/tex]
Collect like terms
[tex]b^2 =400 - 100[/tex]
[tex]b^2 =300[/tex]
Take square roots
[tex]b =17.3[/tex]
