Given triangle is
Using the triangle sum property, we get
[tex]\angle A+\angle B+\angle C=180^o[/tex][tex]\text{ Substitute }\angle B=85^o\text{ and }\angle C=76^o,\text{ we get}[/tex][tex]\angle A+85^o+76^o=180^o[/tex]
[tex]\angle A=180^o-161^o[/tex][tex]\angle A=19^o[/tex]
Given a=1200 yards.
We need to find b and c.
Consider the sine law.
[tex]\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}[/tex]
[tex]\text{ Substitute }\angle A=19^o,\angle B=85^o\text{,}\angle C=76^o,\text{ and a=1200, we get}[/tex]
[tex]\frac{\sin 19^o}{1200}=\frac{\sin 85^o}{b}=\frac{\sin 76^o}{c}[/tex]
[tex]0.00027=\frac{\sin85^o}{b}=\frac{\sin76^o}{c}[/tex]
Consider the following equation to find the value of b.
[tex]0.00027=\frac{\sin85^o}{b}[/tex]
[tex]b=\frac{\sin 85^o}{0.00027}[/tex][tex]b=3689.6\text{ yards.}[/tex]
Consider the following equation to find the value of c.
[tex]0.00027=\frac{\sin76^o}{c}[/tex]
[tex]c=\frac{\sin 76^o}{0.00027}[/tex][tex]c=3593.7\text{ yards.}[/tex]
Hence the distances of the oil platform from each end of the beach are 3689.6 yards and 3592.7 yards.
