First question
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\(cos(B)=sqrt(5)/5\)
Steps:
-Draw triangle ACB, where C is the right angle,
-Given 2 sides: BC=4 and AC=8, Thus the third side AB (Which is the hypotenuse)= 16+64=AB^2, \(AB=\sqrt{80}\)
Notice \(\sqrt{80}=4\sqrt{5}\)
\(cos(B)=\frac{Adjacent side}{hypotenuseside}\)
\(cos(B)=\frac{4}{4\sqrt{5}}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\)
\(tan(A)=\frac{Opposite}{Adjacent}=\frac{4}{8}=\frac{1}{2}\)
\(csc(B)=\frac{4\sqrt{5}}{8}=\frac{\sqrt{5}}{2}\) Notice csc is the reverse of sin (I.e. sin = opposite/hypotenuse, csc (cosecant) is hypotenuse/opposite)
Keep in mind that sec is reverse of cos (cos=adjacent/hypotenuse, while sec=hypotenuse/adjacent)
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Second question,
-Draw triangle ACB, where C is the right angle,
-Now 2 choices, either recall what a 30-60-90 triangle is. or using trigonometry (Let's use trigonometry)
\(sin(60)=\frac{7\sqrt{5}}{x}\) where x is the hypotenuse
\(x=\frac{7\sqrt{5}}{sin(60)}=\frac{14\sqrt{15}}{3}=18.07\)
Now we have 2 sides so let's find the third side by Pythagoras theorem.
\(BC^2=(\frac{14\sqrt{15}}{3})^2-(7\sqrt{5})^2\) Use the calculator
\(BC=\frac{7\sqrt{15}}{3}\)
Perimeter is the sum of the 3 sides of the triangle.
\(7\sqrt{5}+\frac{14\sqrt{15}}{3}+\frac{7\sqrt{15}}{3}\)
Simplifying:
\(7\sqrt{15}+7\sqrt{5}\) (Choice 3)
=42.76335926...
