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1  Feb 21, 2020

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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...

Feb 22, 2020