Exercise 38 Page 560

Practice makes perfect

a Let's start by drawing the described triangle. Notice that ∠ C is a right angle, which means AB must be the hypotenuse of a right triangle.

We know that sin A=0.5. To determine the measure of A, we will take the inverse of sine of both sides of this equation.

sin A=0.5

sin^(-1)(LHS) = sin^(-1)(RHS)

m∠ A=sin^(-1)0.5

300sin^(-1)(0)=0^(∘) 3030sin^(-1)(1/2)=30^(∘) 3045sin^(-1)(sqrt(2)/2)=45^(∘) 3060sin^(-1)(sqrt(3)/2)=60^(∘) 3090sin^(-1)(1)=90^(∘) 30-30sin^(-1)(- 1/2)=- 30^(∘) 30-45sin^(-1)(- sqrt(2)/2)=- 45^(∘) 30-60sin^(-1)(- sqrt(3)/2)=- 60^(∘) 30-90sin^(-1)(- 1)=- 90^(∘)

m∠ A=30^(∘)

b The sine ratio is the opposite side of an angle divided by the hypotenuse.

sin θ =Opposite/HypotenuseNotice that BC is the opposite side to ∠ A. Therefore, we can write the following equation. sin A=BC/10 If we substitute 0.5 for sin A, we can solve for BC.

sin A=BC/10

Substitute

sin A= 0.5

0.5=BC/10

MultEqn

LHS * 10=RHS* 10

5=BC

RearrangeEqn

Rearrange equation

BC=5

c We already know that m∠ C=90^(∘) and that m∠ A=30^(∘). With this information, we can figure out m∠ B by using the Interior Angles Theorem.

90^(∘)+30^(∘)+m∠ B=180^(∘) ⇓ m∠ B=60^(∘) This means we want to figure out cos 60^(∘). This trigonometric ratio actually has an exact value.

cos 60^(∘)

\ifnumequal{60}{0}{\cos\left(0^\circ\right)=1}{}\ifnumequal{60}{30}{\cos\left(30^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{45}{\cos\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{60}{\cos\left(60^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{90}{\cos\left(90^\circ\right)=0}{}\ifnumequal{60}{120}{\cos\left(120^\circ\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{60}{135}{\cos\left(135^\circ\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{150}{\cos\left(150^\circ\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{180}{\cos\left(180^\circ\right)=\text{-} 1}{}\ifnumequal{60}{210}{\cos\left(210^\circ\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{60}{225}{\cos\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{240}{\cos\left(240^\circ\right)=\text{-} \dfrac {1}2}{}\ifnumequal{60}{270}{\cos\left(270^\circ\right)=0}{}\ifnumequal{60}{300}{\cos\left(300^\circ\right)=\dfrac{1}2}{}\ifnumequal{60}{315}{\cos\left(315^\circ\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{330}{\cos\left(330^\circ\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{60}{360}{\cos\left(360^\circ\right)=1}{}

1/2

d Let's have a look at the diagram from Part A. We will add all of the calculated information from previous parts.

a 30-60-90 triangle with a hypotenuse of 10 and a short leg of 5

Notice that this is a 30-60-90 triangle. In such a triangle, the longer leg is sqrt(3) times greater than the shorter leg. Since the shorter leg is 5 units, it must be that AC is 5sqrt(3).

Subchapter links

10.1.1 The Equation of a Circle? p.549-550

10.1.2 Completing the Square for Equations of Circles p.552-553

10.1.3 The Geometric Definition of a Parabola p.558-562

Exercise 38 Page 560

Hints

Check the answer

Practice exercises

Asses your skill