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| 12 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
In the diagram's right triangles, some angle measures and side lengths are shown. Find the missing angle measures and lengths.
Consider the following table of trigonometric ratios for some acute angles.
sinθ | cosθ | |
---|---|---|
θ=15∘ | 0.258819… | 0.965925… |
θ=30∘ | 0.5 | 0.866025… |
θ=45∘ | 0.707106… | 0.707106… |
θ=60∘ | 0.866025… | 0.5 |
θ=75∘ | 0.965925… | 0.258819… |
The sine of an acute angle is equal to the cosine of its complementary angle. Similarly, the cosine of an acute angle is equal to the sine of its complementary angle.
By the Interior Angles Theorem, the sum of the interior angle measures of a triangle is 180∘. For a right triangle, since one angle measures 90∘, the other two angles are acute.
This is true for all pairs of complementary angles.
Write the given expression in terms of sine. Write your answer without the degree symbol.
The sine of an acute angle is equal to the cosine of its complement. Similarly, the cosine of an acute angle is equal to the sine of its complement.
Determine the value of x that makes the equation true.
Complementary angles add up to 90∘.
Substitute expressions
LHS−172x=RHS−172x
Commutative Property of Addition
Factor out x
Subtract fractions
LHS+171=RHS+171
Add fractions
b1⋅a=ba
LHS⋅17=RHS⋅17
Like the sine and cosine, the same relationship exists between the tangent and cotangent.
The tangent of an acute angle is equal to the cotangent of its complementary angle. Similarly, the cotangent of an acute angle is equal to the tangent of its complementary angle. Therefore, the following statements hold true.
Consider a right triangle with side lengths x, y, and z.
Use the tangent and cotangent relationship of complementary angles.
Consider the product of the first term and the last terms of the expression.
In physics, the phenomenon known as refraction of light is described as the change in a light's direction as it passes from one medium to another. Due to refraction, objects in the water may appear to be closer to the water's surface than they actually are. The diagram shows Ignacio's eye, from above the surface of the ocean, viewing a whale that looks to have an apparent depth of 30 feet below the surface of the ocean.
Use the tangent and cotangent relationship of complementary angles.
Begin by labeling points on the diagram.
The relationship discussed throughout this lesson can be extended to include the secant and cosecant ratios.
As may have already been noticed, three of the trigonometric ratios start with the prefixco.
codenotes that β is the co-angle, or complementary angle, of α. The identities seen in this lesson are referred to as cofunction identities.
Cofunction Identities | |
---|---|
sinθ=cos(90∘−θ) | cosθ=sin(90∘−θ) |
tanθ=cot(90∘−θ) | cotθ=tan(90∘−θ) |
secθ=csc(90∘−θ) | cscθ=sec(90∘−θ) |
Consider the following triangle.
In the given right triangle, we know two angles. This means we can find the triangle's third angle by using the Interior Angles Theorem. m∠ θ+67^(∘) +90^(∘)= 180^(∘) ⇓ m∠ θ= 23^(∘) As we can see, the third angle has a measure of 23^(∘). Let's add this information to the diagram.
We want to know the value of sin 67^(∘). This is the ratio of the opposite leg to the hypotenuse. Using the information in the diagram, we can write an equation containing sin 67^(∘).
We can also write a second equation containing cos 23^(∘) by using the cosine ratio.
Since both sin 67^(∘) and cos 23^(∘) equals yx, we can equate these expressions and then use the information in the exercise about cos 23^(∘).
Consider △ABC.
Examining the right triangle, we notice that ∠ A and ∠ C represent the acute angles of the triangle. Therefore, by the Interior Angles Theorem, the sum of their measures equals 90^(∘). m∠ A+m∠ C+90^(∘)=180^(∘) ⇓ m∠ A+m∠ C=90^(∘) Since the cosine of an acute angle equals the sine of the complement of the angle, we know that cos A must equal sin C. Therefore, we get the following equation. cos A&=sin C &⇓ 2x-11/2&=7/2-x Let's solve this equation for x.
From Part A, we know the value of x. Next, let's substitute its value into the expression for sin C.
By using the inverse of the sine ratio we can find the measure of ∠ C. sin C=1/2 ⇕ m∠ C=sin^(- 1)1/2=30^(∘) Let's add m∠ C and the length of BC to the diagram.
By using the tangent ratio we can now find AB.
As we can see, the length of AB is about 4 units.