1. Lesson 18.4 – problem solving with trig
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2. Finding the area of triangles using trig ratios

3. Example 1 – find the area of each triangle to the nearest tenth.
𝐴= (15) sin 34
𝐴= 1 2 3(6) sin 108
𝐴= 1 2 (180) sin 34
𝐴= 1 2 (18) sin 108
𝐴= (90) sin 34
𝐴=(9) sin 108
𝐴≈50.3 𝑚 𝑚 2
𝐴=8.6 𝑐 𝑚 2

4. Solving right triangles
We can solve for missing sides of a right triangle by using Pythagorean Theorem
𝑎 2 + 𝑏 2 = 𝑐 2
We can solve for all angles in a right triangle using trig ratios if we know two side lengths or one side length and an acute angle measure.
Example 2 – Find all side lengths and angle measures of the triangle.
cos 𝐴= 7 9
m∠B=180−39−90
𝐶𝐵 = 9 2
m∠𝐵=51°
𝐶𝐵 2 +49=81
cos − = 𝑚∠𝐴
𝐶𝐵 2 =32
m∠𝐴=39°
𝐶𝐵≈5.7

5. Example 3 - Find all side lengths and angle measures of the triangle.
m∠D=180−27−90
tan 27 = 33 𝐸𝐹
m∠𝐵=63°
tan 27(𝐸𝐹) =33
= 𝐷𝐹 2
𝐸𝐹≈65 𝑚
= 𝐷𝐹 2
5314= 𝐷𝐹 2
𝐷𝐹≈73 𝑚

6. Solving a right triangle in the coordinate plane
We can use the distance formula as well as trig ratios to find the missing side lengths and angle measures of right triangles in the coordinate plane.
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐹𝑜𝑟𝑚𝑢𝑙𝑎= ( 𝑥 2 − 𝑥 1 ) 2 +( 𝑦 2 − 𝑦 1 ) 2

7. Example 4 - Step 1 – Plot the coordinates and sketch the triangle.
Step 2 – Find the lengths of the sides (Use distance formula for diagonal sides).
𝐴𝐵=4
𝐵𝐶=7
𝐴𝐶= (4−(−3 )) 2 +(−1−3 ) 2
𝐴𝐶= (7) 2 +(−4 ) 2
𝐴𝐶=
𝐴𝐶= 65
𝐴𝐶≈8.06
Step 3 – Find the acute angle measures using trig ratios.
tan 𝐴 = 7 4
m∠ 𝐴 = tan −
m∠ 𝐴 ≈60°
m∠B=180−60−90
m∠𝐵=30°

8. Example 5 - 𝐷𝐸= (3−(−4 )) 2 +(4−3 ) 2 𝐷𝐸= 49+1 𝐷𝐸= 50 ≈7.07
𝐷𝐸= (3−(−4 )) 2 +(4−3 ) 2
𝐷𝐸=
𝐷𝐸= 50 ≈7.07
𝐷𝐹= (−4−0) 2 +(3−0 ) 2
𝐷𝐹=
𝐷𝐹= 25 =5
𝐹𝐸= (3−0) 2 +(4−0 ) 2
𝐹𝐸=
𝐹𝐸= 25 =5
tan 𝐷 = 5 5
m∠ 𝐷 = tan −1 (1)
m∠ 𝐷 =45°
m∠E=180−45−90
m∠𝐸=45°

9. Assignment #35
Pg. 974
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