Please help I took a screenshot of the question
2 answers:
Answer:
Step-by-step explanation:
<u>Trigonometric Ratios</u>
The relations between the sides of a right triangle and the angles are called trigonometric ratios.
The longest side of the triangle is called the hypotenuse and the other two sides are the legs.
Selecting any of the acute angles as a reference, it has an adjacent side and an opposite side. The trigonometric ratios are defined upon those sides.
The cosine is defined as:
Considering angle A, we have:
Considering angle B:
Both expressions are exactly the same, thus:
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The required proof is given in the table below:
![\begin{tabular}{|p{4cm}|p{6cm}|} Statement & Reason \\ [1ex] 1. $\overline{BD}$ bisects $\angle ABC$ & 1. Given \\ 2. \angle DBC\cong\angle ABD & 2. De(finition of angle bisector \\ 3. $\overline{AE}$||$\overline{BD}$ & 3. Given \\ 4. \angle AEB\cong\angle DBC & 4. Corresponding angles \\ 5. \angle AEB\cong\angle ABD & 5. Transitive property of equality \\ 6. \angle ABD\cong\angle BAE & 6. Alternate angles \end{tabular}](https://tex.z-dn.net/?f=%20%5Cbegin%7Btabular%7D%7B%7Cp%7B4cm%7D%7Cp%7B6cm%7D%7C%7D%20%0A%20Statement%20%26%20Reason%20%5C%5C%20%5B1ex%5D%20%0A1.%20%24%5Coverline%7BBD%7D%24%20bisects%20%24%5Cangle%20ABC%24%20%26%201.%20Given%20%5C%5C%0A2.%20%5Cangle%20DBC%5Ccong%5Cangle%20ABD%20%26%202.%20De%28finition%20of%20angle%20bisector%20%5C%5C%20%0A3.%20%24%5Coverline%7BAE%7D%24%7C%7C%24%5Coverline%7BBD%7D%24%20%26%203.%20Given%20%5C%5C%20%0A4.%20%5Cangle%20AEB%5Ccong%5Cangle%20DBC%20%26%204.%20Corresponding%20angles%20%5C%5C%0A5.%20%5Cangle%20AEB%5Ccong%5Cangle%20ABD%20%26%205.%20Transitive%20property%20of%20equality%20%5C%5C%20%0A6.%20%5Cangle%20ABD%5Ccong%5Cangle%20BAE%20%26%206.%20Alternate%20angles%0A%5Cend%7Btabular%7D)
![\begin{tabular}{|p{4cm}|p{6cm}|} 7. \angle AEB\cong\angle BAE & 7. Transitive property of equality \\ 8. \overline{EB}\cong\overline{AB} & 8. From 7. $\Delta ABE$ is isosceles \\ 9. EB = AB & 9. De(finition of congruence \\ 10. $\frac{AD}{DC}=\frac{EB}{BC}$ & 10. Triangle proportionality theorem \\ 11. $\frac{AD}{DC}=\frac{AB}{BC}$ & 11. Substitution Property of equality \\[1ex] \end{tabular} ](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%7B%7Cp%7B4cm%7D%7Cp%7B6cm%7D%7C%7D%0A7.%20%5Cangle%20AEB%5Ccong%5Cangle%20BAE%20%26%207.%20Transitive%20property%20of%20equality%20%5C%5C%0A8.%20%5Coverline%7BEB%7D%5Ccong%5Coverline%7BAB%7D%20%26%208.%20From%207.%20%24%5CDelta%20ABE%24%20is%20isosceles%20%5C%5C%0A9.%20EB%20%3D%20AB%20%26%209.%20De%28finition%20of%20congruence%20%5C%5C%2010.%20%24%5Cfrac%7BAD%7D%7BDC%7D%3D%5Cfrac%7BEB%7D%7BBC%7D%24%20%26%2010.%20Triangle%20proportionality%20theorem%20%5C%5C%0A11.%20%24%5Cfrac%7BAD%7D%7BDC%7D%3D%5Cfrac%7BAB%7D%7BBC%7D%24%20%26%2011.%20Substitution%20Property%20of%20equality%20%5C%5C%5B1ex%5D%20%0A%5Cend%7Btabular%7D%0A)