Find the values of the sine, cosine, and tangent for ZA.
1 answer:
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Answer:
1. (x - 3)² = 8
2. (x + 2)² = 3
3. (x + 6)² =
4. (x + 3)² = 27
5. (x + 4)² = 13
6.
Step-by-step explanation:
Completion of Square:
In the following problems the terms in the RHS of the above equation may be missing. We balance the equation. Simplify it and re write it in terms of LHS.
1. x² - 6x + 1 = 0
Taking the constant term to the other side, we get:
x² - 6x = - 1
⇒ x² - 2(3)x = -1
⇒ x² -2(3)x + 9 = - 1 + 9 [Adding 9 to both the sides]
⇒ x² -2(3)x + 3² = 8
⇒ (x - 3)² = 8 is the answer.
2. 3x² + 12x + 3 = 0
Note that the co-effecient of x² is not 1. We make it 1, by dividing the whole equation by 3. And then proceed like the previous problem.
3x² + 12x = -3
Dividing by 3 through out, x² + 4x = - 1
⇒ x² + 2(2) + 4 = -1 + 4
⇒ x² +2(2) + 2² = 3
⇒ (x + 2)² = 3 is the answer.
3. 2x² + 24x = 29
x² + 12x =
⇒ x² + 2(6)x + 36 = + 36
⇒ x² + 2(6)x + 6² =
⇒ (x + 6)² = is the answer.
4. x² + 6x - 18 = 0
x² + 6x = 18
⇒ x² + 2(3)x = 18
⇒ x² + 2(3)x + 9 = 18 + 9
⇒ x² + 2(3)x + 3² = 27
⇒ (x + 3)² = 27 is the answer.
5. x² + 8x + 3 = 0
x² + 8x = -3
⇒ x² + 2(4)x = -3
⇒ x² + 2(4)x + 16 = - 3 + 16
⇒ x² + 2(4)x + 16 = 13
⇒ (x + 4)² = 13 is the answer.
6. 9x² - 30x + 6 = 0
9x² - 30x = - 6
⇒ x² x = - 6
is the answer.
Answer:
15°.
Step-by-step explanation:
1. Angles ADC and CDB are supplementary, thus
m∠ADC+m∠CDB=180°.
Since m∠ADC=115°, you have that m∠CDB=180°-115°=65°.
2. Triangle BCD is isosceles triangle, because it has two congruent sides CB and CD. The base of this triangle is segment BD. Angles that are adjacent to the base of isosceles triangle are congruent, then
m∠CDB=m∠CBD=65°.
The sum of the measures of interior angles of triangle is 180°, therefore,
m∠CDB+m∠CBD+m∠BCD=180° and
m∠BCD=180°-65°-65°=50°.
3. Triangle ABC is isosceles, with base BC. Then
m∠ABC=m∠ACB.
From the previous you have that m∠ABC=65° (angle ABC is exactly angle CBD). So
m∠ACB=65°.
4. Angles BCD and DCA together form angle ACB. This gives you
m∠ACB=m∠ACD+m∠BCD,
m∠ACD=65°-50°=15°.
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