First, tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>), so if cos(<em>θ</em>) = 3/5 > 0 and tan(<em>θ</em>) < 0, then it follows that sin(<em>θ</em>) < 0.
Recall the Pythagorean identity:
sin²(<em>θ</em>) + cos²(<em>θ</em>) = 1
Then
sin(<em>θ</em>) = -√(1 - cos²(<em>θ</em>)) = -4/5
and so
tan(<em>θ</em>) = (-4/5) / (3/5) = -4/3
The remaining trig ratios are just reciprocals of the ones found already:
sec(<em>θ</em>) = 1/cos(<em>θ</em>) = 5/3
csc(<em>θ</em>) = 1/sin(<em>θ</em>) = -5/4
cot(<em>θ</em>) = 1/tan(<em>θ</em>) = -3/4
4/11•7/1=28/11
Answer: 28/11 or 2 6/11
In the parallelogram ABCD, join BD.
Consider the triangle Δ ABD.
It is given that AB > AD.
Since, in a triangle, angle opposite to longer side is larger, we have,
∠ ADB > ∠ ABD. --- (1)
Also, AB || DC and BD is a transversal.
Therefore,
∠ ABD = ∠ BDC
Substitute in (1), we get,
∠ ADB > ∠ BDC.
Answer:
Step-by-step explanation:
Given a general quadratic formula given as ax²bx+c = 0
To generate the general formula to solve the quadratic equation, we can use the completing the square method as shown;
Step 1:
Bringing c to the other side
ax²+bx = -c
Dividing through by coefficient of x² which is 'a' will give:
x²+(b/a)x = -c/a
- Completing the square at the left hand side of the equation by adding the square of half the coefficient x i.e (b/2a)² and adding it to both sides of the equation we have:
x²+(b/a)x+(b/2a)² = -c/a+(b/2a)²
(x+b/2a)² = -c/a+(b/2a)²
(x+b/2a)² = -c/a + b²/4a²
- Taking the square root of both sides
√(x+b/2a)² = ±√-c/a + b²/√4a²
x+b/2a = ±√(-4ac+b²)/√4a²
x+b/2a =±√b²-4ac/2a
- Taking b/2a to the other side
x = -b/2a±√√b²-4ac/2a
Taking the LCM:
x = {-b±√b²-4ac}/2a
This gives the vertex form with how it is used to Solve a quadratic equation.
