I really dk how to do them,i have spend hours doing it
Can somebody please solve and if possible show the steps thx,just 3 questions
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3)Given that tan theta = -✓3 and that 90 degree ≤ delta ≤ 180 degree,find the value of:
a) tan delta b)cos delta
5)Given that tan delta = 2/3 and that delta is reflex find the value of:
a) sin delta b)cos delta
This is the hardest one,I have no idea
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6)Given that tan A = 4/3 and cos B = -1/✓3 ,where A and B are in the same quadrant,find the value of
a) sin delta b)cos A c) sin B d)tan B
for ease of typing, and not having to worry about special symbols on the keyboard, I will
write is as
3. Given that tan k = -√3 and that 90 degree ≤ k ≤ 180 degree,find the value of:
a) sin k
b) cos k
make a sketch of a right-angled triangle and if tan k = -√3/1
ignore the negative sign and label the base angle as k
the opposite side as √3 and the adjacent side as 1
remember in a right-angled triangle r^2 = x^2 + y^2 , where r is the hypotenuse, the opposite is √3 and x = 1
BUT tan k = -√3/1, and from the CAST rule we know that the tangent is negative in quads II or IV
so k must be in II or IV
r^2 = (√3)^2 + 1^2 = 4
r = 2
in II, x = -1, y = √3, r = 2
in IV, x = 1, y = -√3, r = 2
now we can find any of the trig ratios, using our definitions involving opposite, adjacent and hypotenuse
case1 , k is in II
so sin k = opposite/hypotenus = y/r = √3/2
and cos k = adjacent/ hypotenuse = -1/2
case 2, k is in IV
sin k = -√3/2
cos k = 1/2
btw, this is the famous 30-60-90° triangle with corresponding sides 1 - √3 - 2
and you should make a point of memorizing it
I will do the last one, use these ideas to do the 2nd one
6)
Given that tan A = 4/3 and cos B = -1/√3 ,where A and B are in the same quadrant,find the value of
a) sin ???
b) cos A
c) sin B
d)tan B
It said that A and B are in the same quad, and only in III is the tangent positive and the cosine negative, so we are dealing with quadrant III
Again making my triangle for angle A , we have A as the base angle,
tan A = 4/3, so y = -4 and x = -3 (have to be in III)
r^2 = (-4)^2 + (-3)^2 = 25
r = 5
now we can find any trig ratio of angle A
e.g. cosA = -3/5
do the same for B
thank you so much for the explanation teacher Reiny!
oh yea for question 1 a) is sin delta sorry
Sure! I'll be happy to help you solve these trigonometry questions step by step.
To solve these problems, we will be using the trigonometric identities and properties. I will provide you with the necessary explanations and steps.
3) Given that tan theta = -√3 and that 90° ≤ delta ≤ 180°, find the value of:
a) tan delta
b) cos delta
a) tan delta:
To find the value of tan delta, we need to find the angle delta. Since tan theta = -√3 and we know theta, we can find the value of delta using the inverse tan function (also called arctan, denoted as tan^(-1) or atan).
Using the inverse tan, we find:
delta = atan(-√3) ≈ -60°
Now that we have the value of delta, we can find the value of tan delta:
tan delta = tan(-60°) ≈ -1.732
b) cos delta:
To find the value of cos delta, we can use the identity cos^2 delta = 1 - sin^2 delta.
We know that tan delta = -√3, so we can find sin delta using the identity sin delta = tan delta / √(1 + tan^2 delta).
Substituting the values, we get:
sin delta = (-√3) / √(1 + (-√3)^2) ≈ -0.577
Now we can find cos delta using the identity cos delta = ± √(1 - sin^2 delta).
However, since 90° ≤ delta ≤ 180°, cos delta will be negative.
cos delta = - √(1 - (-0.577)^2) ≈ -0.816
5) Given that tan delta = 2/3 and that delta is reflex, find the value of:
a) sin delta
b) cos delta
a) sin delta:
To find the value of sin delta, we can use the identity sin delta = tan delta / √(1 + tan^2 delta).
Given tan delta = 2/3, we can substitute this value into the formula:
sin delta = (2/3) / √(1 + (2/3)^2) ≈ 0.554
b) cos delta:
To find the value of cos delta, we can use the identity cos^2 delta = 1 - sin^2 delta.
Using the value of sin delta from the previous step, we can find cos delta as follows:
cos delta = ± √(1 - (0.554)^2)
However, since delta is reflex, cos delta will be negative.
cos delta = -√(1 - (0.554)^2) ≈ -0.833
6) Given that tan A = 4/3 and cos B = -1/√3, where A and B are in the same quadrant, find the value of:
a) sin delta
b) cos A
c) sin B
d) tan B
a) sin delta:
Unfortunately, the value of delta is not given in this problem. Please provide the information on delta, if available, so that we can proceed with finding the respective trigonometric values.
b) cos A:
To find the value of cos A, we can use the identity cos^2 A = 1 - sin^2 A.
Since we know tan A = 4/3, we can find sin A using the identity sin A = tan A / √(1 + tan^2 A).
Substituting the values, we get:
sin A = (4/3) / √(1 + (4/3)^2) ≈ 0.8
Now we can find cos A using the identity cos A = ± √(1 - sin^2 A).
Since A is in the same quadrant as B, cos A will be positive.
cos A = √(1 - (0.8)^2) ≈ 0.6
c) sin B:
Given that cos B = -1/√3, we can use the identity sin^2 B = 1 - cos^2 B to find sin B.
Using the value of cos B, we get:
sin B = √(1 - (-1/√3)^2) ≈ 0.577
d) tan B:
To find the value of tan B, we can use the identity tan B = sin B / cos B.
Substituting the given values, we find:
tan B = (0.577) / (-1/√3) ≈ -1.732
I hope this helps! Let me know if you have any further questions.