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
3 answers
oh yea for question 1 a) is sin delta sorry
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
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!
1 answer