1. In the right triangle PQR shown below, find the six trigonometric ratios of the angle θ.
2. In the figure shown below, find the six trigonometric ratios of the angle θ.
3. In triangle ABC, right angled at B, 15sinA = 12. Find the other five trigonometric ratios of the angle A.
4. In the figure shown below, find the values of
sinB, secB, cotB, cosC, tanC and cscC
5. Find the length of the missing side (x). Round to the nearest tenth.
6. Find the length of the missing side (x). Round to the nearest tenth.
7. A tower stands vertically on the ground. From a point on the ground, which is 48 m away from the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower.
8. A kite is flying at a height of 75 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
1. Answer :
In the above right angled triangle, note that for the given angle θ, PQ is the ‘opposite’ side and PR is the ‘adjacent’ side.
Then,
sinθ = opposite side/hypotenuse = PQ/QR = 5/13
cosθ = adjacent side/hypotenuse = PR/QR = 12/13
tanθ = opposite side/adjacent side = PQ/PR = 5/12
cscθ = 1/sinθ = 13/5
secθ = 1/cosθ = 13/12
cotθ = 1/tanθ = 12/5
2. Answer :
In the right angled triangle ABC shown above,
AC = 24
BC = 7
By Pythagorean theorem,
AB^{2} = BC^{2} + CA^{2}
AB^{2} = 7^{2} + 24^{2}
AB^{2} = 49 + 576
AB^{2} = 625
AB^{2} = 25^{2}
AB = 25
Now, we can use the three sides to find the six trigonometric ratios of angle θ.
sinθ = opposite side/hypotenuse = BC/AB = 7/25
cosθ = adjacent side/hypotenuse = AC/AB = 24/25
tanθ = opposite side/adjacent side = BC/AC = 7/24
cscθ = 1/sinθ = 25/7
secθ = 1/cosθ = 25/24
cotθ = 1/tanθ = 24/7
3. Answer :
15sinA = 12
sinA = 12/15
sinA = opposite side/hypotenuse
sinA = 12/15
By Pythagorean theorem,
AC^{2} = AB^{2} + BC^{2}
15^{2} = AB^{2} + 12^{2}
225 = AB^{2} + 144
Subtract 144 from each side.
81 = AB^{2}
9^{2} = AB^{2}
9 = AB
Now, we can use the three sides to find the five trigonometric ratios of angle A and six trigonometric ratios of angle C.
cosA = adjacent side/hypotenuse
= AB/AC
= 9/15
= 3/5
tanA = opposite side/adjacent side
= BC/AB
= 12/9
= 4/3
cscA = 1/sinA
= 15/12
= 5/4
secA = 1/cosA
= 5/3
cotA = 1/tanA
= 3/4
4. Answer :
In the right ΔABD, by Pythagorean Theorem,
AB^{2} = AD^{2} + BD^{2}
13^{2} = AD^{2} + 5^{2}
169 = AD^{2} + 25
Subtract 25 from each side.
144 = AD^{2}
12^{2} = AD^{2}
12 = AD
In the right ΔACD, by Pythagorean Theorem,
AC^{2} = AD^{2} + CD^{2}
AC^{2} = 12^{2} + 16^{2}
AC^{2} = 144 + 256
AC^{2} = 400
AC^{2} = 20^{2}
AC = 20
Then,
sinB = opposite side/hypotenuse = AD/AB = 12/13
secB = hypotenuse/adjacent side = AB/BD = 13/5
cotB = adjacent side/opposite side = BD/AD = 5/12
cosC = adjacent side/hypotenuse
= CD/AC
= 16/20
= 4/5
tanC = opposite side/adjacent side
= AD/CD
= 12/16
= 3/4
cscC = hypotenuse/opposite side
= AC/AD
= 20/12
= 5/3
5. Answer :
In the triangle shown above, the length of the hypotenuse is x and for the angle 51°, the side which has the length 10 is adjacent side.
We know the length of adjacent side and we have to find the length of hypotenuse.
So, we have to use the trigonometric ratio in which we have hypotenuse and adjacent side.
sec51° = hypotenuse/adjacent side
sec51° = x/10
Multiply each side by 10.
10 ⋅ sec51° = x
Use calculator.
15.9 ≈ x
6. Answer :
In the triangle shown above, the length of the hypotenuse is 17 and for the angle 59°, the side which has the length x is opposite side.
We know the length of hypotenuse and we have to find the length of opposite side.
So, we have to use the trigonometric ratio in which we have opposite side and hypotenuse.
sin59° = opposite side/hypotenuse
sin59° = x/17
Multiply each side by 17.
17 ⋅ sin59° = x
Use calculator.
14.6 ≈ x
7. Answer :
Let PQ be the height of the tower.
Take PQ = h and QR is the distance between the tower and the point R.
In right triangle PQR above, considering ∠PRQ = 30°, PQ being the opposite side and QR being the adjacent side.
We know the length of the adjacent side (= 48 m) and we have to find the length of the opposite side (height of the tower).
We know that
tanθ = opposite side/adjacent side
In right triangle PQR,
tan∠PRQ = PQ/QR
tan30° = h/48
From trigonometric ratio table, we have tan30° = √3/3.
√3/3 = h/48
Multiply both sides by 48.
48√3/3 = h
16√3 = h
The height of the tower is 16√3 m.
8. Answer :
Let AB be the height of the kite above the ground. Then, AB = 75. Let AC be the length of the string.
In right triangle ABC above, considering ∠ACB = 60°, AB being the opposite side and AC being the hypotenuse.
We know the length of the opposite side (= 75 m) and we have to find the length of the hypotenuse (length of the string).
We know that
sinθ = opposite side/hypotenuse
In right triangle PQR,
sin∠ACB = AB/AC
sin60° = 75/AC
From trigonometric ratio table, we have sin60° = √3/2.
√3/2 = 75/AC
Take reciprocal on both sides.
2/√3 = AC/75
Multiply both sides by 75.
150/√3 = AC
150√3/3 = AC
50√3 = AC
The length of the string is 50√3.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Sep 29, 22 04:11 AM
Sep 29, 22 04:08 AM
Sep 29, 22 12:02 AM