Advertisements
Advertisements
Question
In ΔPQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Advertisements
Solution
Given that, PR + QR = 25
PQ = 5
Let PR be x.
Therefore, QR = 25 − x
Applying Pythagoras theorem in ΔPQR, we obtain
PR2 = PQ2 + QR2
x2 = (5)2 + (25 − x)2
x2 = 25 + 625 + x2 − 50x
50x = 650
x = 13
Therefore, PR = 13 cm
QR = (25 − 13) cm
= 12 cm
sin P = `("QR")/("PR")=12/13`
cos P = `("PQ")/("PR")=5/13`
tan P = `("QR")/("PQ")=12/5`
APPEARS IN
RELATED QUESTIONS
In ΔABC right angled at B, AB = 24 cm, BC = 7 m. Determine:
sin A, cos A
State whether the following are true or false. Justify your answer.
sec A = `12/5` for some value of angle A.
State whether the following are true or false. Justify your answer.
cot A is the product of cot and A.
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
`cot theta = 12/5`
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
`cos theta = 12/2`
If `cot theta = 1/sqrt3` show that `(1 - cos^2 theta)/(2 - sin^2 theta) = 3/5`
if `tan theta = 12/13` Find `(2 sin theta cos theta)/(cos^2 theta - sin^2 theta)`
Evaluate the following
cos2 30° + cos2 45° + cos2 60° + cos2 90°
Evaluate the following
tan2 30° + tan2 60° + tan2 45°
Evaluate the following
`sin^2 30° cos^2 45 ° + 4 tan^2 30° + 1/2 sin^2 90° − 2 cos^2 90° + 1/24 cos^2 0°`
Evaluate the Following
4(sin4 30° + cos2 60°) − 3(cos2 45° − sin2 90°) − sin2 60°
Find the value of x in the following :
`2 sin x/2 = 1`
The value of sin² 30° – cos² 30° is ______.
3 sin² 20° – 2 tan² 45° + 3 sin² 70° is equal to ______.
If sin 2A = `1/2` tan² 45° where A is an acute angle, then the value of A is ______.
If sin A = `1/2`, then the value of cot A is ______.
Given that sinα = `1/2` and cosβ = `1/2`, then the value of (α + β) is ______.
Prove that sec θ + tan θ = `cos θ/(1 - sin θ)`.
Proof: L.H.S. = sec θ + tan θ
= `1/square + square/square`
= `square/square` ......`(∵ sec θ = 1/square, tan θ = square/square)`
= `((1 + sin θ) square)/(cos θ square)` ......[Multiplying `square` with the numerator and denominator]
= `(1^2 - square)/(cos θ square)`
= `square/(cos θ square)`
= `cos θ/(1 - sin θ)` = R.H.S.
∴ L.H.S. = R.H.S.
∴ sec θ + tan θ = `cos θ/(1 - sin θ)`
In ΔBC, right angled at C, if tan A = `8/7`, then the value of cot B is ______.
