Given that,
PR + QR = 25
PQ = 5 Let PR be x.
Therefore, QR = 25 - x
Applying Pythagoras theorem in \(Δ\)PQR, we obtain
\(\text{PR}^ 2 = \text{PQ}^ 2 + \text{QR}^ 2 \)
\(x^2= (5)^ 2 + (25 - x)^ 2 \)
\(x^2= 25 + 625 + x^ 2 - 50x \)
\(50x=650\)
\(x=13\)
Therefore, PR = 13 cm
QR = (25 - 13) cm = 12 cm
\(\text{sin p} =\frac{\text{ Opposite Side}}{\text{Hypotenuse }}= \frac{QR}{PR} = \frac{12}{13}\)
\(\text{sin p} = \frac{\text{Opposite Side}}{\text{Hypotenuse }}= \frac{QR}{PR} =\frac{ 12}{13}\)
\(\text{tan p} =\frac{\text{Opposite Side}}{\text{Adjacent side }}= \frac{QR}{PQ} = \frac{12}{5}\)
Assertion (A): The sum of the first fifteen terms of the AP $ 21, 18, 15, 12, \dots $ is zero.
Reason (R): The sum of the first $ n $ terms of an AP with first term $ a $ and common difference $ d $ is given by: $ S_n = \frac{n}{2} \left[ a + (n - 1) d \right]. $
Assertion (A): The sum of the first fifteen terms of the AP $21, 18, 15, 12, \dots$ is zero.
Reason (R): The sum of the first $n$ terms of an AP with first term $a$ and common difference $d$ is given by: $S_n = \frac{n}{2} \left[ a + (n - 1) d \right].$
The relationship between the sides and angles of a right-angle triangle is described by trigonometry functions, sometimes known as circular functions. These trigonometric functions derive the relationship between the angles and sides of a triangle. In trigonometry, there are three primary functions of sine (sin), cosine (cos), tangent (tan). The other three main functions can be derived from the primary functions as cotangent (cot), secant (sec), and cosecant (cosec).
sin x = a/h
cos x = b/h
tan x = a/b
Tan x can also be represented as sin x/cos x
sec x = 1/cosx = h/b
cosec x = 1/sinx = h/a
cot x = 1/tan x = b/a