Question

In $ \vartriangle PQR $ right angled at Q, $ PR + QR = 25{\text{ cm}} $ and $ PQ = 5{\text{ cm}} $ . Determine the values of $ \sin {\text{P}},\cos {\text{P}} $ and $ {\text{tanP}} $ .

Answer 1

Hint: From the given equations we have the Pythagoras theorem to solve the above given equations as the given triangle is a right angled triangle. We will get the values of sides of the triangle. From the trigonometric formulae we can find the values of $ \sin {\text{P}},\cos {\text{P}} $ and $ {\text{tanP}} $ .

Complete step-by-step answer:
Given $ \vartriangle PQR $ right angled at Q.

Right angled triangle is defined as a triangle in which one angle is a right angle. The relation between the sides and angles of a right triangle is the basis for trigonometry. The side opposite the right angle is called the hypotenuse. The sides adjacent to the right angle are called legs.
PQ+QR=25cm….(1)
PQ=5cm
By Pythagoras theorem,
 $ {\text{P}}{{\text{Q}}^2} + {\text{Q}}{{\text{R}}^2} = {\text{P}}{{\text{R}}^2} $
Pythagoras theorem is a theorem attributed that the square the square on the hypotenuse of a right-angled triangle is equal in area to the sum of the squares on the other two sides
 $ \Rightarrow 25 = {\text{P}}{{\text{R}}^2} - {\text{Q}}{{\text{R}}^2} $
 $ \Rightarrow 25 = \left( {{\text{PR + QR}}} \right)\left( {{\text{PR - QR}}} \right) $
 $ \Rightarrow {\text{PR - QR = }}\dfrac{{25}}{{25}} $
 $ \Rightarrow {\text{PR - QR = 1}} $ … (2)
We now add (1) and (2)
 $ \Rightarrow 2{\text{PR = 26}} $
 $ \Rightarrow {\text{PR = 13}} $
Therefore, QR = 12 and PR = 13
Trigonometric values are based on three major trigonometric ratios, Sine, Cosine and Tangent. Sine or $ \sin \theta $ = Side opposite to $ \theta $ is to Hypotenuse. Cosines or $ \operatorname{Cos} \theta $ = Adjacent side to $ \theta $ is to Hypotenuse. Tangent or $ \tan \theta $ = Side opposite to $ \theta $ is to Adjacent side to $ \theta $ .
 $ \Rightarrow \sin {\text{P = }}\dfrac{{{\text{QR}}}}{{{\text{PR}}}} $
 $ \Rightarrow \sin {\text{P = }}\dfrac{{12}}{{13}} $
And $ \cos {\text{P = }}\dfrac{{{\text{PQ}}}}{{{\text{PR}}}} $
 $ \Rightarrow \cos {\text{P = }}\dfrac{5}{{13}} $
And $ {\text{tan P = }}\dfrac{{\sin {\text{P}}}}{{\cos {\text{P}}}} $
 $ \Rightarrow {\text{tan P = }}\dfrac{{12}}{5} $

Note: There are six trigonometric ratios, sine, cosine, tangent, cotangent, secant and cosecant. We can find the remaining trigonometric ratios which are cotangent, secant and cosecant. Cotangent or $ \cot \theta $ = Adjacent side to $ \theta $ is to Side opposite to $ \theta $ . Secant or $ \operatorname{Sec} \theta $ = Hypotenuse is to Adjacent side to $ \theta $ . Cosecant or $ \operatorname{Cos} ec\theta $ = Hypotenuse is to Side opposite to $ \theta $