Question

In figure, ps=3cm,qs=4cm, angle prq=theta angle psq =90 pq is perpendicular to rq , rq=9cm. Evaluate tan theta

Answer 1

In triangle PSQ,

(PQ)^2=(PS)^2+(SQ)^2

(PQ)^2=(3)^2+(4)^2

(PQ)^2=9+16

(PQ)^2=25

(PQ) =5

tan theta =opp./adj.

[tan theta =5/9]

Answer 2

Answer:

The value of tanθ is $\frac{5}{9}$.

Step-by-step explanation:

From the figure it is clear that

$\angle PSQ=90^{\circ}$

It means triangle PSQ is a right angled triangle.

Using Pythagoras theorem in triangle PQS, we get

$hypotenuse^2=base^2+perpendicular^2$

$PQ^2=PS^2+QS^2$

$PQ^2=3^2+4^2$

$PQ^2=25$

$PQ=5$

The length of PQ is 5 cm.

$\angle PQR=90^{\circ}$

It means triangle PQR is a right angled triangle. In a right angle triangle

$\tan \theta=\frac{perpendicular}{base}$

$\tan \theta=\frac{PQ}{QR}$

$\tan \theta=\frac{5}{9}$

Therefore the value of tanθ is $\frac{5}{9}$.