Question

from the figure find the value of 25(sin^2 theta +2 cos^2 theta - tan theta)​

Answer 1

Answer:

okay plz give me marks

from the figure we have sin theta is equal to PQ/PR =8/10=4/5

COS =QR/ PR =6/10=3/5

TAN =

Answer 2

The required figure is attached with the answer.

Step-by-step explanation:

From the given figure, we can find

$sin\theta=\dfrac{perpendicular}{hypotenuse}=\dfrac{PQ}{PR}=\dfrac{8}{10}$

$cos\theta=\dfrac{base}{hypotenuse}=\dfrac{QR}{PR}=\dfrac{6}{10}$

$tan\theta=\dfrac{perpendicular}{base}=\dfrac{PQ}{QR}=\dfrac{8}{6}$

Now, $25(sin^{2}\theta+2cos^{2}\theta-tan\theta)$

$=25\{(\dfrac{8}{10})^{2}+ 2(\dfrac{6}{10})^{2}-\dfrac{8}{6}\}$

$=25\{\dfrac{64}{100}+\dfrac{72}{100}-\dfrac{8}{6}\}$

$=25\{\dfrac{192+216-400}{300}\}$

$=25\times\dfrac{8}{300}$

$=\dfrac{2}{3}$

Final answer:

$25(sin^{2}\theta+2cos^{2}\theta-tan\theta)=\dfrac{2}{3}$