Finding theta

Here's a solution which uses Componendo-et-dividendo , which was discussed last week.

We have $\frac{ \cos \theta } { \sin \theta } = \frac{7}{2}$ . Applying Componendo et Dividendo statement 3 with $k = 3$ , we get that

$\frac { \cos \theta + 3 \sin \theta } { \cos \theta - 3 \sin \theta} = \frac{ 7 + 3 \times 2 } { 7 - 3 \times 2 } = 13.$

Tangθ=senθ/cosθ..Com isso temos que a tangente vale 7/2..Assim: Senθ/cosθ = 7/2

Isolando cosθ,temos: cosθ=7senθ/2 dessa forma substituímos na expressão em questão: cosθ+3sinθ/cosθ−3sinθ

7senθ/2 + 3 senθ dividido por 7senθ-3senθ...Assim temos: 13senθ/1senθ = 13

$1.$ Multiply both the numerator and denominator by the conjugate of the denominator, $cosθ + 3sinθ$ .

You should now have this: $\frac{cos^2θ + 6cosθsinθ + 9sin^2θ}{cosθ^2 - 9sin^2θ}$ .

$2.$ Divide both the numerator and denominator by $cos^2θ$ .

You should now have this: $\frac {1 + 6tanθ + 9tan^2θ}{1 - 9tan^2θ}$ .

$3.$ Fill in the value of $tanθ$ , $\frac{2}{7}$ , and simplify.

Your final answer should be $13$ .

(cos theta + 3 sin theta)/ (cos theta - 3 sin theta )
= cos theta (1+3 sin theta /cos theta ) / cos theta (1 - 3 sin theta /cos theta)
= (1+3 tan theta/ 1-3 tan theta)
=(1+3(2/7))/(1-3(2/7))=13/1=13

tanθ = sinθ / cosθ thus sinθ = 2; cosθ = 7

Using this we get 7+6/7-6 = 13/1

sin A/cos A =tan A therefore sin A/cos A=2/7 gives that 3sin A/cos A=6/7
so cos A/3sin A=7/6 and (cosA +3sinA)/(cos A-3sin A)=(6+7)/(7-6)=13 (using compoundo dividendo)

tanθ=sinθ/cosθ ...Here tanθ=sinθ/cosθ=2/7.........so 3*sinθ/cos=6/7 ................and we can also write cosθ/sinθ=7/6..........by adding and subtracting , cosθ+sinθ/cosθ-sinθ=7+6/7-6=13

as tanθ=2/7 then cosθ = (7/√53) and sinθ = (2/√53) then sup in given format we get the answer = 13

tan theta=2/7 (given)(1) but we know that..., tan theta=sin theta/cos theta (2) therefore, from (1) and (2) sin theta = 2 & cos theta = 7 now substitute the value of sin theta & cos theta in... cos theta+3 sin theta/cos theta - 3 sin theta = 7+3 2/ 7-3 2 =7+6/ 7-6 =13/1 =13 therefore the correct answer is 13 :)

We have big problem on the hypotenuse, $\sqrt{53}$ . So, let's multiply by, say, $\sqrt{7}$ to make it a perfect square?

So, we have $a = 2 , b = 7, c = \sqrt{53}$ ;

we use this: { $a= 2\sqrt{7} , b = 7\sqrt{7} , c = 19.$ }

Equating sin and cos, we have:

$\frac{\frac{7\sqrt{7}}{19} + \frac{6\sqrt{7}}{19}}{\frac{7\sqrt{7}}{19} - \frac{6\sqrt{6}}{19}}$ .

= $\frac{\frac{13\sqrt{7}}{19}}{\frac{\sqrt{7}}{19}}$

= $\frac{247\sqrt{7}}{19\sqrt{7}}$

Cancelling gives...

= $\frac{247}{19}$

= $13$