Question

If sin(0)=24/26 what are the other two trigonomic ratios

Answer 1

See below

Explanation:

We are here given that ,

`\longrightarrow \sin\theta =(24)/(26)`

And we would like to find out other trigonometric ratios ( since it is not mentioned about which two I will find out all 6 ) . As we know that ,

`\longrightarrow \sin\theta =(perpendicular)/(hypotenuse) =(p)/(h)`

So that ,

`\longrightarrow (p)/(h) =(24)/(26)`

Let us assume the given ratio to be 24x : 26x .

So we can say that ,

• hypotenuse = 26x
• perpendicular = 24x

Now let's find out the value of base , so that we can find out the other ratios .

By Pythagoras Theorem ,

`\longrightarrow (perpendicular) ^2+(base)^2=(hypotenuse) ^2`

Substitute the respective values ,

`\longrightarrow (26x)^2=(24x)^2+b^2\\`

`\longrightarrow 676x^2 = 576x^2+b^2\\`

`\longrightarrow b^2=100x^2\\`

`\longrightarrow b =√(100x^2)\\`

`\longrightarrow \boldsymbol{ b = 10x }`

Now we may find the other ratios as ,

`\longrightarrow \cos\theta = (base)/(hypotenuse)=(10x)/(26x)=\boxed{\boldsymbol{(5)/(13)}}\\`

`\longrightarrow \tan\theta =(perpendicular)/(base)=(24x)/(10x)=\boxed{\boldsymbol{(12)/(5)}}\\`

`\longrightarrow \cosec\theta =(hypotenuse)/(perpendicular)=(26x)/(24x)=\boxed{\boldsymbol{(13)/(12)}}\\`

`\longrightarrow \sec\theta =(hypotenuse)/(base)=(26x)/(10x)=\boxed{\boldsymbol{(13)/(5)}} \\`

`\longrightarrow \cot\theta =(base)/(perpendicular)=(10x)/(24x)=\boxed{\boldsymbol{(5)/(12)}}`

And we are done !

Answer 2

`\begin{aligned}{&\cos \theta=(10)/(26) \\\bf{&\tan \theta=(24)/(10)}\end{aligned}`

Explanation:

Since the angle is in the first quadrant, the values of all trigonometric functions are positive. Also. the formulas for the trigonometric functions we need to find from the problem are shown below:

`\begin{aligned}\cos \theta &=(x)/(r) \\\tan \theta &=(y)/(x)\end{aligned}`

Based on the formulas listed, the value of y and r are y = 24 and r = 26 respectively. This means that in order to determine the value of the other functions, we need In compute for the value of x using the Pythagorean Theorem:

`\begin{aligned}r^(2) &=x^(2)+y^(2) \\r^(2)-y^(2) &=x^(2)+y^(2)-y^(2) \quad\left[\text { \text{Subtracting \ both \ sides \ of \ the \ equation\ by} } y^(2)\right] \\r^(2)-y^(2) &=x^(2) \\\sqrt{r^(2)-y^(2)} &=\sqrt{x^(2)} \\\sqrt{r^(2)-y^(2)} &=x \\\sqrt{26^(2)-24^(2)} &=x \\10 &=x\end{aligned}`Then, substituting this to the formula to determine the answers yields:

`\begin{aligned}\cos \theta &=(x)/(r) \\&=(10)/(26) \\\tan \theta &=(y)/(x) \\&=(24)/(10)\end{aligned}`

Therefore, the answers are:

`\begin{aligned}&\cos \theta=(10)/(26) \\&\tan \theta=(24)/(10)\end{aligned}`