Maths Trig question

Not quite sure how to work this question out, does anyone have any ideas?
θ is obtuse and sin θ= 7/25. Write down the values of cos θ and tan θ


Thanks))

Not quite sure how to work this question out, does anyone have any ideas?
θ is obtuse and sin θ= 7/25. Write down the values of cos θ and tan θ


Thanks))

Well what do you already know about an obtuse angle? It's between 270 and 360 degrees.

Try working out 7/25, and then pressing the /shift,sin/ button on your calculator to find θ. Once you've done that, you can find cos θ and tan θ.

Hope that helps

Not quite sure how to work this question out, does anyone have any ideas?
θ is obtuse and sin θ= 7/25. Write down the values of cos θ and tan θ


Thanks))

...

It's my go-to method for these types of questions, but the one thing you're forgetting is the fact that one should consider that since $\theta$ is obtuse, we have $90< \theta < 180$ and in this region we have $\sin \theta > 0$, $\cos \theta < 0$ and so $\tan \theta < 0$ which is important for getting the signs right, as the diagrams do not take this into consideration due to lengths being positive.

Well what do you already know about an obtuse angle? It's between 270 and 360 degrees.

Try working out 7/25, and then pressing the /shift,sin/ button on your calculator to find θ. Once you've done that, you can find cos θ and tan θ.

Hope that helps

This is a classic C3 question, that i come across often. It is not as hard as it seems.

1.Draw out a triangle
2.Pick out θ
3. Since Sine is Opposite over Adjacent. Lable the opposite of θ and the Adjacent of θ with your values 7 and 25.
4.Use Pythagoras to work out the side that doesn't have a number on it.
5.Since cosine is Adjacent over hypotenuse. you can work out the value of cos θ
6. to work out tanθ, you already know it is sinθ/cosθ
there fore you can divide your value for sinθ with your value for cosθ

thats how you do it

This method is wrong, θ is obtuse so you can't apply Pythagoras to the triangle as it won't be a right angle triangle. (Not to mention Sine is opposite over hypotenuse not opposite over adjacent).