Question

Drag the tiles to the boxes to form correct pairs.

Answer 1

Answer:

1) cos (140°-(50°+30°)) = 1/2

2)  $sin255^{\circ}=\frac{-\sqrt{2}-\sqrt{6} }{\sqrt{4} }$

3)  $cos150^{\circ}=-\frac{ \sqrt{3} }{2}$

4) ,  $tan 240^{\circ}=\sqrt{3}$

Step-by-step explanation:

1) cos (140°-(50°+30°))

Solving:

cos (140°-(50°+30°))

=cos(140°-80°)

=cos(60°)

=1/2

So, cos (140°-(50°+30°)) = 1/2

2) sin 255°

We can write sin 255° as sin(180°+75°)

Solving sing formula sin(a+b)=sin(a)cos(y)+cos(x)sin(y)

$sin(180^{\circ}+75^{\circ})\\=sin(180^{\circ})cos(75^{\circ})+cos(180^{\circ})sin(75^{\circ})\\=0(cos(75^{\circ}))+(-1)(\frac{\sqrt{2}+\sqrt{6} }{\sqrt{4} } )\\=0+(-1)(\frac{\sqrt{2}+\sqrt{6} }{\sqrt{4} } )\\\=\frac{-\sqrt{2}-\sqrt{6} }{\sqrt{4} }$

So, $sin255^{\circ}=\frac{-\sqrt{2}-\sqrt{6} }{\sqrt{4} }$

3) cos 150°

We can solve this using the identity: cos(x)=sin(90°-x)

Solving:

cos 150°= sin(90°-150°)

=sin(-60°)

We know sin(-x)=-sin(x)

=-sin(60°)

=$-\frac{ \sqrt{3} }{2}$

So, we get $cos150^{\circ}=-\frac{ \sqrt{3} }{2}$

4) tan 240°

We know tan(180°+x)=tan(x)

tan 240° can be written as tan(180°+60°)

Using the above identity:

tan(180°+60°)=tan(60°) = $\sqrt{3}$

So,  $tan 240^{\circ}=\sqrt{3}$

Answer 2

Answer:

here ya go

Step-by-step explanation: