Block 1

Gradients and Angles

What is to be learned?

• A formula connecting gradient and angle

Gradient & Angle

tan θ

= opposite/adjacent

= (y2-y1) (x2-x1)

= mAB

•

B(x2,y2)

A(x1,y1)

Xθ

θ

Opposite

Adjacent gradient of line = tangent of angle

m = Tanθ

mGH = tan34° = 0.67

NB: line looks like

G

H

The line GH makes an angle of 34° with the X-axis. Find its gradient.

X34°

m = Tan θ

mCD = tan110° = -2.75

NB: line looks like

D

C

The line CD makes an angle of 110° with the X-axis. Find its gradient.

X110°

m = Tan θ

Gradients and Angles

If a line makes an angle of θ with the positive direction of the X-axis

m = Tan θ

A

B

Xθ

mGH = tan116.6° = -2.

NB: line looks like

G

H

The line GH makes an angle of 116.6° with the X-axis. Find its gradient.

X116.6°

tan θ = (y2-y1) (x2-x1)

= (10 + 2) ( 17 - 5) = 12/12

= 1

θ = tan-1(1) = 45°.

NB: line looks like

P

Q

P is (5, -2) and Q is (17,10).

What angle does PQ make with the X-axis?

X

tan = (y2-y1) (x2-x1)

= (7 + 9) ( 7 - 5) = 16/2

= 8

θ = tan-1(8) = 83°.

NB: line looks like

P

Q

P is (5, -9) and Q is (7,7).

What angle does PQ make with the X-axis?

X

a0180 – a

180 + a 360 - a

iii

iii iv

CT

ASTanx = -0.4

Tan-1(0.4) = 220

+ve or –ve?Tan -ve in ii and iv

x = 180 - 22 or 360-22 = 1580

Always put a positive number here

= 3380

Making sense of this!(1580 or 3380)

M = -0.4

xθ

θ = 1580

3380?