Finding angles with inverse trigonometry

Three quick questions from earlier lessons.

In a right triangle, $\tan A = 1$ . What is $A$ ?

Inverse trigonometric functions turn ratios back into angles.

If

\sin A = r

then

A = \sin^{-1} r

; similarly

\cos^{-1}

and

\tan^{-1}

. Here

^{-1}

means inverse function, not reciprocal. Use degree mode.

Legs 5 (opposite) and 8 (adjacent).

\tan A = \tfrac{5}{8} = 0.625

TOA.

A = \tan^{-1}(0.625) \approx 32.0^\circ

Inverse tangent.

Your turn: $\sin A = \tfrac{1}{2}$ . Find $A$ .

30^\circ

A special angle.

A right triangle has legs 10 (opposite angle $A$ ) and 12 (adjacent). Find angle $A$ in degrees (1 decimal place).
A right triangle has legs 8 (opposite angle $A$ ) and 9 (adjacent). Find angle $A$ in degrees (1 decimal place).

What does $\sin^{-1}(0.7)$ mean?
1. The angle whose sine is 0.7
2. $\dfrac{1}{\sin 0.7}$
3. $-\sin(0.7)$

In a right triangle the side opposite angle $A$ is 5 and the hypotenuse is 15. Find $A$ in degrees (1 decimal place).
In a right triangle the side opposite angle $A$ is 6 and the hypotenuse is 10. Find $A$ in degrees (1 decimal place).

Three questions to check you have got it.

A right triangle has legs 10 (opposite angle $A$ ) and 12 (adjacent). Find angle $A$ in degrees (1 decimal place).
A right triangle has legs 8 (opposite angle $A$ ) and 9 (adjacent). Find angle $A$ in degrees (1 decimal place).
A right triangle has legs 12 (opposite angle $A$ ) and 9 (adjacent). Find angle $A$ in degrees (1 decimal place).

Finding angles

$A = \sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ of the side ratio.

Watch out: inverse, not reciprocal; degree mode.

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