Solve right triangles

To solve a right triangle means to calculate the sides and angles from the information given.

If a triangle has a right-angle then in order to determine all the sides and angles it is sufficient to know either two sides or one side and one angle other than the right-angle.

For the solution, the mathematical tools needed may include Pythagoras's theorem, the trigonometric functions (sine, cosine and tangent) and the inverse trigonometric functions.

Worked Examples

Question 1

A right triangle is given in which the hypotenuse has length 1553 and another side length is 495 units. Find the third side and the two non right-angles.

By Pythagoras, the missing side is $\sqrt{1553^2-495^2}=1472$√15532−4952=1472.

According to the given information, $\cos A=\frac{495}{1553}\approx0.31874$cosA=4951553​≈0.31874. To find the angle whose cosine is $0.31874$0.31874 we use the inverse cosine function:

$A=\cos^{-1}0.31874\approx71.4^\circ$A=cos−10.31874≈71.4°

We could use the inverse sine function to find angle $B$B but it is simpler to use the fact that $A+B=90^\circ$A+B=90°. Thus,

$B=90^\circ-71.4^\circ=18.6^\circ$B=90°−71.4°=18.6°.

Question 2

In the given triangle, find angle $A$A and the sides $a$a and $c$c.

Since angle $A$A is the complement of the $18.6^\circ$18.6° angle, we see that $A=90^\circ-18.6^\circ=71.4^\circ$A=90°−18.6°=71.4°.

Focussing on the given angle, we can use the fact that $\tan18.6^\circ=\frac{495}{a}$tan18.6°=495a​. Thus,

$a=\frac{495}{\tan18.6^\circ}\approx1471$a=495tan18.6°​≈1471.

We could now use Pythagoras's theorem to find $c$c, or we could use the sine function: $\sin18.6^\circ=\frac{495}{c}$sin18.6°=495c​. So,

$c=\frac{495}{\sin18.6^\circ}\approx1552$c=495sin18.6°​≈1552.

The differences between these results and those in the first example can be explained as round-off errors.

More Worked Examples

QUESTION 3

QUESTION 4

QUESTION 5