Question

Find the length of the altitude of an isosceles triangle with a 41° base angle and a base 38 cm long. Round your answer to the nearest tenth.

Answer 1

Let's draw a picture of our problem:

From our figure we have the following right triangle:

so, in order to get the altitute (height) of our isosceles triangle, denoted by h, we can use a trigonometric function .

If we apply the tangent function, we have

`\tan (41)=(h)/(19)`

then, by moving 19 to the left hand side, we get

`19\cdot\tan (41)=h`

which gives

`\begin{gathered} h=19\cdot\tan (41)\text{ cm} \\ h=19\cdot(0.8693)\text{ cm} \\ h=16.52\text{ cm} \end{gathered}`

Therefore, by rounding down to the nearest tenth, the altitude is equal to 16.5 centimeters