Question

Find the altitude of an isosceles triangle with a vertex angle (that is, the non-base angle) of 70° and a base of 246. Draw a diagram and find the length of the altitude

Answer 1

Altitude = 175.6 units

Answer 2

Answer:

AD=175.66 units

Step-by-step explanation:

It is given that ABC is an isosceles triangle in which base is 246 and the non base angle is 70°.

Thus, applying the angle sum property in ΔABC, we get

∠A+∠B+∠C=180°

⇒70°+2∠B=180°

⇒∠B=55°

Thus, ∠B=∠C=55°.

Now, Draw AD perpendicular to BC such that BD=DC=123, then applying trigonometry in ΔADC, we have

$\frac{AD}{DC}=tan55^{\circ}$

⇒$\frac{AD}{123}=1.428$

⇒$AD=123(1.428)$

⇒$AD=175.66 units$

Thus, the length of the altitude is 175.66.