How do you find the angles of a triangle given side a = 10, side b = 10, base = 15?

Sep 17, 2016

see below.

Explanation:

Given is that 2 sides are of same length i.e. 10cm. This tells us that this is an isosceles triangle.

{ Isosceles triangles have 2 same angles and one different angle .} so using above picture let AC and BC be 10cm. And AB be 15 cm.

Now find CD using Pythagoras theorem, {then later the angle will be found}

${10}^{2} = {7.5}^{2} + C {D}^{2}$ {7.5 is half of 15 i.e length DB}

$C {D}^{2} = {10}^{2} - {7.5}^{2}$

$C D = 6.6$

Now apply trigonometric ratio(s)

$\tan B = \frac{C D}{B D}$

$\tan B = \frac{6.6}{7.5}$

$B = {\tan}^{-} 1 \left(\frac{6.6}{7.5}\right)$

$B = 41.34$

Hence $\textcolor{red}{\text{angle B is 41.34 degrees}}$ .

Therefore $\textcolor{b l u e}{\text{angle A will also be 41.34 degrees.}}$

Now its simple,

180^o=41.34^o + 41.34^o +/_C

$\angle C = {180}^{o} - {41.34}^{o} - {41.34}^{o}$

$\angle C = {97.32}^{o}$

hence $\textcolor{g r e e n}{\text{angle C is 97.32 degrees}}$