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

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Answer 1 · 2016-09-17T08:21:15

see below.

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 .}

enter image source here

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}

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

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

$C D = 6.6$ $CD=6.6$

Now apply trigonometric ratio(s)

$tan B = \frac{C D}{B D}$ $tan B = (CD)/(BD)$

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

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

$B = 41.34$ $B=41.34$

Hence $angle B is 41.34 degrees$ $color(red)"angle B is 41.34 degrees"$ .

Therefore $angle A will also be 41.34 degrees.$ $color(blue)("angle A will also be 41.34 degrees.")$

Now its simple,

$180^{o} = {41.34}^{o} + {41.34}^{o} + ∠ C$ $180^o=41.34^o + 41.34^o +/_C$

$∠ C = 180^{o} - {41.34}^{o} - {41.34}^{o}$ $/_C=180^o-41.34^o-41.34^o$

$∠ C = {97.32}^{o}$ $/_C=97.32^o$

hence $angle C is 97.32 degrees$ $color(green)"angle C is 97.32 degrees"$