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

1 Answer
Sep 17, 2016

Answer:

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 .}

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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+CD^2# {7.5 is half of 15 i.e length DB}

#CD^2=10^2-7.5^2#

#CD=6.6 #

Now apply trigonometric ratio(s)

#tan B = (CD)/(BD)#

#tanB=6.6/7.5#

#B=tan^-1(6.6/7.5)#

#B=41.34#

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

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

Now its simple,

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

#/_C=180^o-41.34^o-41.34^o#

#/_C=97.32^o#

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