Question

find the length of AD. Given : <ABC=60°, <DBC=45° and BC =40 cm.​

Answer 1

Step-by-step explanation:

i need figure ok send me

Answer 2

Step-by-step explanation:

Answer:

BC = 40 cm

∠ DBC = 45°

∠ ABC = 60°

• In Triangle DBC :

$:\implies\sf \tan(\theta)=\dfrac{Perpendicular}{Base}\\\\\\:\implies\sf \tan(45)=\dfrac{DC}{BC}\\\\\\:\implies\sf 1 =\dfrac{DC}{40}\\\\\\:\implies\sf 40 = DC\qquad...eq\:(I)$

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• In Triangle ABC :

$:\implies\sf \tan(\theta)=\dfrac{Perpendicular}{Base}\\\\\\:\implies\sf \tan(60)=\dfrac{AC}{BC}\\\\\\:\implies\sf \sqrt{3} =\dfrac{AC}{40}\\\\\\:\implies\sf 40 \sqrt{3} = AC\\\\\\:\implies\sf40 \sqrt{3} = AD+DC\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Putting value of DC from eq. ( I )}}\\\\:\implies\sf 40 \sqrt{3} =AD+40\\\\\\:\implies\sf 40 \sqrt{3} - 40 = AD\\\\\\:\implies\sf 40( \sqrt{3} - 1) = AD\\\\\\:\implies\underline{\boxed{\sf AD = 40( \sqrt{3} - 1)\:cm}}$