Please solve q 80 ?

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Please solve q 80 ?

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Answer 1 · 2018-05-17T15:11:51

option (4) is acceptable.

Explanation:

enter image source here

Given that, $A B = A C = B D$ $AB=AC=BD$ and $A C ⊥ B D$ $AC_|_BD$ .

$\to A B = A C$ $rarrAB=AC$

$\to ∠ B = ∠ C$ $rarr/_B=/_C$

$\to 90 - a + 90 - d = d$ $rarr90-a+90-d=d$

$\to a = 180 - 2 d$ $rarra=180-2d$ .....[1]

Also, $\to A B = B D$ $rarrAB=BD$

$\to ∠ A = ∠ D$ $rarr/_A=/_D$

$\to a + b = 90 - b$ $rarra+b=90-b$

$\to a = 90 - 2 b$ $rarra=90-2b$ ....[2]

From [1] and [2], we have,

$\to 180 - 2 d = 90 - 2 b$ $rarr180-2d=90-2b$

$\to d - b = 45$ $rarrd-b=45$ ....[3]

Now, $∠ C + ∠ D$ $/_C+/_D$

$= ∠ B C A + ∠ B D A = 90 - b + d = 90 + 45 = 135$ $=/_BCA+/_BDA=90-b+d=90+45=135$

Answer 2 · 2018-05-19T02:28:43

$∠ C + ∠ D = 135^{\circ}$ $angleC+angleD=135^@$

Explanation:

enter image source here
Let $∠ A B D = x$ $angleABD=x$ ,
given $A B = B D, \Rightarrow ∠ B D A = \frac{180 - x}{2}$ $AB=BD, => angleBDA=(180-x)/2$
let $∠ C B D = y$ $angleCBD=y$ ,
given $A B = A C, \Rightarrow ∠ A C B = ∠ A B C = x + y$ $AB=AC, => angleACB=angleABC=x+y$ ,
In $DeltaCBE, => color(red)(x+2y=90^@)$
$angleC+angleD=angleACB+angleBDA=x+y+(180-x)/2$
$=x+y+90-x/2$
$=x/2+y+90$
$=(x+2y)/2+90$
$=90/2+90=45+90=135^@$

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Please solve q 80 ?

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