Question

In the below figure, ∆ABC is a right triangle, AC = CD, CDEF is a rectangle, and ∠BAC = 50°. Then, the measurement of ∠BDE is​

Answer 1

Answer:

angle bde is 40 degree

Step-by-step explanation:

if angle bac is 50 degree, then angle dac is also 50 degree. so.. by the formula that sum of the angles of a triangle is 180 degree, angle dca is 80 degree. we can understand that angle bca is right angled, so angle dce=90-80=10 degree. now, angle bac+bca+abc=180 degree. so, angle abc is 80 degree. angle edc is 90 degree{angle if a rectangle}. so adc+dce+bde=180. and adc=50 and edc=90. so bde is 40 degree.

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Answer 2

Given:

In right triangle ABC, $AC=CD$, CDEF is a rectangle and $\[\angle BAC = {50^ \circ }\]$.

To find: $\[\angle BDE\]$.

Solution:

Observe that, $AC=CD$.

$\[ \Rightarrow \]$ $\[\Delta CDA\]$ is an isosceles triangle.

$\[\angle BAC = {50^ \circ }\]$.

$\[\begin{array}{l} \Rightarrow \angle CDA = \angle BAC\\ \Rightarrow \angle CDA = {50^ \circ }\end{array}\]$

Understand that, CDEF is a rectangle and all the individual interior angle are always equal to 90 degrees.

$\[ \Rightarrow \angle EDC = {90^ \circ }\]$

Find the measure of $\[\angle BDE\]$.

$\[\angle BDE + \angle EDC + \angle CDA = {180^ \circ }\]$ {sum of straight line angle is 180 degrees}

$\[\begin{array}{l} \Rightarrow \angle BDE + {90^ \circ } + {50^ \circ } = {180^ \circ }\\ \Rightarrow \angle BDE + {140^ \circ } = {180^ \circ }\\ \Rightarrow \angle BDE = {180^ \circ } - {140^ \circ }\\ \Rightarrow \angle BDE = {40^ \circ }\end{array}\]$

Hence, the measure of $\[\angle BDE\]$ is $\[{40^ \circ }\]$.