
ABCD is a rhombus such that \[\angle \text{ACB}={{40}^{\circ }}\ then \\angle \text{ADB}\] is ?
A. \[{{40}^{\circ }}\]
B. \[{{50}^{\circ }}\]
C. \[{{45}^{\circ }}\]
D. None of these
Answer
552.6k+ views
Hint:To find the value of \[\angle \text{ADB}\], we first find the value of \[\angle \text{CBD}\] by using the formula of sum of all the angles equal to \[{{180}^{\circ }}\] and after finding the value of angle \[\angle \text{CBD}\], we use the alternative angle method where we find the value of \[\angle \text{ADB}\].
Complete step by step solution:
Let us draw the diagram and according to the diagram given, we will find the value of the angle \[\angle \text{CBD}\]. To find the \[\angle \text{CBD}\], we first check the triangle \[CBO\]which contains the angle \[\angle OCB\] and \[\angle COB\] i.e. \[{{40}^{\circ }}\] and \[{{90}^{\circ }}\] respectively.
Hence, the value of the angle \[\angle \text{CBD}\], we use the formula of the sum of the angle equal to \[{{180}^{\circ }}\] where we put the values in the formula as:
\[\Rightarrow \angle C+\angle O+\angle B={{180}^{\circ }}\]
\[\Rightarrow {{40}^{\circ }}+{{90}^{\circ }}+\angle B={{180}^{\circ }}\]
\[\Rightarrow \angle B={{180}^{\circ }}-{{130}^{\circ }}\]
\[\Rightarrow \angle B={{50}^{\circ }}\]
Now that we have the value of \[\angle B\], we can find the value of the \[\angle D\] by using the alternate angle method where the angle \[CBD\] and \[BDA\] are the same. Therefore, the value angle \[BDA\] is also given as \[{{50}^{\circ }}\].
Note: Rhombus like square have equal sides but the outer angles are not same hence, the outer angle of a rhombus can’t be deemed as \[{{90}^{\circ }}\] and similarly student may go wrong if they try to take the outer angles as \[{{90}^{\circ }}\] and start their questions with it.
Complete step by step solution:
Let us draw the diagram and according to the diagram given, we will find the value of the angle \[\angle \text{CBD}\]. To find the \[\angle \text{CBD}\], we first check the triangle \[CBO\]which contains the angle \[\angle OCB\] and \[\angle COB\] i.e. \[{{40}^{\circ }}\] and \[{{90}^{\circ }}\] respectively.
Hence, the value of the angle \[\angle \text{CBD}\], we use the formula of the sum of the angle equal to \[{{180}^{\circ }}\] where we put the values in the formula as:
\[\Rightarrow \angle C+\angle O+\angle B={{180}^{\circ }}\]
\[\Rightarrow {{40}^{\circ }}+{{90}^{\circ }}+\angle B={{180}^{\circ }}\]
\[\Rightarrow \angle B={{180}^{\circ }}-{{130}^{\circ }}\]
\[\Rightarrow \angle B={{50}^{\circ }}\]
Now that we have the value of \[\angle B\], we can find the value of the \[\angle D\] by using the alternate angle method where the angle \[CBD\] and \[BDA\] are the same. Therefore, the value angle \[BDA\] is also given as \[{{50}^{\circ }}\].
Note: Rhombus like square have equal sides but the outer angles are not same hence, the outer angle of a rhombus can’t be deemed as \[{{90}^{\circ }}\] and similarly student may go wrong if they try to take the outer angles as \[{{90}^{\circ }}\] and start their questions with it.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

