Question
In figure, O is the centre of the circle. Find $\angle\text{BAC}.$
✓
Answer
We have $\angle\text{AOB}=80^\circ$ And $\angle\text{AOC}=110^\circ$ Therefore, $\angle\text{AOB}+\angle\text{AOC}+\angle\text{BOC}=360^\circ$ [Complete angle]$\Rightarrow80^\circ+100^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow\angle\text{BOC}=360^\circ-80^\circ-110^\circ$
$\Rightarrow\angle\text{BOC}=170^\circ$
By degree measure theorem$\angle\text{BOC}=2\angle\text{BAC}$
$\Rightarrow170^\circ=2\angle\text{BAC}$
$\Rightarrow\angle\text{BAC}=\frac{170^\circ}{2}=85^\circ$
$\Rightarrow\angle\text{BOC}=360^\circ-80^\circ-110^\circ$
$\Rightarrow\angle\text{BOC}=170^\circ$
By degree measure theorem$\angle\text{BOC}=2\angle\text{BAC}$
$\Rightarrow170^\circ=2\angle\text{BAC}$
$\Rightarrow\angle\text{BAC}=\frac{170^\circ}{2}=85^\circ$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If the median of $33,28,20,25,34$, $x$ is 29 , find the maximum possible value of $x$.
→A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients?
The vertical angle of an isosceles triangle is 100º. Find its base angles.
The following data gives the production of food grains (In thousand tonnes) for some years:
Represent the above data with the help of a bar graph.
|
Year
|
1995
|
1996
|
1997
|
1998
|
19999
|
2000
|
|
Production (in thousand tonnes)
|
120
|
150
|
140
|
180
|
170
|
190
|
It is required to make a closed cylindrical tank of height 1m and the base diameter of 140cm from a metal sheet. How many square meters of the sheet are required for the same?
In the adjoining figure, PQRS is a trapezium in which PQ || SR and M is the midpoint of PS. A line segment MN || PQ meets QR at N. Show that N is the midpoint of QR.
Draw the graph of the following linear equations in two variables:
x + y = 4
x + y = 4
The base BC of $\triangle\text{ABC}$ is divided at D such that $\text{BD}=\frac{1}{2}\text{DC}.$ Prove that $\text{ar}(\triangle\text{ABD})=\frac{1}{3}\times\text{ar}(\triangle\text{ABC}).$
→Locate $\sqrt{8}$ on the number line.
→The daily minimum temperatures in degrees Celsius recorded in a certain arctic region are as follows:
-12.5, -10.8, -18.6, -8.4, -10.8, -4.2, -4.8, -6.7, -13.2, -11.8, -2.3, 1.2, 2.6, 0, -2.4, 0, 3.2, 2.7, 3.4, 0, -2.4, -2.4, 0, 3.2, 2.7, 3.4, 0, -2.4, -5.8, -8.9, -14.6, -12.3, -11.5, -7.8, -2.9.
Represent them as frequency distribution table taking -19.9 to -15 as the first class interval.
-12.5, -10.8, -18.6, -8.4, -10.8, -4.2, -4.8, -6.7, -13.2, -11.8, -2.3, 1.2, 2.6, 0, -2.4, 0, 3.2, 2.7, 3.4, 0, -2.4, -2.4, 0, 3.2, 2.7, 3.4, 0, -2.4, -5.8, -8.9, -14.6, -12.3, -11.5, -7.8, -2.9.
Represent them as frequency distribution table taking -19.9 to -15 as the first class interval.