Question
Prove that any three points on a circle cannot be collinear.
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Find the slopes of the lines passing through the given points.
(1) A (2, 3), B (4, 7)
(2) P (-3, 1), Q (5, -2)
(3) C (5, -2), D (7, 3)
(4) L (-2, -3), M (-6, -8)
(5) E(-4, -2), F (6, 3)
(6) T (0, -3), S (0, 4)
(1) A (2, 3), B (4, 7)
(2) P (-3, 1), Q (5, -2)
(3) C (5, -2), D (7, 3)
(4) L (-2, -3), M (-6, -8)
(5) E(-4, -2), F (6, 3)
(6) T (0, -3), S (0, 4)
In a $\triangle\text{ABC},\angle\text{B}=90^\circ,\text{AB}=24\text{cm}$ and BC = 7cm.
Find:
→Find:
- $\sin\text{A}$
- $\cos\text{A}$
- $\sin\text{C}$
- $\cos\text{C}.$
A person deposits ₹ $x$ in savings bank account at the rate of $5 \%$ per annum and ₹ y in fixed deposit at $10 \%$ per annum. At the end of one year, he gets ₹ 400 as total interest. If he deposits ₹ y in savings bank account and ₹ $x$ in fixed deposit, he would get ₹ 350 as total interest. Find the total amount he deposited.
→Solve the following quadratic equations by factorization:
$3\Big(\frac{7\text{x}+1}{5\text{x}-3}\Big)-4\Big(\frac{5\text{x}-3}{7\text{x}+1}\Big)=11,$ $\text{x}\neq\frac{3}{5},-\frac{1}{7}$
→$3\Big(\frac{7\text{x}+1}{5\text{x}-3}\Big)-4\Big(\frac{5\text{x}-3}{7\text{x}+1}\Big)=11,$ $\text{x}\neq\frac{3}{5},-\frac{1}{7}$
$\frac{14}{x+y}+\frac{3}{x-y}=5 ; \frac{21}{x+y}-\frac{2}{x-y}=1$
→In the following figure, shows a sector of a circle, centre O, containing an angle $\theta^\circ.$ Prove that: Perimeter of the shaded region is $\Big(\tan\theta+\sec\theta+\frac{\pi\theta}{180}-1\Big)$
→Find the distance between the points $\Big(\frac{-8}{5},2\Big)$ and $\Big(\frac{2}{5},2\Big).$
→Solve: $x+y=8 ; x-y=2$
→Find the quotient and the remainder when:
$f(x) = x^4 - 5x + 6$ is divided by $g(x) = 2 - x^2$
→$f(x) = x^4 - 5x + 6$ is divided by $g(x) = 2 - x^2$
The angle of elevation of the top of building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60°. If the tower is 60m high, Find the height of the building.