We can directly use the formula r=Δ/s where r is the radius of in-circle, Δ is the area of the triangle and s is the semi-perimeter. Since its a right-angled triangle, we can use Pythagoras Theorem to find the third side.AC^2 = AB^2 + BC^2AC = 17 cm semi−perimeter = s = (AB+BC+CA)/2 = 20cm Δ = (base∗height)/2 = 60 sq.cm. … Continue reading In a right triangle ABC, right angled at B, BC = 15 cm, and AB = 8 cm. A circle is inscribed in triangle ABC. What is the radius of the circle?
What is the y intercept for the equation y=-10x+14?
Intercept form of a line is given by x/a + y/b = 1 y=−10x+14y ..(transposing −10x to LHS)10x+y=14 ..(dividing both sides by 14)(x/1.4)+(y/14)=1 This is the intercept form of a line.So, the y-intercept point of the line is (0,14); and the length of the y-intercept is 14 units.
How do you prove that SinA+Sin(120+A) +Sin(240+A) =0?
Identities to be used: sin(90+A)=cosAsin(270−A)=−cosAcos(A+B)=cosAcosB−sinAsinBcos(A−B)=cosAcosB+sinAsinB ATQ sinA+sin(120+A)+sin(240+A)=sinA+sin(90+(30+A))+sin(270−(30−A))=sinA+cos(30+A)−cos(30−A)=sinA+cos30.cosA−sin30.sinA−cos30.cosA−sin30.sinA=sinA−2sin30.sinA=sinA−sinA=0 Hence, Proved.
M CUBE: MatheMatics by Maheshwari 