  Question

$$3$$ circle of radii $$1,2$$ and $$3$$ and centres at $$A,B,C$$ respectively, touch each other. Another circle whose centre is $$P$$ touches all these $$3$$ circles externally and has radius $$r$$. Also $$\angle PAB=\theta$$ and $$\angle PAC=\alpha$$

A
cosθ=3r3(1+r)  B
cosα=2r2(1+r)  C
r=623  D
r=623  Solution

The correct options are A $$\cos \theta=\cfrac{3-r}{3(1+r)}$$ B $$\cos \alpha=\cfrac{2-r}{2(1+r)}$$ C $$r=\cfrac{6}{23}$$$$\triangle ABC$$ is right angleApplying cosine rule in $$\triangle PAB$$$$\cos{\theta}=\cfrac{{3}^{2}+{(1+r)}^{2}-{(2+r)}^{2}}{2.3(1+r)}$$$$=\cfrac{3-r}{3(1+r)}$$Again applying cosine rule in $$\triangle PAC$$$$\cos {\alpha}=\cfrac { { \left( 1+r \right) }^{ 2 }+{ 4 }^{ 2 }-{ \left( 3+r \right) }^{ 2 } }{ 2.4(1+r) } =\cfrac { 2-r }{ 2(1+r) }$$$$\because \quad \alpha +\theta ={ 90 }^{ o }\quad \quad \alpha ={ 90 }^{ o }-\theta \quad \Rightarrow \quad \cos { \alpha } =\sin { \theta } \quad$$$${ \left( \cfrac { 3-r }{ 3(r+1) } \right) }^{ 2 }+{ \left( \cfrac { 2-r }{ 2(r+1) } \right) }^{ 2 }=1$$ Maths

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