Tangent Segment Problem

Geometry Level 4

A right triangle overlapped a circle wherein its center is the same as its vertex of right angle. Line A B AB is the hypotenuse and is tangent to the circle and D D is the point of tangency. If ∣ D B ∣ = 15 |DB| = 15 and B B is 3 5 ∘ 5 0 ′ 35^\circ 50' , what is ∣ A D ∣ |AD| ?


The answer is 7.8216.

Jade Mijares by Jade Mijares , 30, Philippines

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1 solution

I n r t . Δ s D B C , H y . = B C , a n d A B C , H y . = A B , A B = B C C o s ( 3 5 o 5 0 ′ ) = B D C o s ( 3 5 o 5 0 ′ ) ∗ 1 C o s ( 3 5 o 5 0 ′ ) = 15 C o s 2 ( 3 5 o 5 0 ′ ) . S o D A = A B − D B = 15 ∗ ( 1 C o s 2 ( 3 5 o 5 0 ′ ) − 1 ) . ∴ D A = 15 ∗ t a n 2 ( 3 5 o 5 0 ′ ) = 7.8216. N o t e : − C D i s a l s o t h e a l t i t u d e o f A B C o n H y . A B . In~rt.~\Delta~ s~~ ~DBC,~Hy. =~BC,~~~and~ABC,~Hy.~= ~AB,\\ AB=\dfrac{BC}{Cos(35^o~50')}=\dfrac{BD}{Cos(35^o~50')}*\dfrac 1 {Cos(35^o~50')}\\ =\dfrac{15}{Cos^2(35^o~50')}.\\ So~DA=AB-DB=15*\left (\dfrac 1 {Cos^2(35^o~50')}-1 \right ).\\ \therefore~DA=15* tan^2(35^o~50')=\large \color{#D61F06}{7.8216} .\\ Note:-~~CD~is~also~the~altitude~of~ABC~on~Hy.~AB.

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