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$M$ is the midpoint of $\overline{AB}$ and $N$ is the midpoint of $\overline{AC}$, and $T$ is the intersection of $\overline{BN}$ and $\overline{CM}$, as shown. If $\overline{BN}\perp\overline{AC}$, $BN = 12$, and $AC = 14$, then find $CT$.

1 Answer

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Where the above condition are given, it is to be noted that the value of CT is 8.07 units.

To determine CT, first we solve for the internal angles of ΔBCN.
From the given stats, we know that:

Height of the Triangle BN is 12 Units and the base is
AC/2

= 14/2

= 7 units.

One of the acute angles can be derived as Tan (θ) = Opposite / Adjacent

In this case, θ is the acute angle, the opposite side is the height (12 unit) and the adjacent side is the base (7 unit)

thus, Tan (θ) = 12/7

θ = tan⁻¹ (12/7)

= 59.743562836470735161391454608421

≈ 59.74°

Since, MC bisects AB, that means it also bisects ∠C. Since ∠C is 59.74°, therefore, ∠TCN = 59.74°/2

= 29.871781418235367580695727304211

≈ 29.87

Since Δ TCN is a right triangle, and we have two angles and a base, we can figure out the base which is the TN.

First lets find the hypotenuse

Hypotenuse (CT) = 7/cos (29.87)
CT ≈ 8.0723500503056508204904073699311

CT ≈ 8.07

Thus, the base TN = √TC² - NC²
= √(8.07² - 7²)

= 4.0203

Thus, TN = 4.0203 or approximately 4 units.

Full Question:

Although part of your question is missing, you might be referring to this full question:

M is the mid-point of segment AB and N is the midpoint of AC, and T is the intersection of BN and CM, in triangle ABC. If BN ⊥ AC, BN = 12, and AC = 14, then find CT.

$M$ is the midpoint of $\overline{AB}$ and $N$ is the midpoint of $\overline{AC}$, and-example-1
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