Mathematics: In ΔBCD, the measure of ∠D=90°, the measure of ∠C=42°, and CD = 7.5 feet. Find the length of DB to the nearest tenth of a foot.

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In ΔBCD, the measure of ∠D=90°, the measure of ∠C=42°,

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FrankyV3220

27.01.2022

We have been given that in ΔBCD, the measure of ∠D=90°, the measure of ∠C=42°, and CD = 7.5 feet. We are asked to find the length of DB to nearest tenth of foot.

First of all, we will draw a right triangle using our given information.

We can see from the attachment that DB is opposite side to angle C and CD is adjacent side to angle.

We know that tangent relates opposite side of right triangle to adjacent side of right triangle.

$\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}$

$\text{tan}(\angle C)=\frac{DB}{CD}$

$\text{tan}(42^{\circ})=\frac{DB}{7.5}$

$7.5\cdot\text{tan}(42^{\circ})=\frac{DB}{7.5}\cdot 7.5$

$7.5\cdot\text{tan}(42^{\circ})=DB$

$7.5\cdot0.900404044298=DB$

$DB=7.5\cdot0.900404044298$

$DB=6.753030332235\approx 6.8$

Therefore, the length of DB is approximately 6.8 feet.

In ΔBCD, the measure of ∠D=90°, the measure of ∠C=42°, and CD = 7.5 feet. Find the length of DB to t

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In ΔBCD, the measure of ∠D=90°, the measure of ∠C=42°, and CD = 7.5 feet. Find the length of DB to the nearest tenth of a foot.

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