Question

In △CDE , CD=10 , DE=7 , and m∠E=62∘ . What is m∠D to the nearest tenth of a degree?

Answer 1

We have been given that in △CDE , CD=10 , DE=7 , and m∠E=62 degrees. We are asked to find the measure of angle D.

We will use law of sines to solve our given problem.

$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$, where, a, b and c are opposite sides to angles A, B and C respectively.

First of all we will find measure of angle C as:

$\frac{7}{\sin(C)}=\frac{10}{\sin(62^{\circ})}$

$\sin(C)=\frac{7\cdot \sin(62^{\circ})}{10}$

$\sin(C)=\frac{7\cdot 0.882947592859}{10}$

$\sin(C)=0.6180633150013$

Now we will use arcsin to solve for C as:

$C=\sin^{-1}(0.6180633150013)$

$C=38.174844995363^{\circ}$

$C\approx 38.2^{\circ}$

Now we will use angle sum property to find measure of angle D.

$\angle D+\angle C+\angle E=180^{\circ}$

$\angle D+38.2^{\circ}+62^{\circ}=180^{\circ}$

$\angle D+100.2^{\circ}=180^{\circ}$

$\angle D+100.2^{\circ}-100.2^{\circ}=180^{\circ}-100.2^{\circ}$

$\angle D=79.8^{\circ}$

Therefore, the measure of angle D is approximately $79.8^{\circ}$.

Answer 2

  Measure of angle D will be 79.8°.

Sine rule in a triangle,

•  Sine rule,

          $\frac{c}{\text{sinC}}= \frac{d}{\text{sinD}}= \frac{e}{\text{sinE}}$

        Here, c, d and e are the sides and C, D, E are the angles opposite

        to these sides of the given triangle.

Given in the question,

In ΔCDE,

CD = 10 units

DE = 7 units

m∠E = 62°

By applying sine rule,

$\frac{7}{\text{sinC}}= \frac{d}{\text{sinD}}= \frac{10}{\text{sin(62)}}$

$\frac{7}{\text{sinC}}= \frac{10}{\text{sin(62)}}$

$\text{sin(C)}=\frac{\text{sin(62)}\times 7}{10}$

          $=0.618063$

C = $\text{sin}^{-1}(0.618063)$

C = 38.175°

   ≈ 38.2°

By triangle sum theorem of a triangle,

m∠C + m∠D + m∠E = 180°

38.2 + m∠D + 62° = 180°

m∠D = 79.8°

       Therefore, measure of angle D will be 79.8°.

Learn more about the sine rule here,

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