# Help please, this triangle question makes me crazy.

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### 4 Answers

In the diagram label `m /_ CAD = x, m /_ ACD=25-x ` . (This is because the measures of angles A and C in the right triangle must add to 90 and we have already accounted for 65 degrees leaving 25 degrees as the sum of the acute angles of triangle ADC.)

Then the measure of angle ADC is 155 degrees.

Using the law of cosines we can get AC:

`"AC"^2=9^2+8^2 - 2(9)(8)cos155 `

`"AC"~~16.5984 `

For triangle ADC, we can use the Law of Sines to get the acute angles:

`(sin A)/(8)=(sin 155)/(16.5984) ==> A~~11.75^@ `

`(sin B)/(9)=(sin 155)/(16.5984) ==> B~~13.25^@ `

Thus in the right triangle, `m /_ A = 36.75^@, m /_ B = 53.25^@ `

We can use right triangle trigonometry to find the missing sides:

`sin 36.75^@=("BC")/16.5984 ==> "BC"~~9.93 `

`sin 53.25^@="AB"/16.5984 ==> AB~~13.39 `

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`m /_ CAD ~~ 11.75^@, "AB"~~13.39, "BC"~~9.93 `

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**Sources:**

Dear Vamkitten,

kindly make corrections in my solution,

A + C = 25

therefore, D = 180-25 = 155 = <ADC

using this value of <ADC, we will get AC = 16.6 cm

Angle A or **<CAD = 11.78 degrees**

and angle C = 25-A = 13.23 degrees

<BAC = 25 + 11.78 = 36.78 degrees

and using trigonometry,

**AB = 13.3 cm**

**BC = 9.94 cm**

sorry about the calculation error.

Hi ,** i got the answers** , as i had to **solve them clearly** and for better understanding i had to put it on the paper , as i see you were not comfortable with the above answers , so i had to take this step . Hope you understand.

plz check the detailed solution in the attachments given below . :)

The **third solution is repeated** in the 3 rd picture just for clarity ,

**Images:**

Draw DY and DX perpendiculars from D to BY and BA to meet at y and x,

Now, DYBX is a rectangle. Triangles, BYD and DXA are right angled.

So,XB = DY = BDsin 40 = 8sin40.

BY = XD = 9sin25.

Therefore, AB = AX + XB = ADcos25 + 8sin40 = 9cos 25 + 8sin40 = **13.3 cm **cm

Similarly, BC = BY + YC = 9 sin25 + 8 cos40 = **9.93 cm**

Angle at ADC:

Consider the quadrilateral ADCB:

angle, ADC = x , angle DCB = 40, angle, ABC = 90. Angle, BCD = 25.

Total = x+40+90+25 = 360

x = 205.

Therefore outer angle ADC = 360 - 205 = **155 deg**